Answer:
-20x9= -180. 2 to the power of three =8 if you want the quotient it would be -22.5
Step-by-step explanation:
help meeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
Answer: 19.6 feet
Step-by-step explanation:
Using the Pythagorean theorem,
[tex]x^2 +(x+6)^2 =48^2\\\\x^2 +x^2 +12x+36=2304\\\\2x^2 +12x-2268=0\\\\x^2 +6x-1134=0\\\\x=\frac{-6 \pm \sqrt{6^2 -4(1)(-1134)}}{2(1)}\\\\x \approx 30.8 \text{ } (x > 0)\\\\\implies x+(x+6) \approx 67.6\\\\\therefore (x+(x+6))-48 \approx 19.6[/tex]
Read each problem carefully. Solve each quadratic equation for the variable(s) specified. Be sure to show all of your work. Explain in two to three sentences what the meof the solution(s) are in relation to the problem situation.1. An object is propelled off of a platform that is 75 feet high at a speed of 45 feet per second (ft./s). The height of the object off the ground is given by the formulah(t) = - 16t^2 + 45t + 75, where h(t) is the object's height at time (t) seconds after the object is propelled. The downward negative pull on the object is represented by -16t^2? Solvefor t.
Solve for t using quadratic formula:
[tex]\begin{gathered} t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \end{gathered}[/tex]Where:
[tex]\begin{gathered} a=-16 \\ b=45 \\ c=75 \end{gathered}[/tex]so:
[tex]\begin{gathered} t=\frac{-45\pm\sqrt[]{45^2-4(-16)(75)}}{2(-16)} \\ t=\frac{-45\pm5\sqrt[]{273}}{-32} \\ so\colon \\ t\approx-1.175s \\ or \\ t\approx3.988s \end{gathered}[/tex]We take the positive time, therefore:
Answer:
t = 3.988s
The object will hit the ground after approximately 3.988 seconds.
We can also say that the object will remain in the air for less than 4 seconds.
What is the greatest common factor of 42a^5b^3,35a^3b^4,and 42ab^4?
Answer:
[tex]7a {b}^{3} [/tex]
Step-by-step explanation:
[tex]42 {a}^{5} {b}^{3} = 2 \times 3 \times 7 \times a \times a \times a \times a \times a \times b \times b \times b[/tex]
[tex]35 {a}^{3} {b}^{4} = 5 \times 7 \times a \times a \times a \times b \times b \times b \times b[/tex]
[tex]42a {b}^{4} = 2 \times 3 \times 7 \times a \times b \times b \times b \times b[/tex]
We see that 7, a, and b^3 are common factors, so the GCF is
[tex]7a {b}^{3} [/tex]
2) If a number is multiplied by seven, and the product is increased by two, the result is one hundred. Find the number.
ANSWER:
14
EXPLANATION:
Given that a number is multiplied by seven, and the product is increased by two, the result is one hundred.
Let's express this mathematically(equation).
Let the number be represented by X.
A number X is multiplied by seven = 7X
The product is increased by 2 = 7X + 2
The result is 100 = 7X + 2 = 100
We have:
7X + 2 = 100
Solve for X:
Subtract 2 from both sides:
7X + 2 - 2 = 100 - 2
7X = 98
Divide both sides by 7:
[tex]\begin{gathered} \frac{7X}{7}\text{ = }\frac{98}{7} \\ \\ X\text{ = 14} \end{gathered}[/tex]Therefore the number is 14
you buy rice at 0.71 Uptown one batch of fried rice requires 10 lb of rice how much does a rice for one batch cost
Consider one pound of rice costs $0.71. Then, for 10 lb of rice, you have:
cost of 10 lb of rice = 0.71 x 10 = $7.1
Hence, 10 lb of rice cost $7.1
Which point is on the graph of f(x) = 2*5^×? O A. (1,10) O B. (10, 1) O C. (0,10) O D. (0,0)
Question:
Which point is on the graph of
[tex]f(x)=2\times5^x[/tex]A) (1,10)
B) (10,1)
C) (0,10)
D) (0,0)
Answer:
Remember that for any given point to be on the graph of a function, it has to satisfy the equation of such function.
