To find the term with quotient in 63 in from the given option divide each quotient with 63. If the quatiend is divisible that term contain the quotient 63.
The quotient in the first option is 4650. Divide the quotient with 63.
[tex]\frac{4650}{63}=72.38[/tex]The final answer contains decimal places. Thus, there first option does not contain 63 as quotient.
The quotient in the second term is 2867. Divide the quotient with 63.
[tex]\frac{2867}{63}=45.190[/tex]The final answer contains decimal places. Thus, there second option does not contain 63 as quotient.
The quotient in the second term is 3276. Divide the quotient with 63.
[tex]\frac{3276}{63}=52[/tex]The final answer does not contain any decimal places. Thus, the third option contains 63 as quotient.
Thus, the correct option is option 3) 3276÷52.
The image of the point (-2,2) under a translation is (-3,5). Find the coordinatesof the image of the point (-3,1) under the same translation.Submit Answer
We have an inital coordinate given (-2,2) and after a translation we got a new coordinate called (-3,5)
We can find the transformation like this:
-3= -2-1
5= 2+3
So then the transformation is given by:
[tex]T\rightarrow(x-1,y+3)[/tex]If we apply this transformation to the (-3,1) coordinate we got:
[tex](-3-1=-4,1+3=4)[/tex]So then the final answer would be (-4,4)
1.Given triangle ABC shown, graph its image after a dilation centered at the origin with a scale factor of two. Give the coordinates of the images of the vertices below.2. List all pairs of parallel line segments from problem 1.
for the question 1: the image of the vertices will be A'=2(-4,4)=(-8,8); B'=2(-4,-3)=(-8,-6); C'=2(4,-3)=(8,-6)
for the question 2: there will be no pairs of parallel lines in the triangle
what is the inequality of 7x ≤14 on a numberline
To find the inequality on a number line, we need to solve the inequality for x:
[tex]7x\leq14[/tex]Divide both sides by 7 to isolate the x variable:
[tex]\frac{7x}{7}\leq\frac{14}{7}[/tex][tex]x\leq\frac{14}{7}[/tex]Then:
[tex]x\leq2[/tex]Therefore, the inequality represents that x can be equal to or less than 2
Kapp and Stark go for a cross country run along a mountain trail. This graph models the elevation over time for their runwhich statement accurately describes Kapp and Starks run shown in the graph?
Hi! Let's analyze the sentences attached in the image:
a. They spent more minutes maintaining a constant elevation than decreasing.False. They just maintain a constant elevation at 11100feet (15min) and at 2800feet (15min), while they spent 90 minutes decreasing.
b. They spent more minutes maintaining a constant elevation than increasing.False. They just maintain a constant elevation at 11100feet (15min) and at 2800feet (15min), while they spent 60 minutes increasing.
c. They spent more time decreasing the elevation than increasing.True. They spent 90 minutes decreasing and 60 minutes increasing. So, 90>60.
d. They spent more time increasing the elevation than decreasing.False. They spent 60 minutes increasing and 90 minutes decreasing, so 60<90.
Set m to 0.0 to create a horizontal line. Then set x, to 3.0 and y, to -2.0.
