The temperature in degrees celsius of the surface of the moon is -24.6111.
Fahrenheit and Celsius are directly proportionate to one another due to their relationship. When the temperature rises on the Celsius scale, it likewise rises on the Fahrenheit scale. Similar to how the Celsius scale, the Fahrenheit scale similarly drops in temperature when the Celsius scale does.
The ships sensor array detects that the temperature on the surface of the on is -12.3 F.
To convert the Fahrenheit to Celsius we will use the given formula.
[tex]C=\frac{5}{9}(F-32)[/tex]
Given F=-12.3
Substituting F in the equation, we get
[tex]C=\frac{5}{9}(-12.3-32)[/tex]
[tex]C=\frac{5}{9}(-44.3)[/tex]
[tex]C=\frac{-221.5}{9}[/tex]
[tex]C=-24.6111[/tex]
Therefore, the temperature -12.3 f to Celsius is -24.6111 C on the surface of the moon.
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Elaine drives her car 50 miles and has an average of a certain speed. If the average speed had been 4mph more, she could have traveled 58 miles in the same length of time. What was her average speed?
Hello there. To solve this question, we'll have to remember some properties about average speed.
When we're talking about moving in a straight line, the average speed is given by the ratio between the displacement ΔS and the interval of time Δt, namely
[tex]v=\frac{ΔS}{Δt}[/tex]In this case, say we have an average speed of v and the initial displacement ΔS = 50 miles in a certain interval of time Δt, such that
[tex]v=\frac{50}{Δt}[/tex]We know that if the average speed had been 4 mph more, then Elaine could have traveled 58 miles in the same length of time.
This means that v + 4 (that is, the average speed plus 4 mph) is equal to the ratio:
[tex]v+4=\frac{58}{Δt}[/tex]To solve this for v, we can start assuming that the interval of time Δt is not equal to zero, so do the average speed in the second equation.
Divide the first equation by the second, such that
[tex]\begin{gathered} \frac{v}{v+4}=\frac{\frac{50}{Δt}}{\frac{58}{Δt}} \\ \\ \frac{v}{v+4}=\frac{50}{58} \end{gathered}[/tex]Cross multiply the numbers, that is:
[tex]\begin{gathered} 58v=50\cdot(v+4) \\ 58v=50v+200 \end{gathered}[/tex]Subtract 50v on both sides of the equation
[tex]8v=200[/tex]Divide both sides by a factor of 8
[tex]v=25\text{ mph}[/tex]So this is her average speed.
How do u figure out what x is in a normal distribution question
Data:
• Mean (μ) = 50
,• Standard deviation (σ) = 3
,• P( ,x >=47 ,)
Procedure:
1. Since μ = 50 and σ = 3:
[tex]P(x\le47)=P(X-\mu<47-50)=P(\frac{x-\mu}{\sigma}<\frac{47-50}{3})[/tex][tex]Z=\frac{x-\mu}{\sigma}[/tex][tex]\frac{47-50}{3}=-1[/tex]2. Replacing the values:
[tex]P(x\le47)=P(Z\le-1)[/tex]With this, we do not have to figure out what x is.
3. Using the standard normal table:
[tex]P(Z\le-1)=0.1587\approx0.16[/tex]Answer: A. 0.16
What does the constant 0.98 reveal about the rate of change of the quantity?
The given function is:
[tex]f(x)=780(0.98)^{10t}[/tex]It is required to state what the constant 0.98 reveal about the rate of change of the quantity by completing the given statement:
The function is exponentially at a rate of every .
Notice from the given exponential function that the factor is 0.98.
Since it is less than 1, it implies that it is an exponential DECAY.
Subtract the factor 0.98 from 1 to get the decay rate:
[tex]1-0.98=0.02=0.02\times100\%=2\%[/tex]Unit of t: decades
The exponent is 10t.
The reciprocal is 1/10.
Hence, the time frame of rate is: 1/10 of a decade.
This is equivalent to a year.