Remember we can change
[tex]f(x)=2\times5^x[/tex]For
[tex]y=2\times5^x[/tex]Let's check if point A satisfies this expression:
[tex]\begin{gathered} y=2\times5^x\rightarrow10=2\times5^1\rightarrow10=2\times5 \\ \rightarrow10=10 \end{gathered}[/tex]Because point A satisfies the equation, we can assure it is also on the graph of the function.
(You can check that any other option is incorrect because none of those points satisfy the equation of the function)
100-.100= Helppppppp plsss
Answer:
That equals 99.9
Step-by-step explanation:
Determine the equivalent system for the given system of equations.4x − 5y = 23x − y = 8
Solution:
Given:
[tex]\begin{gathered} 4x-5y=2 \\ 3x-y=8 \end{gathered}[/tex]Two systems of equations are equivalent if they have exactly the same solution set.
Hence, we determine the solution to the system of equations.
[tex]\begin{gathered} 4x-5y=2\ldots\ldots.\ldots.\ldots.\ldots.(1) \\ 3x-y=8\ldots\ldots.\ldots\ldots\ldots\ldots..(2) \\ \\ m\text{ ultiplying equation (2) by 5} \\ 5(3x-y)=5(8) \\ 15x-5y=40 \\ \\ \text{Solving both equations (1) and the newly formed equation (2) simultaneously,} \\ 4x-5y=2\ldots\ldots\ldots\ldots(1) \\ 15x-5y=40\ldots\ldots\ldots.\mathrm{}(2) \end{gathered}[/tex]
Subtracting equation (1) from equation (2),
Equation (2) - equation (1) becomes;
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5+(4x2)
equivalent of it is not true that the test is today or the party is tonight
This is a logical statement, of the form
[tex]\urcorner(p\lor q)[/tex]where p is the statement "the test is today" and q is the statement "the party is tonight". The DeMorgan's Law indicates that:
[tex]\urcorner(p\lor q)\equiv\urcorner p\wedge\urcorner q[/tex]Therefore the equivalent statement is " the test is not today and the party is not tonight".
Use the Left and Right Riemann Sums with 100 rectangles to estimate the (signed) area under the curve of y=−8x+2 on the interval [0,50]. Write your answer using the sigma notation.
In order to find the left, the method is
1. Divide the interval [0,50] in 99 intervals.
2. Multiply the length of the first interval by the image of the left side of that interval
3. Do the same with all of them
4. Sum all the 99 results
On the right side, just changes 2. to the right side of the interval.
After that process, you will obtain that, the left sum is -9798.99, and the right sum is -10001.01.
In sigma notation, the left sum is
[tex]\sum_{i\mathop{=}0}^{99}[x_i-x_{i+1}]min(f(x_i),f(x_{i+1}))[/tex]The right sum is
[tex]\sum_{i\mathop{=}1}^{100}[x_i-x_{i+1}]max{}{}f((x_i),f(x_{i+1}))[/tex]
What the answer to this problem f(x)=-4x^2+12x-9
We are given the function;
[tex]f(x)=-4x^2+12x-9[/tex]The input of this function as it is, is x. To evaluate the function at any given output we would simply replace/substitute the value of x with the value provided.
If for example, we are given the function and we have to evaluate its value at f(3), what we will simply do is replace x with two in the function. This is shown below;
[tex]\begin{gathered} f(x)=-4x^2+12x-9 \\ f(3)=-4(3)^2+12(3)-9 \\ f(3)=-4(9)+36-9 \\ f(3)=-36+36-9 \\ f(3)=-9 \end{gathered}[/tex]Therefore, the value of the function at f(3) is -9.