We have the following:
the equation in the slope form is:
[tex]y=mx+b[/tex]m = 0 and goes through (3, -2)
therefore:
[tex]\begin{gathered} -2=3\cdot0+b \\ b=-2 \end{gathered}[/tex]now,
[tex]\begin{gathered} m=0 \\ \text{point = (3,-2)} \\ y=-2 \end{gathered}[/tex]Can you please help me out with a question
The arc length formula is:
[tex]L=\frac{\theta}{360}\cdot2\pi r[/tex]Where
θ is the angle
r is the radius
Given,
θ = 75°
r = 15
Now, we find the arc length (L) of Arc AC by substituting the information we know [ Remembering to use 3.14159 as π ]:
[tex]\begin{gathered} L=\frac{\theta}{360}\cdot2\pi r \\ L=\frac{75}{360}\cdot2(3.14159)(15) \\ L=\frac{5}{24}\cdot94.2477 \\ L=19.6349 \end{gathered}[/tex]Rounding to the nearest thousandth (3 decimal places), we have:
Arc Length = 19.635 unitsGiven the equation of the function, write the equation of the inverse, g(x). f(x) = 3x -1
1. Replace f(x) with y:
y = 3x - 1
2. Replace every "x" with "y" and every "y" with "x":
x = 3y - 1
3. Solve for y:
Add 1 to both sides:
x + 1 = 3y -1 + 1
x + 1 = 3y
Divide both sides by 3:
(x + 1)/3 = 3y/3
y = (x + 1)/3
4. Replace y with f−1(x) :
f-1 (x) = (x + 1)/3
The volume of the cylinder is approximately 7,959.9 cubic inches. The radius is ___ inches.Use π = 3.14.
The figure given is a cylinder.
The volume of a cylinder is given by the formula:
[tex]V=\pi r^2h[/tex]From the data given
The height is given to be 15 inches
The volume is also given to be 7,959.9 cubic inches
pi is 3.14
Upon substituting the values into the equation to solve for r, we will obtain
[tex]\begin{gathered} 7959.9=3.14\times r^2\times15 \\ 7959.9=47.1r^2 \end{gathered}[/tex][tex]\frac{47.1r^2}{47.1}=\frac{7959.9}{47.1}[/tex][tex]r^2=169[/tex][tex]\begin{gathered} r=\sqrt[]{169} \\ r=13\text{ inches} \end{gathered}[/tex]Radius is 13 inches
what is 9×9 can you pls tell me
ANSWER
9X9 is a product of two integers numbers.
It's equal to 81.
Use the approximate half-life formula for the case described below. Discuss whether the formula is valid for the case described.Urban encroachment is causing the area of a forest to decline at the rate of 9% per year. What is the half-life of the forest? What fraction of the forest will remain in 30 years?(Type an integer or decimal rounded to the nearest hundredth as needed.)
Answer:
Half-life = 7.35 years
After 30 years 0.06 of the forest will remain
Explanation:
Half-life is the amount of time it takes the forest to decline to half its initial value.
Now we are told that the forest declines at a rate of 9% per year. This means the amount left next year is 100% - 9% = 91% of the previous. Therefore, if we call the initial amount A, then the amount left after t years will be
[tex]P(t)=A(\frac{91\%}{100\%})^t[/tex][tex]\Rightarrow P(t)=A(0.91)^t[/tex]Now, when the forest declines to half its initial value, we have
[tex]\frac{A}{2}=A(0.91)^t[/tex]Canceling A from both sides gives
[tex]\frac{1}{2}=0.91^t[/tex]Taking the logarithm (of base 0.91) of both sides gives
[tex]\log_{0.91}(\frac{1}{2})=t[/tex][tex]t=7.35\text{ years.}[/tex]For each angle θ listed below, find the reference angle α, and then find sin θ. Round sin θ to four decimal places, if necessary.θ = 255° ? ?
A reference angle is the angle created by the terminal arm and X-axis, and must be in the same quadrant as the terminal arm.
The given angle is 255°. It is located at quadrant III, then we can find the reference angle by subtracting 180°:
[tex]255\degree-180\degree=75\degree[/tex]The sin of 75 is:
[tex]\sin 75\degree=0.9659[/tex]In quadrant III, the sine is negative, then the sin of 255° is equal to the sine of 75° but negative. So:
[tex]\sin 255\degree=-0.9659[/tex]The answer is option C. sin75=0.9659 sin255=-0.9659
Answer:
The answer is option C. sin75=0.9659 sin255=-0.9659
Step-by-step explanation:
Which expression can be used to name the angle below?AE"There are 3 possible answersO ZUAEOZAO ZUEAZUNo answer text provided,ZEAUO ZAUE
An angle can be named in several ways:
*With the capital letter representing its vertex
*With three capital letters: the two extreme letters represent the sides and the middle one the vertex.
