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Mekhanik [1.2K]
1 year ago
9

What is it help pleaseee

Mathematics
1 answer:
timurjin [86]1 year ago
6 0

Answer:

25.13cm

Step-by-step explanation:

formula for circumference of a circle is (2r)π with r being the radius

since the problem gave us the diameter already we can change the formula to 8π (2r is equal to the diameter since diameter is twice as long as the radius)

this gives us the circumference of 25.13

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A building is 5 feet tall. the base of the ladder is 8 feet from the building. how tall must a ladder be to reach the top of the
iris [78.8K]

The ladder must be 9.4 ft to reach the top of the building

Here, we want to get the length of the ladder that will reach the top of the building

Firstly, we need a diagrammatic representation

We have this as;

As we can see, we have a right triangle with the hypotenuse being the length of the ladder

We simply will make use of Pythagoras' theorem which states that the square of the hypotenuse is equal to the sum of the squares of the two other sides

Thus, we have;

\begin{gathered} x^2=5^2+8^2 \\ x^2=\text{ 25 + 64} \\ x^2\text{ = 89} \\ x=\text{ }\sqrt[]{89} \\ x\text{ = 9.4 ft} \end{gathered}

5 0
1 year ago
Which expression is equivalent to 4^7 x 4^-5​
melisa1 [442]
<h2>4⁷×4⁵</h2><h3>{a^m×a^n=a^(m×n)}</h3><h3>=4^(⁷×-⁵)</h3><h3>=4^-35</h3>

please mark this answer as brainlist

3 0
3 years ago
Evaluate the double integral.
Fynjy0 [20]

Answer:

\iint_D 8y^2 \ dA = \dfrac{88}{3}

Step-by-step explanation:

The equation of the line through the point (x_o,y_o) & (x_1,y_1) can be represented by:

y-y_o = m(x - x_o)

Making m the subject;

m = \dfrac{y_1 - y_0}{x_1-x_0}

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

m= \dfrac{2-1}{1-0}

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

m = \dfrac{1-2}{4-1}

m = \dfrac{-1}{3}

∴

y-2 = -\dfrac{1}{3}(x-1)

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

y = 2

Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy

\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg)   dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ([y^2(-4y+8)] \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( -4y^3+8y^2 \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1

\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1

\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [  \dfrac{-45+56}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [  \dfrac{11}{3}\bigg]

\iint_D 8y^2 \ dA = \dfrac{88}{3}

4 0
2 years ago
GIVING BRAINLIEST IF YOU CAN HELP PLEASE
alex41 [277]

Answer:

The rate of change will be $25

Step-by-step explanation:

Since she started from $45 we have to count again so on the chart it goes 5, 10, 15, 20, and 25!

<h2 />
3 0
2 years ago
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Point w id located at (-2,3) on a coordinate plane. Point W is reflected over the x-axis to create point W" is then reflected ov
Anna11 [10]

Answer:

W'' = (2, -3)

Step-by-step explanation:

Reflection over the x-axis negates the y-coordinate and leaves the x-coordinate alone. The W becomes ...

... W' = (-2, -3)

Reflection over the y-axis negates the x-coordinate and leaves the y-coordinate alone. The W' becomes ...

... W'' = (2, -3)

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2 years ago
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