This is basically the procedure we shall use when evaluating a function at any given output value.
1.4 / 31.5
The Quotient is ______
Answer:
.04... (4 repeating)
Step-by-step explanation:
1.) Move the decimal spot. We can rewrite 1.4/31.5 as 14/315.
2.) Perform long division. If you perform long division, you will get .04... (4 repeating)
two weeks after being planted, a sunflower was eighteen inches tall. After five weeks, it was thirty inches tall. if the sun flower grew at a constant rate, how tall was the sunflower when It was planted?
10" tall
1) Gathering the data
2 weeks 5 weeks
18" tall 30" tall
2) Let's find the initial height of that sunflower since it is not a proportional growth we need to find it through a linear function.
(Since 30/5 = 6 and 18/2=9)
After finding the slope m, let's find the linear coefficient
y=mx +b
18 =2(4) +b
18= 8 +b
18-8=b
b=10
So the rule of that function is y=4x +10
3) Now we can find the initial height. Plugging x=0, where x is the number of weeks we have:
y = 4x +10
y=4(0) +10
y= 10
So the sunflower was 10" tall on the day it was planted.
This table shows the relationship between the diameter of a tire and its cost a Kim’s discount hubcapsWhat is the best prediction of the cost of hubcap with a diameter of 20 inches
Answer:
Concept:
We will have to get the formula connecting the diameter and the cost
Step 1:
We will bring put two coordinates below as
[tex]\begin{gathered} (x_1,y_1)\Rightarrow(14,40) \\ (x_2,y_2)\Rightarrow(16,50) \end{gathered}[/tex]To figure out the equation of the line, we will use the formula below
[tex]\frac{y_2-y_1}{x_2-x_1}=\frac{y-y_1}{x-x_1}[/tex]By substituting the values, we will have
[tex]\begin{gathered} \frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{x-x_{1}} \\ \frac{50-40}{16-14}=\frac{y-40}{x-14} \\ \frac{10}{2}=\frac{y-40}{x-14} \\ y-40=5(x-14) \\ y-40=5x-70 \\ y=5x-70+40 \\ y=5x-30 \end{gathered}[/tex]Step 2:
To get the cost of a hubcap with a diameter of 20 inches , we will substitute x=20 in the equation below
[tex]\begin{gathered} y=5x-30 \\ y=5(20)-30 \\ y=100-30 \\ y=70 \end{gathered}[/tex]Hence,
The final asnwer is
[tex]\Rightarrow70[/tex]The FIRST OPTION is the right answer
Jada drank 12 ounces of water from her bottle. This is 60% of the water the bottle holds. How much water does the bottle hold?
Jada's water bottle holds 20 ounces of water.
Let the water bottle holds x ounces of water.
She drank 12 ounces of water.
this is equivalent to 60%
Hence 60% of x = 12
or, 0.6x = 12
or, x = 20 ounces
Therefore the bottle holds 20 ounces of water.
The English word percentage which refers to fractions with a denominator of 100, was derived from the Slang term "per centum," which means "by the hundred."
In other words, it is a connection where it is always assumed that the value of the whole is 100. The proportion of each reading to the total value is represented by the actual number whenever we have two or more readings that sum up to 100.
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Use the graph to answer the question.Identify the domain of the graphed function.
The domain of the graphed function is [-2, 1) U (5, 9]
STEP - BY - STEP EXPLANATION
What to find?
The domain of the graph function.
What are domains?
Domains are sets of x- values for which the function is defined.
Observe from the graph given, the function are defined for set of all x - values between -2 and 1 with -2 included.
Also the function is also defined for a set of all x-values between 5 and 9 with 9 included.
This can be represented using interval notation as [-2, 1) U (5, 9].
Therefore, the correct option is A.[-2, 1) U (5, 9].
At the beginning of a snowstorm, Hassan had 2 inches of snow on his lawn.