In the figure you can see that the vertex of the angle is A and that the sides are E and U, then, the expressions you can use to name the shown angle are:
[tex]\begin{gathered} \angle A \\ \angle EAU \\ \angle UAE \end{gathered}[/tex]A rectangular field is 300 meters long and 150 meters wide.What is the area of the field in square kilometers? Do notround your answer.km²XG?Conversion facts for length1000 millimeters (mm) = 1 meter (m)100 centimeters (cm) = 1 meter (m)10 decimeters (dm) = 1 meter (m)1 decameter (dam) = 10 meters (m)1 hectometer (hm)100 meters (m)1 kilometer (km)1000 meters (m)I need help with this math problem.
Given:
A rectangular field is 300 meters long and 150 meters wide.
Required:
Find the area of the field in square kilometer.
Explanation:
The area of the rectangle is given by the formula:
[tex]\begin{gathered} A=length\times width \\ A=300\times150 \\ A=45000\text{ m}^2 \end{gathered}[/tex][tex]1m=\frac{1}{1000}km[/tex][tex]\begin{gathered} A=45000\times\frac{1}{1000000} \\ A=0.045\text{ km}^2 \end{gathered}[/tex]Final Answer:
The area of the
What is the probability of flipping a coin 11 times and getting heads 5 times?Round your answer to the nearest tenth of a percent.O A. 16.1%B. 8.1%O C. 22.6%O D. 0.5%SUBMIT
If you flip a coint there are two possible outcomes, "head"
1. A company supplies pins to a customer. It uses an automatic lathe to produce the pins. Due to factors such as vibration, temperature and wear and tear, the lengths of the pins and normally distributed with a mean of 25.30 mm and a standard deviation of 0.45 mm. The customer will only buy pins with lengths in the interval 25.00 ± 0.50 mm.
The percentage of the pins that will be acceptable to the customer is 63.16%.
What will the percentage be?Based on the information, the probabilty that the pin lies is between 24.5 and 25.5. This will be illustrated as P(24.5 <x < 25.5).
So convert this into Z score, will be:
P(24.5 <x < 25.5):
= P((24.5-25.3)/0.45 <Z < (25.5-25.3)/0.45)
Solving this, we will get
P(-16/9<Z < 4/9) = P(-1.77 <Z < 0.44),
By looking at the z table and solving for Z by using P (Z<0.44)-P(Z>-1.77) will be:
= 0.67 -(1-0.9616)
= 0.6316
= 63.16%
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Complete question
A company supplies pins to a customer. It uses an automatic lathe to produce the pins. Due to factors such as vibration, temperature and wear and tear, the lengths of the pins and normally distributed with a mean of 25.30 mm and a standard deviation of 0.45 mm. The customer will only buy pins with lengths in the interval 25.00 ± 0.50 mm. What percentage of the pins will be acceptable to the customer?
I need help graphing a problem I have the answer I just need help learning to graph
Given the inequality:
[tex]\frac{a}{10}-6>-12[/tex]Solving the inequality as follows:
Multiply both sides by 10
[tex]\begin{gathered} 10\cdot\frac{a}{10}-10\cdot6>10\cdot(-12) \\ a-60>-120 \end{gathered}[/tex]Add (60) to both sides:
[tex]\begin{gathered} a-60+60>-120+60 \\ a>-60 \end{gathered}[/tex]The solution on the number line will be as follows:
Study Surface Area and Volume of Pyramid and Cone1. How to find lateral area and total surface area of pyramid?
Explanation:
Lateral area of pyramid
Lateral area of squared pyramid is the sum of areas of its side faces.