The complete statement is, therefore:
The function is decaying exponentially at a rate of 2% every year.
Create a list of steps, in order, that will solve the following equation.5(x-3)² + 4 = 129Solution steps:
We want to solve the equation
[tex]5\cdot(x-3)^2+4=129[/tex]To solve this equation, we must isolate the variable on one side of the equation. So, noticed that the term with the x has multiple things (it is raised to the second power, multiplied by five, etc) so we begin by subtracting 4 on both sides. (Step 1)
So we get
[tex]5\cdot(x-3)^2=129\text{ -4 =125}[/tex]Now we divide both sides by 5 (Step 2), so we get
[tex](x-3)^2=\frac{125}{5}=25[/tex]now we take the square root on both sides (Step 3). Then we get
[tex]\sqrt[]{(x-3)^2}=\sqrt[]{25}=5=x\text{ -3}[/tex]Finally, we add 3 on both sides (Step 4). Then we get
[tex]x=5+3=8[/tex]Question
Hong hikes at least 1 hour but not more than 4 hours. She hikes at an average rate of 2.7 mph. The function f(t)=2.7t represents the distance she hikes in t hours.
What is the practical range of the function?
Responses
all real numbers from 1 to 4, inclusive
all multiples of 2.7 between 2.7 and 10.8, inclusive
all real numbers
all real numbers from 2.7 to 10.8, inclusive
The practical range of the function is D. all real numbers from 2.7 to 10.8, inclusive.
What is a range?A function's range refers to all of the possible values for y. The formula for determining a function's range is y = f. (x). A function's range is the set of all its outputs. After we have substituted the domain, the range of a function is the complete set of all possible resulting values of the dependent variable (y, usually).
In this case, since the range is the value that satisfies the given function. For function f(t) = 2.7t the practical range of the function can be solved by substituting the lowest time and the highest possible time which are 1 and 4.
At t = 1 f(t) = 2.7 and at t = 4 f(t) = 10.8. so the range is all real numbers from 2.7 to 10.8, inclusive.
Therefore, the correct option is D.
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Use the drawing tools to form the correct answer on the graph.Graph the composite function &(/(e)¡(=)-2I• 5g(I) =1 - 1
Solution:
Given:
The functions are given below as
[tex]\begin{gathered} f(x)=-2x-5 \\ g(x)=x-1 \end{gathered}[/tex]To find:
[tex]g(f(x)[/tex]To figure out the value of the composite function, we will replace x with (-2x-5) in g(x)
[tex]\begin{gathered} g(f(x))=-2x-5-1 \\ g(f(x))=-2x-6 \end{gathered}[/tex]Hence,
Using a graphing tool, we will have the composite function be
A zero-coupon bond is a bond that is sold now at a discount and will pay its face value at the time when it matures; no interest payments are made.A zero-coupon bond can be redeemed in 20 years for $10,000. How much should you be willing to pay for it now if you want the following returns?(a) 8% compounded daily(b) 8% compounded continuously
EXPLANATION:
We are given a zero-coupon bond that will be worth $10,000 if redeemed in 20 years time at an annual rate of 8% compounded;
(a) Daily
(b) Continuously
The formula for compounding annually is given as follows;
[tex]A=P(1+r)^t[/tex]Here the variables are;
[tex]\begin{gathered} P=initial\text{ investment} \\ A=Amount\text{ after the period given} \\ r=rate\text{ of interest} \\ t=time\text{ period \lparen in years\rparen} \end{gathered}[/tex]Note that this zero-coupon bond will yield an amount of $10,000 after 20 years at the rate of 8%. This means we already have;
[tex]\begin{gathered} A=10,000 \\ r=0.08 \\ t=20 \end{gathered}[/tex](a) For interest compounded daily, we would use the adjusted formula which is;
[tex]A=P(1+\frac{r}{365})^{t\times365}[/tex]This assumes that there are 365 days in a year.
