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3 years ago

9

Which of these skills do you need more help with select all that apply

1 answer:

Oksi-84 [34.3K]3 years ago

5 0

Dude this question is asking you which of these do you need help with.
So just choose all thee answers that you don't know how to do.

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A ramp has a height of 7 feet and

algol13

Answer:

59

Step-by-step explanation:

The answer is 59 because when you add 24 and 7 it is 31 and 31 +59 makes a 90° angle.

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4 years ago

Read 2 more answers

The equation -2x + 2y = -8 and the graph below each represent linear functions.

tankabanditka [31]

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The answer is B

Step-by-step explanation:

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4 years ago

Question 2 Which fraction is greater and why? 1/51 or 1/52

zheka24 [161]

Answer: 1/51

Step-by-step explanation: 1/51 is cut into bigger pieces. Its like cutting 2 pizza's of the exact same size. If you cut one into 5 pieces and another one into 10. If you take a piece of pizza out of both pizza's then the one cut into 5 pieces would be a ligger size.

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3 years ago

PLEASE HELP!! WILL MARK BRAINLIEST AND THANK YOU IF CORRECT

nexus9112 [7]

Answer:

i think it true if right mark me as brainiest and thank

Step-by-step explanation:

8 0

3 years ago

g The fencing of the left border costs $10 per foot, while the fencing of the lower border costs $2 per foot. (No fencing is req

klio [65]

Answer:

$Area = 1690ft^2\\x = 130ft, y = 26ft$

Step-by-step explanation:

Given the costs we can form an equation:

$10y + 2x =520 - Eq(A)$

and fencing is triangular such that the area enclosed can be written as:

$A = \dfrac{xy}{2} -Eq(B)$

First need to convert the above equation so that it is only in terms of one variable. [either x or y]

To make the equation only in terms of $x$ we can substitute $y$ from Eq(A) i.e, $y =\frac{520-2x}{10}$ , to Eq(B)

$A = \dfrac{x}{2} \dfrac{520-2x}{10}$

simplify

$A = \dfrac{520x - 2x^2}{20}$

$A = -\dfrac{1}{10}x^2 + 26x$

Now, in order to find the maximum area enclosed we can find $\frac{dA}{dx}$ and equate it zero.

$\dfrac{dA}{dx} = -\dfrac{1}{5}x + 26$

$0 = -\dfrac{1}{5}x + 26$

$-26 = -\dfrac{1}{5}x$

$x = 130$

we have the length of one dimension: specifically, the lower fence will be $x =130ft$

we can use this value of $x$ to find the corresponding value of $y$ . From Eq(A)

$10y + 2x =520$

$10y +2(130) = 520$

$y = \dfrac{520 - 2(130)}{10}$

$y = 26$

the length of the left fence will be $y =26ft$

The enclosed area by the fence will be

$A = \dfrac{xy}{2}$

$A = \dfrac{(130)(26)}{2}$

$A = 1690$

Hence the maximum area that can enclosed by the fences provided the costs will be $1690ft^2$

You can even check the cost of the dimensions whether they all add up to $520 or not.

Use Eq(A)

$10y + 2x =520$

$10(26) + 2(130) =520$

and indeed it does!

5 0

3 years ago