[tex]\begin{gathered} L=2al \\ \text{a is base length} \\ l\text{ is slant height and it is given as,} \\ l=\sqrt[]{\frac{a^2}{4}+h^2} \\ \text{Here, h is height of pyramid.} \end{gathered}[/tex]Total surface area of pyramid
It is sum of the areas of its lateral faces and its base.
[tex]\begin{gathered} \text{Total surface area=}\frac{1}{2}pl+B \\ p\text{ is perimeter of base} \\ l\text{ is slant height} \\ B\text{ is base area} \end{gathered}[/tex]How do you perform the indicated operation?(4y + 11)(3y² -2y -7)
we use distributive property
[tex](4y\times3y^2)+(4y\times-2y)+(4y\times-7)+(11\times3y^2)+(11\times-2y)+(11\times-7)[/tex][tex]\begin{gathered} (12y^3)+(-8y^2)+(-28y)+(33y^2)+(-22y)+(-77) \\ 12y^3-8y^2+33y^2-28y-22y-77 \\ 12y^3+25y^2-50y-77 \end{gathered}[/tex]At KEY Middle School, there are 240 girls and 160 boys. What percent of all the students are girls?
Given;
Number of boys = 160
Number of girls = 240
Total number of students = 160 + 240 = 400
To find the percentage of all students that are girls, use the formula below:
[tex]\begin{gathered} \text{ \% of girls = }\frac{Number\text{ of girls}}{Total\text{ number}}\times\frac{100}{1} \\ \\ \text{ } \end{gathered}[/tex]Therefore, we have:
[tex]\begin{gathered} \text{ \% of girls = }\frac{240}{400}\times\frac{100}{1} \\ \\ \text{ = }0.6\text{ }\times\text{ 100 = 60\%} \end{gathered}[/tex]The percent of all the students that are girls is 60 percent.
ANSWER:
60%
What is 7×312 using mental math
Step-by-step explanation:
2184 simple ..... .............
Answer:
Its 2184
Step-by-step Explained
Find the point-slope equation of the line using the point (7, 4) and slope of2Use the slash key (/) to indicate a fraction.
The general equation of a line is given as:
y = mx + c where
m = slope
c = intercept on y axis.
We are given a point (x,y) so we use the relation below to develop the equation.
[tex]\begin{gathered} \frac{y-y_1}{x-x_1}=slope \\ \text{where:} \\ y_1=\text{ 4} \\ x_1=7 \end{gathered}[/tex][tex]\frac{y-4}{x-7}=2[/tex]Crossmultiplying, we have:
2x - 14 = y - 4
Adding 4 to both sides,
y = 2x - 14 + 4
y = 2x - 10
The figure shows the first three in a sequence of squares. The first square in the sequence has a side length of 3 units, and each square after that has a side length that is 2 units longer than the previous square.What is the explicit equation for f (n) that represents the areas of the squares in the sequence? f (n) = 2(n − 1)2 + 3 f (n) = (3 + 2(n − 1))2 f (n) = (3 + 2n)2 f (n) = 3n2
SOLUTION:
Since the sequence of side lengths are;
[tex]3,3+2n,3+4n,...[/tex]Their areas would be the sequence;
[tex]9,(3+2n)^2,(3+4n)^2,...[/tex]Thus, the explicit formula for the area is;
[tex]f(n)=(3+2(n-1))^2[/tex]f (n) = (3 + 2(n − 1))² is the explicit equation for f (n) that represents the areas of the squares in the sequence
What is Sequence?a sequence is an enumerated collection of objects in which repetitions are allowed and order matters.
Given,
The figure shows the first three in a sequence of squares.
First three in a sequence of squares. The first square in the sequence has a side length of 3 units
Each square after that has a side length that is 2 units longer than the previous square.
3,3+2n,3+4n....
The area of square is square of its length
The areas would be the sequence
3²,(3+2n)²,(3+4n)²....