We now have;
[tex]10000=P(1+\frac{0.08}{365})^{20\times365}[/tex][tex]10000=P(1.00021917808)^{7300}[/tex][tex]10000=P(4.95216415047)[/tex]Now we divide both sides by 4.95216415047;
[tex]P=\frac{10000}{4.95216415047}[/tex][tex]P=2019.31916959[/tex]We can round this to 2 decimal places and we'll have;
[tex]P=2019.32[/tex](b) For interest compounded continuously, we would use the special formula which is;
[tex]A=Pe^{rt}[/tex]Note that the variable e is a mathematical constant whose value is approximately;
[tex]e=2.7183\text{ \lparen to }4\text{ }decimal\text{ }places)[/tex][tex]10000=Pe^{0.08\times365}[/tex][tex]10000=Pe^{29.2}[/tex]With the use of a calculator we have the following value;
[tex]\frac{10000}{e^{29.2}}=P[/tex]I need help with this question... the correct answer choice
Solution:
The original parallelogram is as shown below with its coordinates;
The transformation that carries the parallelogram below onto itself is a rotation of 180 degrees counterclockwise about the origin.
When rotating a point 180 degrees counterclockwise about the origin, the point (x,y) will become (-x,-y)
This means to transform 180 degrees counterclockwise, we negate the x and y-coordinates of the original form.
The transformation is as shown below;
Therefore, the correct answer is a rotation of 180 degrees counterclockwise about the origin.
Use substitution to determine the solution of the system of equations. Write the solution as an ordered pair.x + 2y = 14y = 3x – 14solution =
the first step:
we will rewrite the second equation to look like the first one
y=3x-14
lets subtract 3x from both sides
y-3x=-14
-3x+y=-14
our system of equations will look like
x+2y=14
-3x+y=-14
however, for a substitution method we can use:
x+2(3x-14)=14
i multiply 2 with the parenthesis
x+6x-28=14
7x=28+14
7x=42, x=42/7=6
x=6
y=3x-14
y=(3x6)-14=18-14=4
solution= 6,4
Please solve the problem in the attachment and provide the steps, the reason why your answer is correct and why all the other answer choices are incorrect.
Answer:
A
[tex]A\text{. Rectangles also have four right angles}[/tex]Explanation:
We want to find a counterexample to disprove the conjecture below;
- A square is a figure with four right angles.
To disprove this, we need to find a shape that also has four right angles but is not a square.
So, from the option the only shape that also has four right angles is a rectangle.
Therefore, the counterexample to disprove the conjecture is;
[tex]A\text{. Rectangles also have four right angles}[/tex]Use sigma notation to represent the sum of the first eight terms of the following sequence: 4,7. 10,
To answer this question, we need to check the kind of sequence here. We have that the first three elements of the sequence are:
[tex]4,7,10[/tex]If we have the difference between the second element and the first element, and the difference between the third element and the second element, we have:
[tex]7-4=3,10-7=3[/tex]Thus, we have a common difference of 3. This is an arithmetic sequence. Then, we also have that:
[tex]a_n=a_1+(n-1)d[/tex]This is the formula for finding a general term in an arithmetic sequence, where:
• an is any term, n, in the sequence.
,• a1 is the first term in the sequence.
,• n the number of the term in the sequence.
,• d is the common difference of the sequence.
In this way, we have:
a1 = 4
d = 3
Can you help me please?A. How can Marc provide proof that his mighty shot actually hung in the air for 15 seconds? Or is this just another one of his lies?B. How long did the ball actually hang in the air?
The given formula for Marc's shot is:
[tex]h(x)=-16x^2+200x[/tex]a. To prove that the shot actually hung in the air for 15 seconds, we need to replace x=15 in the formula and solve for h, as follows:
[tex]\begin{gathered} h(15)=-16(15)^2+200(15) \\ h(15)=-16\times225+3000 \\ h(15)=-3600+3000 \\ h(15)=-600 \end{gathered}[/tex]As the height is negative, it means after 15 seconds the ball already hit the ground, because the ground is located at h=0. Then this result proves that this is just another one of Marc's lies.
b. To find how long the ball actually hung in the air, we need to find the x-values that makes h=0, as follows:
[tex]0=-16x^2+200x[/tex]We have a polynomial in the form: ax^2+bx+c=0, where a=-16, b=200 and c=0.