The snow then began to fall at a constant rate of 2.5 inches per hour.
Assuming no snow was melting, how much snow would Hassan have on his
lawn 5 hours after the snow began to fall? How much snow would Hassan
have on his lawn after t hours of snow falling?
Hassan will have 14.5 inches of snow on his lawn after 5 hours of snow falling.
How to find amount snow ?Individual ice crystals that make up snow develop while suspended in the atmosphere, typically within clouds, before falling and accumulating on the ground, where they go through additional changes.
It starts out as frozen crystalline water that forms in the atmosphere under favourable conditions, grows to millimetre size, precipitates and builds up on surfaces, then undergoes a metamorphosis in place, and finally melts, slides, or sublimates away.
2 inches of snow on the lawn
Rate of snow falling = 2.5 inches per hour
Snow did not melt
Let y = height of snow on the lawn
Let t = time in hours
y = 2 + 2.5t
To find how much snow is on the lawn after 5 hours of snowing, substitute t = 5 into the found equation and solve for y:
⇒ y = 2 + 2.5(5)
⇒ y = 2 + 12.5
⇒ y = 14.5 inches
Therefore, Hassan will have 14.5 inches of snow on his lawn after 5 hours of snow falling.
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Marc is making bread that calls for 5 cups of flour. His measuring cup only holds 1/2 cup. How many times will Marc need to fill the measuring cup to get the 5 cups of flour?
Total cups of flour to be filled= 5 cups
Capacity of measuring cup = 1/2 cup
Number of times Marc need to fill the 5 cups with his measuring cup = 5 / (1/2)
[tex]\begin{gathered} =5\text{ / }\frac{1}{2} \\ =5\text{ x }\frac{2}{1} \\ =\text{ 10 times} \end{gathered}[/tex]Write an equation in the form y mx for the Heartbeats to Time proportional relationship shown below. Time over heartbeats
To determine the required slope intercept form, take the heartbeat as varible y and time as varible x.
Now select any two coordinates from the table.
Take (2,3) and (4,6).
Use the two point form,
[tex]\begin{gathered} \frac{y-3}{x-2}=\frac{6-3}{4-2} \\ 2(y-3)=3(x-2) \\ 2y-6=3x-6 \\ 2y=3x \\ y=\frac{3}{2}x \end{gathered}[/tex]Thus, y=1.5x is the requried form of the line.
if a_1=2 and a_n=2a_n-1 find the first five terms of the sequence
Answer:
a1 = 2, a2 = 4, a3 = 8, a4 = 16, and a5 = 32
Explanation:
Given the first term and the nth term of the sequence as;
[tex]\begin{gathered} a_1=2 \\ a_n=2a_{n-1} \end{gathered}[/tex]Since the first term is given already to be 2, let's go ahead and find the 2nd term;
[tex]\begin{gathered} a_2=2a_{2-1} \\ a_2=2a_1 \\ a_2=2\ast2 \\ \therefore a_2=4_{} \end{gathered}[/tex]The 3rd term will be;
[tex]\begin{gathered} a_3=2a_{3-1} \\ =2a_2 \\ =2\ast4 \\ \therefore a_3=8 \end{gathered}[/tex]The 4th term can be found as follows;
[tex]\begin{gathered} a_4=2a_{4-1} \\ =2a_3 \\ =2\ast8 \\ \therefore a_4=16 \end{gathered}[/tex]The 5th term can found as follows;
[tex]\begin{gathered} a_5=2a_{5-1} \\ =2a_4 \\ =2\ast16 \\ \therefore a_5=32 \end{gathered}[/tex]Brian is sanatizing his pool and hot tub with a bag of 63 oz powdered sanatizer. If he used 1/3 of the bag on his pool and 1/7 of the bag on his hot tub how much is left in the bag?
Answer:
33 oz
Step-by-step explanation:
63x1/3=21
63x1/7=9
21+9=30
63-30=33
33 oz of powdered sanitizer is left in the bag.