Thus, the explicit formula for the area is;
f (n) = (3 + 2(n − 1))²
Hence f (n) = (3 + 2(n − 1))² is the explicit equation for f (n) that represents the areas of the squares in the sequence
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Can someone Help me with Trigonometry, will mark Brainliest if correct ;) show your work and draw a diagram of the answer pls
So we will use trigonometry to solve this because it is a right triangle. The hypotenuse is the ladder (h) and the two smallest sides are the floor and the vertical wall (w).
That angular ladder does with the ground= A
sin A = opposite / hypotenuse
[tex]\begin{gathered} \sin \text{ A = }\frac{14.8}{15}=0.986 \\ A=\sin ^{-1}(0.986)=80.4\text{degrees} \end{gathered}[/tex]No, the ladder will not be safe
Now let's make it safe:
The lenght of the ladder (w) is constant, so it remains 15
So now let's ask in an inequality what height will be safe (70degrees or less)
[tex]\begin{gathered} A=\sin ^{-1}(\frac{w}{15})\leq70 \\ \sin (\sin ^{-1}(\frac{w}{15}))\leq\sin (70) \\ \frac{w}{15}\leq0.9396 \\ (15)\frac{w}{15}\leq0.9396(15) \\ w\leq14.09 \end{gathered}[/tex]What does that mean? As long as you position the ladder against the wall so that the height from the ground to the top of the ladder is <14.09 ft
Use the spinner to find the theoretical probability of the event 6 2 The theoretical probability of spinning a multiple of 2 is
The spinner has six possible outcomes, 3 of them are multiples of 2 (2, 4 and 6). Then the probability is:
[tex]P=\frac{3}{6}=\frac{1}{2}[/tex]Therefore the probability is 1/2.
Answer 23 and the 24 ples explain. draw the problem or calculate it.
Given :
The slope = -3
Y- intercept = 7
The general equation of the line is :
[tex]y=m\cdot x+b[/tex]Where m is the slope and b is y- intercept
So,
[tex]\begin{gathered} m=-3 \\ b=7 \end{gathered}[/tex]Substitute with m and b in the general form
so, the equation of the line will be :
[tex]y=-3x+7[/tex]
Find the area the sector.arc circle 7A. 1083π4 in²B. 1083π8 in²C. 57π4 in²D. 38π in²
Solution:
Given:
A circle with the sector details;
[tex]\begin{gathered} r=19\text{ }in \\ \theta=135^0 \end{gathered}[/tex]The area of a sector is given by;
[tex]\begin{gathered} A=\frac{\theta}{360}\times\pi r^2 \\ A=\frac{135}{360}\times\pi\times19^2 \\ A=\frac{1083\pi}{8}\text{ }in^2 \end{gathered}[/tex]Therefore, the area of the sector is;
[tex]\frac{1083\pi}{8}\text{ }in^2[/tex]Lucy sold some items at a garage sale. She spent 7/12 of her earnings on a new bike. She uses 3/5 of the remainder to purchase a gift for her mom. What traction of her total earnings was spent on her mom's gift?
First we have to find the fraction that represents the remainder after buying the bike, subtracting 7/12 from the total, represented by 12/12
The result is 5/12
Then we have to multiply 5/12 by 3/5 to find our final answer
[tex]\begin{gathered} \frac{5}{12}\cdot\frac{3}{5}=\frac{15}{60} \\ \frac{15}{60}=\frac{5}{20}=\frac{1}{4}\text{ Simplifying our fraction} \end{gathered}[/tex]The fraction of her total earnings spent on her mom's gift was 1/4
Find the equation for the line that passes through the point (2,4) and that is parallel to the line with the equation x=-2
Given:
The passing point of line is (2,4)
The line is parallel to x = - 2
Any equation parallel to x= A has an equation of the form x = B.
Now the equation passing through (2,4) and parallel to x = - 2 is given by :
[tex]x=2[/tex]This is the required answer.