We can use the quadratic formula to solve for x:
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-200\pm\sqrt[]{(200)^2-4(-16)(0)}}{2(-16)} \\ x=\frac{-200\pm\sqrt[]{40000+0}}{-32} \\ x=\frac{-200\pm\sqrt[]{40000}}{-32} \\ x=\frac{-200\pm200}{-32} \\ x=\frac{-200+200}{-32}=\frac{0}{-32}=0\text{ and }x=\frac{-200-200}{-32}=\frac{-400}{-32}=12.5 \end{gathered}[/tex]Then the two x-values are x=0 and x=12.5.
The starting time is 0 and the end time when the ball hit the ground is x=12.5.
The ball actually hung in the air 12.5 seconds.
Hi! I just wanted to know may you please help me on knowing how to solve an inequality?
When given an inequality, we solve it as if we were solving an equation but there is one important consideration: If we divide or multiply by a negative number the inequality gets inverted.
Recall that to solve an equation we add, subtract, multiply, etc. to isolate the variable, in the case of inequalities we do the same process.
Example 1. x+58>36, to solve the inequality for x, we have to put the variable by itself on one side of the inequality, to do that we have to eliminate whatever is next to the variable, in the case 58 is adding, therefore, we must subtract it, but to not alter the inequality we do the same to both sides of the inequality.
Adding -58 to both sides of the inequality we get:
[tex]\begin{gathered} x+58-58>36-58, \\ x>-22. \end{gathered}[/tex]Therefore, the solution to this inequality is x>-22.
Example 2. -3x+3<4:
Adding -3 to both sides of the inequality we get:
[tex]\begin{gathered} -3x+3-3<4-3, \\ -3x<1. \end{gathered}[/tex]Dividing by -3 we get:
[tex]\begin{gathered} \frac{-3x}{-3}>\frac{1}{-3}, \\ x>-\frac{1}{3}\text{.} \end{gathered}[/tex]An electronics store purchases laptops for $425.00. They use a markup rate of 60%. How much do they sell the laptop to their customers?
Given that An electronics store purchases laptops for $425.00.
markup rate of 60%.
Selling price is:
[tex]\begin{gathered} SP=425+(0.60)425 \\ Sp=425+255 \\ SP=680 \end{gathered}[/tex]they sell the laptop to their customers at $680.
For how many books produced will the costs from the two methods be the same
Answer:
[tex]4780\text{ books}[/tex]Explanation:
Here, we want to get the number of books for which the cost of the two methods will be the same
What we have to do here is to get the cost of each method, then equate to find the number of books
Let the number of books be b
For the first method, we have it that:
[tex]\begin{gathered} 70976\text{ + }9.75(b) \\ =\text{ 70976 + 9.75b} \end{gathered}[/tex]For the second method, we have it that:
[tex]\begin{gathered} 16006\text{ + }21.25(b) \\ =\text{ 16006 + 21.25b} \end{gathered}[/tex]To get the number of books, we have to equate both
Mathematically, that would be:
[tex]\begin{gathered} 70976\text{ + 9.75b = 16006 + 21.25b} \\ 70976-16006\text{ = 21.25b-9.75b} \\ 54970\text{ = 11.5b} \\ b\text{ = }\frac{54970}{11.5} \\ b\text{ = 4,780} \end{gathered}[/tex]Six less than x equals twenty-two.
What do I need to do?
Write the word expression in algebraic expression?
16.- x - 6 = 22
x = 22 + 6
x = 28
17.-
y/4 = -10
y = 4(-10)
y = -40
19.- h - 7 = -9
h = -9 + 7
h = -2
20.- x + y = -7
if x = 11
11 + y = -7
y = -7 - 11
y = -18
The tables of ordered pairs represent some points on the graphs of Lines F and G.