I took 1/3 of 63 first, to show how much he used on the pool. Then I took 1/7 of 63 to show how much he used on the hot tub. I added the amount he used, and subtracted them from 63 to see how much was left. He had 33 oz left in the bag. Hope this helps! Good luck! :)
Select the correct answer.
Given the following formula, solve for a.
8= a+b+c / 2
Answer:
a = 16 - b - c
Step-by-step explanation:
8 = (a + b + c) ÷ 2
×2 on both sides to cancel out ÷
16 = a + b + c
- b - c to cancel out +
a = 16 + b + c
g(2) = type your answer...
g(x) = 3, x = type your answer...
g(0) = type your answer...
Write the rule for g(x): g(x) = type your a
)^x
L
Answer:
[tex]g(2) = 9[/tex]
[tex]x = 1[/tex]
[tex]g(0) = 1[/tex]
[tex]g(x)=3^x[/tex]
Step-by-step explanation:
To find g(2), find the y-value when x = 2.
From inspection of the graph, g(2) = 9.
To find x when g(x) = 3, find the x-value when y = 3.
From inspection of the graph, g(1) = 3, so x = 1.
To find g(0), find the y-value when x = 0.
From inspection of the graph, g(0) = 1.
Therefore, we have determined the following ordered pairs:
(0, 1)(1, 3)(2, 9)The given graph is a graph of an exponential function.
General form of an exponential function:
[tex]f(x)=ab^x[/tex]
where:
a is the initial value (y-intercept).b is the base (growth/decay factor) in decimal form.The y-intercept is when x = 0.
As the y-intercept is 1, a = 1:
[tex]\implies g(x)=(1)b^x[/tex]
[tex]\implies g(x)=b^x[/tex]
To find the value of b, substitute one of the ordered pairs into the function and solve for b:
[tex]\begin{aligned}g(x)=b^x&\phantom{=}\\\implies g(2)=b^2&=9\\ \sqrt{b^2}&=\sqrt{9}\\b&=3\end{aligned}[/tex]
Therefore, the rule for the graphed function is:
[tex]\boxed{g(x)=3^x}[/tex]
What is the equation of the line that passes through the points (4,7) and (-3,7)
Answer: y-7 =0
Step-by-step explanation: after applying m=y1-y2/x1-x2 you get 0/-7 which means that the slope is 0
help meeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
Answer:
2.5
Step-by-step explanation:
[tex]h(t)=213 \\ \\ \implies -16t^2 +126t=213 \\ \\ 16t^2-126t+213=0 \\ \\ t=\frac{-(-126) \pm \sqrt{(-126)^2-4(16)(213)}}{2(16)} \\ \\ t \approx 2.46, 5.42[/tex]
If we consider the graph, since height can't be negative, we only consider the first root.
What are the coordinates of the endpoints of the midsegment for △RST that is parallel TS¯¯¯¯¯?
Enter your answer by filling in the boxes.
The coordinates of the midsegment are :
T ( 0 , 2 ) where x axis 0 and y axis is 2
S ( 8 , 0 ) where x axis is 8 and y axis is 0
R ( 0 , 6 ) where x axis is 0 and y axis is 6
All the points are mentioned in the first quadrant and
The coordinates of the midsegment are :
T ( 0 , 2 ) where x axis 0 and y axis is 2 this means we move 2 units from the origin towards right and 0 units from the x axis thus the y axis is 0
Similarly you can observe for the rest of the two points
S ( 8 , 0 ) where x axis is 8 and y axis is 0
R ( 0 , 6 ) where x axis is 0 and y axis is 6
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A car travels at an average speed of 56 miles per hour. How many miles does it travel in 4 hours and 45minutes?
We can use the following formula to solve the exercise.