Find the equation of the normal in the form ax + by + c = 0 at the point where x = 4, for thecurve8=y = 2x2 - 4x3 - - 1х
We are given the equation of a curve;
[tex]2x^2-4x^{\frac{3}{2}}-\frac{8}{x}-1[/tex]To solve this we begin by taking the derivative of this curve. Note that the slope of this curve is its first derivative.
We now have;
[tex]\begin{gathered} \frac{d}{dx}(2x^2-4x^{\frac{3}{2}}-\frac{8}{x}-1 \\ =4x-6x^{\frac{1}{2}}-\frac{8}{x^2} \end{gathered}[/tex]At this point we should note that the slope (gradient) is the value of this first derivative when x = 4.
We can now plug in this value and we'll have;
[tex]\begin{gathered} f^{\prime}(x)=4x-6x^{\frac{1}{2}}-\frac{8}{x^2} \\ At\text{ } \\ x=4,\text{ we would have;} \\ f^{\prime}(4)=4(4)-6(4)^{\frac{1}{2}}-\frac{8}{4^2} \\ f^{\prime}(4)=16-6(2)-\frac{8}{16} \\ f^{\prime}(4)=16-12-\frac{1}{2} \\ f^{\prime}(4)=3\frac{1}{2} \\ OR \\ f^{\prime}(4)=\frac{7}{2} \end{gathered}[/tex]Now we can see the slope of the curve. The slope of the normal line perpendicular to the tangent of the curve is a negative inverse of this.
The negative inverse of 7/2 would be;
[tex]\begin{gathered} \text{Gradient}=\frac{7}{2} \\ \text{Gradient of perpendicular}=-\frac{2}{7} \end{gathered}[/tex]Now to use this value to derive the equation in the form
[tex]ax+by+c=0[/tex]We start by expresing this in the form;
[tex]y=mx+b[/tex]We now have;
[tex]y=-\frac{2x}{7}+b[/tex]We can convert this to the standard form as indicated earlier;
[tex]\begin{gathered} From\text{ the original equation; when} \\ x=4 \\ y=2(4)^2-4(4)^{\frac{3}{2}}-\frac{8}{4}-1 \\ y=32-4(8)-2-1 \\ y=32-32-2-1 \\ y=-3 \end{gathered}[/tex]With the points
[tex](4,-3)[/tex]We now have, the equation;
[tex]\begin{gathered} y=mx+b \\ -3=-\frac{2(4)}{7}+b \\ -3=-\frac{8}{7}+b \end{gathered}[/tex]We now collect like terms;
[tex]\begin{gathered} b=\frac{8}{7}-3 \\ b=-\frac{13}{7} \end{gathered}[/tex]We now have the y-intercept as calculated above.
We can now write up our equation is the standard form as indicated from the beginning;
[tex]\begin{gathered} ax+by+c=0 \\ (x,y)=(4,-3) \\ c=-\frac{13}{7} \end{gathered}[/tex][tex]\begin{gathered} 4a+(-3)b+(-\frac{13}{7})=0 \\ 4a-3b-\frac{13}{7}=0 \end{gathered}[/tex]Note that A, B and C must be integers. Therefore, we multiply all through by 7;
ANSWER:
[tex]28a-21b-13=0[/tex]NEED HELP!! Graph each function.Find the asymptote. Tell how the graph is transformedfrom the graph of its parentfunction.2. f(x)=3log4 (x + 6)1.f(x)= log₂x +43.f(x)=log (x+5)5.f(x)=2.5log2 (x+7)-34. f(x) = 3 + ln x6. f(x)=-0.8 In (x-1.5) +2
1)
The given function is expressed as
f(x) = log2x + 4
where
2 is the base of the logarithm
The graph is shown below
If a function, f(x) is translated d units upwards, it becomes f(x) + d
For the given function, the parent function is f(x) = log2x where 2 is the base.
f(x) = log2x + 4 means that the parent function was translated or shifted by 4 units upwards
On the left, the graph gets close to x = 0 but it doesn't touch it. Thus,
Vertical asymptote is x = 0