Line F
x y
2 7
4 10.5
7 15.75
11 22.75
Line G
x y
-3 4
-2 0
1 -12
4 -24
Which system of equations represents Lines F and G?
1. y=1.75x+3.5
y=-4x-8
2. same as 1 but -8 is -2
3. 1.75 and 3.5 are switched
4. 2 and 3 combined
In linear equation, Equation for line F is y = 1.75x + 3.5 and equation for line G is y = -4x-8.
What is a linear equation example?
Ax+By=C is the typical form for linear equations involving two variables. A linear equation in standard form is, for instance, 2x+3y=5.Finding both intercepts of an equation in this format is rather simple (x and y).y = 1.75x + 3.5 (For line F)
let's take the point (2,7) and put in the equation,
y = 1.75*2 + 3.5
= 3.5 +0.35
= 7
which is true.
Hence, (2,7) satisfies the equation.
y = -4x-8 (For line G)
lets take the point (-3,4) and put in the equation,
y = (-4)*(3) - 8
= 12 - 8
= 4
which is true.
Hence, (-3,4) satisfies the equation.
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Deion measure the volume of a sink basin by modeling it as a hemisphere. Deion measures its diameter to be 28 3/4 inches. Find the sink’s volume in cubic inches. Round your answer to the nearest tenth if necessary.
28.75We are asked to determine the volume of a hemisphere of diameter 28 3/4 in.
A hemisphere is half a sphere therefore, its volume is half the volume of a sphere:
[tex]V=\frac{1}{2}(\frac{4\pi r^3}{3})=\frac{2\pi r^3}{3}[/tex]Where "r" is the radius. Since the radius is half the diameter we have that:
[tex]r=\frac{D}{2}=\frac{28\frac{3}{4}}{2}[/tex]We will convert the mixed fraction into a standard fraction using the following:
[tex]28\frac{3}{4}=28+\frac{3}{4}=28.75[/tex]Substituting in the formula for the radius:
[tex]r=\frac{28.75in}{2}=14.38in[/tex]Now, we substitute the value of the radius in the formula for the volume:
[tex]V=\frac{2\pi(14.38in)^3}{3}[/tex]Solving the operations:
[tex]V=6221.3in^3[/tex]Therefore, the volume is 6221.3 cubic inches.
The figure shows the layout of a symmetrical pool in a water park. What is the area of this pool rounded to the tens place? Use the value = 3.14.
EXPLANATION:
Given;
We are given a symmetrical pool as indicated in the attached picture.
The pool consists of two sectors and two triangles and each pair has the same dimensions.
The dimensions are as follows;
[tex]\begin{gathered} Sector: \\ Radius=30 \\ Central\text{ }angle=2.21\text{ }radians \end{gathered}[/tex][tex]\begin{gathered} Triangle: \\ Slant\text{ }height=30 \\ Vertical\text{ }height=25 \\ Base=20 \end{gathered}[/tex]Required;
We are required to calculate the area of the pool.
Step-by-step solution;
We shall begin by calculating the area of the sector and the formula for the area of a sector is;
[tex]\begin{gathered} Area\text{ }of\text{ }a\text{ }sector: \\ Area=\frac{\theta}{2\pi}\times\pi r^2 \end{gathered}[/tex]Where the variables are;
[tex]\begin{gathered} \theta=2.21\text{ }radians \\ r=30 \\ \pi=3.14 \end{gathered}[/tex]We now substitute and we have the following;
[tex]Area=\frac{2.21}{2\pi}\times\pi\times30^2[/tex][tex]Area=\frac{2.21}{2}\times900[/tex][tex]Area=994.5ft^2[/tex]Since there are two sectors of the same dimensions, the area of both sectors therefore would be;
[tex]Area\text{ }of\text{ }sectors=994.5\times2[/tex][tex]Area\text{ }of\text{ }sectors=1989ft^2[/tex]Next we shall calculate the area of the triangles.
