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AfilCa [17]
2 years ago
11

Ellis is painting wooden fenceposts before putti by then in his yard. They are each 5 feet tall and have a diameter of 1 foot. T

here are 10 fenceposts in all. How much paint will Ellis need to paint all the surfaces of the 10 fenceposts? Use 3.14 for pie, and round your answer to the nearest hundredth. Provide an explanation and proof for your answer to receive full credit.
Mathematics
1 answer:
lawyer [7]2 years ago
3 0

Answer:

39.25

Step-by-step explanation:

diameter. is 1 therefore radius is 1/2

formula is πr²h

where

h is height=5

=3.14x½²x5

=3.925 for one paint

therefore for 10 paints, we have

=3.925x10

=39.25

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Order the following rational numbers from least to greatest:3•4 -0.01 1/4 12% -26.1
Angelina_Jolie [31]

Answer:

-26.1 , -0.01 , 12% , 1/4 , 3-4

Step-by-step explanation:

7 0
3 years ago
B - (- 7) =-1<br> what is b?
irina1246 [14]

B-(-7)= -1

First negative multiply by negative is POSITIVE.

So: B+7= -1

Subtract 7 from both sides

B= -7-1

B= -8.

Answer: -8.

To Check Answer: Plug in the variable of B in the original equation.

So: -8-(-7)= -1

-8+7= -1

-1= -1

If the Solution works out the answer is right.

4 0
3 years ago
Read 2 more answers
The graphs of the equation y = 4x + 1 and y - kx = 10 are perpendicular when k = _______?
Ann [662]

Answer:

k=-\frac{1}{4}

Step-by-step explanation:

we have

Line 1

y=4x+1

Equation in slope intercept form

The slope is equal to

m_1=4

Line 2

y-kx=10

y=kx+10

Equation in slope intercept form

The slope is equal to

m_2=k

we know that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of the slopes is equal to -1)

so

m_1*m_2=-1

substitute

(4)(k)=-1

k=-\frac{1}{4}

3 0
3 years ago
M so tried and braindead so plz hlp me
padilas [110]
The answer is D

explanation:
4 0
3 years ago
Read 2 more answers
Find sin(a)&amp;cos(B), tan(a)&amp;cot(B), and sec(a)&amp;csc(B).​
Reil [10]

Answer:

Part A) sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}

Part B) tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}

Part C) sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}

Step-by-step explanation:

Part A) Find sin(\alpha)\ and\ cos(\beta)

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

\alpha+\beta=90^o ---> by complementary angles

so

sin(\alpha)=cos(\beta)

Find the value of sin(\alpha) in the right triangle of the figure

sin(\alpha)=\frac{8}{14} ---> opposite side divided by the hypotenuse

simplify

sin(\alpha)=\frac{4}{7}

therefore

sin(\alpha)=\frac{4}{7}

cos(\beta)=\frac{4}{7}

Part B) Find tan(\alpha)\ and\ cot(\beta)

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

\alpha+\beta=90^o ---> by complementary angles

so

tan(\alpha)=cot(\beta)

<em>Find the value of the length side adjacent to the angle alpha</em>

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132

x=\sqrt{132}\ units

simplify

x=2\sqrt{33}\ units

Find the value of tan(\alpha) in the right triangle of the figure

tan(\alpha)=\frac{8}{2\sqrt{33}} ---> opposite side divided by the adjacent side angle alpha

simplify

tan(\alpha)=\frac{4}{\sqrt{33}}

therefore

tan(\alpha)=\frac{4}{\sqrt{33}}

tan(\beta)=\frac{4}{\sqrt{33}}

Part C) Find sec(\alpha)\ and\ csc(\beta)

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

\alpha+\beta=90^o ---> by complementary angles

so

sec(\alpha)=csc(\beta)

Find the value of sec(\alpha) in the right triangle of the figure

sec(\alpha)=\frac{1}{cos(\alpha)}

Find the value of cos(\alpha)

cos(\alpha)=\frac{2\sqrt{33}}{14} ---> adjacent side divided by the hypotenuse

simplify

cos(\alpha)=\frac{\sqrt{33}}{7}

therefore

sec(\alpha)=\frac{7}{\sqrt{33}}

csc(\beta)=\frac{7}{\sqrt{33}}

6 0
3 years ago
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