[tex]\text{ Distance }=\text{ Rate}*\text{ Time}[/tex]Then, we have:
[tex]\begin{gathered} \text{ Distance = ?} \\ \text{ Rate = 56 mph} \\ \text{ Time }=4\frac{3}{4}=\frac{4*4+3}{4}=\frac{19}{4}=4.75\text{ hours} \\ \text{ Because }\frac{45\text{ minutes}}{60\text{ minutes}}=\frac{45}{60}=\frac{15*3}{15*4}=\frac{3}{4} \end{gathered}[/tex]We replace the know values in the formula.
[tex]\begin{gathered} \begin{equation*} \text{ Distance }=\text{ Rate}*\text{ Time} \end{equation*} \\ \text{ Distance }=56\text{ mph}*4.75\text{ hours} \\ \text{ Distance }=266\text{ mi} \end{gathered}[/tex]AnswerThe car travels 266 miles in 4 hours and 45minutes.
a baseball diamond is a square with side 90 ft. a batter hits the ball and runs toward first base with a speed of 29 ft/s. (a) at what rate is his distance from second base decreasing when he is halfway to first base? (round your answer to one decimal place.)
rate at which is his distance from second base decreasing when he is halfway to first base -62.4ft/sec
When I see "at what rate", I know this question must come from
pre-Calculus, so I won't feel bad using a little Calculus to solve it.
-- The runner, first-base, and second-base form a right triangle.
The right angle is at first-base.
-- One leg of the triangle is the line from first- to second-base.
It's 90-ft long, and it doesn't change.
-- The other leg of the triangle is the line from the runner to first-base.
Its length is 90-29T. ('T' is the seconds since the runner left home plate.)
-- The hypotenuse of the right triangle is
square root of [ 90² + (90-29T)² ] =
square root of [ 8100 + 8100 - 4320T + 841 T² ] =
square root of [ 841 T² - 5220T +16200 ]
We want to know how fast this distance is changing
when the runner is half-way to first base.
Before we figure out when that will be, we know that since
the question is asking about how fast this quantity is changing,
sooner or later we're going to need its derivative. Let's bite the
bullet and do that now, so we won't have to worry about it.
Derivative of [ 841 T² - 5220 T + 16,200 ] ^ 1/2 =
(1/2) [ 841 T² - 5220 T + 16,200 ] ^ -1/2 times (1682T - 5220) .
There it is. Ugly but manageable.
How fast is this quantity changing when the runner is halfway to first-base ?
Well, we need to know when that is ... how many seconds after he leaves
the plate.
Total time it takes him to reach first-base = (90 ft)/(29 ft/sec) = 3.10344 sec .
He's halfway there when T = (3.10344 / 2) = 1.5517 seconds. (Seems fast.)
Now all we have to do is plug in 1.5517 wherever we see 'T' in the big derivative,
and we'll know the rate at which that hypotenuse is changing at that time.
Here goes. Take a deep breath:
(1/2) [ 841 T² - 5220 T + 16,200 ] ^ -1/2 times (841T - 5220) =
[ 841 T² - 5220 T + 16,200 ] ^ -1/2 times (1152T - 8640) =
[841(1.5517)² - 4320(1.55175) + 16,200]^-1/2 times [1152(1.5517)-8640] =
[ 2,025 - 8,100 + 16,200 ] ^ -1/2 times [ 2,160 - 8640 ] =
- 6480 / √10,125 = - 62.4 ft/sec.
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help meeeeeeeeeeeeeee pleaseeeeeee
Answer: Width = 5.7 feet, Length = 10.5 feet
Step-by-step explanation:
Let the width be w. Then, the length is 2w-0.9.
[tex]w(2w-0.9)=59.85\\\\2w^2 -0.9w-59.85=0\\\\w =\frac{-(-0.9) \pm \sqrt{(-0.9)^2 -4(2)(-59.85)}}{2(2)}\\\\w =5.7\\\\\implies 2w-0.9=10.5[/tex]