Note the formula for calculating the area of a triangle;
[tex]\begin{gathered} Area\text{ }of\text{ }a\text{ }triangle: \\ Area=\frac{1}{2}bh \end{gathered}[/tex]Note the variables are;
[tex]\begin{gathered} b=20 \\ h=25 \end{gathered}[/tex]The area therefore is;
[tex]Area=\frac{1}{2}\times20\times25[/tex][tex]Area=\frac{20\times25}{2}[/tex][tex]Area=250[/tex]For two triangles the area would now be;
[tex]Area\text{ }of\text{ }triangles=250\times2[/tex][tex]Area\text{ }of\text{ }triangles\text{ }equals=500ft^2[/tex]Therefore, the area of the pool would be;
[tex]\begin{gathered} Area\text{ }of\text{ }pool: \\ Area=sectors+triangles \end{gathered}[/tex][tex]\begin{gathered} Area=1989+500 \\ Area=2489ft^2 \end{gathered}[/tex]Rounded to the tens place, we would now have,
ANSWER:
[tex]Area=2,490ft^2[/tex]Option D is the correct answer
6 a. Sketch a reflection triangleAABC about the line y - X.Label the image AA'B'C'. I know youdon't have graph paper, just sketch.b. What are the coordinates of C'?
Notice that we are asked to find the image of a triangle that
what type of angle is angle 98 degrees
Answer:
Obtuse
Step-by-step explanation:
Since it is larger than 90 degrees but less than 180 degrees, an angle of 98 degrees is considered to be obtuse.
Hope this helps! :)
find the measure of arc DB mDB = __ degrees simply
ANSWER:
90°
STEP-BY-STEP EXPLANATION:
Chord TD separates the circle into two equal 180° angles, so angles Since the angle 90°
Write an explanation of how you solved the problem. Write the explanation so that another student could follow your thought process.
Solution
- The solution steps are given below:
[tex]\begin{gathered} \sqrt{125} \\ 125\text{ can be written as:} \\ 125=5\times25 \\ \\ \text{ Thus, we have:} \\ \sqrt{125}=\sqrt{5\times25} \\ \\ \text{ Using the surd rule that:} \\ \sqrt{ab}=\sqrt{a}\times\sqrt{b} \\ \text{ We have:} \\ \\ \sqrt{5\times25}=\sqrt{5}\times\sqrt{25} \\ \\ \therefore\sqrt{125}=\sqrt{5}\times5 \\ \\ =5\sqrt{5} \end{gathered}[/tex]Final Answer
The answer is
[tex]5\sqrt{5}[/tex]I did 4/cos(62°) but it didn't give me any of the answer options
we have that
tan(62)=x/4 ------> by opposite side divided by the adjacent side
solve for x
x=4*tan(62)
x=7.5 unitsWhat is the best estimation of the equation [-å? Drag the numbers into the boxes. Numbers may beused once, twice, or not at all.1142011/21/8
Answer:
[tex]1-\frac{1}{2}=\frac{1}{2}[/tex]Explanation:
Given the below expression;
[tex]\frac{7}{8}-\frac{6}{11}[/tex]We can see that 7/8 is closer to 1 and that 6/11 is closer to 1/2, so we'll now have;
[tex]1-\frac{1}{2}=\frac{2-1}{2}=\frac{1}{2}[/tex]So the best estimation of the equation is 1 - 1/2 = 1/2
He starts by finding the sum of the exterior angles of a Pentagon, which isand then he solves to find that x is
The sum of the exterior angle of all polygons is same.
This is equal to 360
So, when we add all the exterior angles, this is equal to 360
Thus, mathematically;
[tex]\begin{gathered} x\text{ +2x + 78 + 77 + 3x+3 = 360} \\ \\ 6x\text{ + 158 = 360} \\ \\ 6x\text{ = 360-158} \\ \\ 6x\text{ = 202} \\ \\ x\text{ = }\frac{202}{6} \\ \\ x\text{ = 33}\frac{2}{3} \end{gathered}[/tex]Harmie's average s-day natural gas usage rate is 5.g therms AR are sdays What day- and 8-day natural gas usage rates? are her
We know that Harmie's average 5-day natural gas usage rate is 5.9 therms/5 days.
