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3 years ago

11

Rectangles always have congruent diagonals. False True False

1 answer:

Aleksandr [31]3 years ago

6 0

False because there is no diagonal part in the rectangle

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PLEASE HELP ME GRAPH THIS

spin [16.1K]

Answer:

$\huge\boxed{(12,80),(10,65)}$

Step-by-step explanation:

The line given has a slope of 7.5 and a point at (12, 80)

Because of the slope, simply subtract 2 from the x and subtract 15 from the y to get another point, at (10, 65)

Thus, simply put your points at (10, 65) and (12, 80)

Hope it helps and let me know if you want me to elaborate.

3 0

3 years ago

Tony makes a phone call at a pay phone. The charge is 25 cents for the first four minutes,

Andrews [41]

Answer:

it would be a max of 22 mins

Step-by-step explanation:

25C=4 mins

10C*10 mins =1$

$1.25

2.10-1.25=.85

10C*8=80C

so

4+10+8=22

7 0

3 years ago

Sarah had 28 peaches left at her roadside fruit stand. She went to the orchard and picked more peaches to stock up the stand. Th

Ann [662]

Answer:

32

Step-by-step explanation:

If she started with 28, ad had 50 after she picked, you wold subtract 28 from 50 to find your answer.

4 0

3 years ago

-3/4(8n+ 12) = 3(n - 3)

miskamm [114]

Answer:

n=0

Step-by-step explanation:

8 0

4 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