We have to calculate the 1 day average usage rate and 8 day average usage rate.
We can calculate this by transforming the denominator from 5 days, as it is in the information given, to 1 day and 8 days respectively:
[tex]\frac{5.9\text{ therms}}{5\text{ days}}\cdot\frac{1\text{ day}}{1\text{ day}}=\frac{5.9}{5}\cdot\frac{\text{therms}}{1\text{ day}}=\frac{1.18\text{ therms}}{1\text{ day}}[/tex][tex]\frac{5.9\text{ therms}}{5\text{ days}}\cdot\frac{8\text{ days}}{8\text{ days}}=(\frac{5.9\cdot8}{5})\cdot\frac{\text{therms}}{8\text{ days}}=\frac{9.44\text{ therms}}{8\text{ days}}[/tex]Answer: 1.18 therms / 1 day, 9.44 therms / 8 days [Option 1]
Write down the first five terms of the sequence an=(n+4)!2n2+6n+7a1 = a2 = a3 = a4 = a5 =
Step 1
Given;
[tex]\begin{gathered} a_n=\frac{(n+4)!}{2n^2+6n+7} \\ n=1,2,3,4,5 \end{gathered}[/tex]Step 2
[tex]a_1=\frac{(1+4)!}{2(1)^2+6(1)+7}=\frac{120}{15}=8[/tex][tex]a_2=\frac{(2+4)!}{2(2)^2+6(2)+7}=\frac{720}{27}=\frac{80}{3}[/tex][tex]a_3=\frac{(3+4)!}{2(3)^2+6(3)+7}=\frac{5040}{43}[/tex][tex]a_4=\frac{(4+4)!}{2(4)^2+6(4)+7}=\frac{8!}{63}=640[/tex][tex]a_5=\frac{(5+4)!}{2(5)^2+6(5)+7}=\frac{9!}{87}=\frac{120960}{29}[/tex]Let f(x) = (5)2+1. Which is equal to f(-3)?
Let's solve f(x) = (5)2+1 for f(-3):
Replacing x by -3, we have:
f (-3) = - 11
But there is no value for x, in the given function, there isn't a coefficient for x.
Can you please check the problem?
Hi I need help with this i’m in a hurry so can you please just tell me the answer lol sorry i’m just in a little rush
Step 1
In the example why is the area of one triangle multiplied by 2.
This is because the hexagon is divided into one rectangle and 2 congruent triangles. Therefore, the area of the two triangles will be the same since they are congruent triangles. In order to get the area of the hexagon, the area of one of the triangles is mutiplied by 2 and added to the area of the rectangle.
Step 2
Find the dimension of one of the shaded triangle from Bev's pattern.
[tex]\begin{gathered} \\ \text{For Bev's triangle;} \\ \text{base}=4 \\ \text{height}=3 \\ Slantheight^2=(\text{ }\frac{base}{2})^2+height^2 \\ Slantheight^2=(\frac{4}{2})^2+3^2 \\ Slantheight^{}=\sqrt[]{2^2+9} \\ Slantheight=\sqrt[]{13}\text{unit} \end{gathered}[/tex]The dimensions will therefore be;
[tex]\begin{gathered} \text{base= 4unit} \\ \text{slant height=}\sqrt[]{13}unit \\ \text{slant height=}\sqrt[]{13}unit \end{gathered}[/tex]What can you say about the shaded area of all the shaded triangles in Bev's pattern.
[tex]\begin{gathered} \text{Area of given triangle=6unit}^2 \\ \text{Area of Bev's triangle=}\frac{1}{2}\times4\times3=6unit^2 \end{gathered}[/tex]The area of all shaded triangles in Bev's pattern are equal. This is because all the shaded triangles have the same dimensions and can be said to be congruent. Hence, they will have the same area.