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3 years ago

14

The length of a rectangle is shown below

1 answer:

blondinia [14]3 years ago

8 0

Answer:

A. C(2,-2) , D(-1,-2)

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Rewrite the following equation in slope-intercept form. 17x + 15y = 5 Write your answer using integers, proper fractions, and im

omeli [17]

Answer:
y = $\frac{-17}{15} x$ + $\frac{1}{3}$

Explanation:
The slope-intercept form of the equation has the following formula:
y = mx + c
where:
m is the slope
c is the y-intercept

The given is:
17x + 15y = 5

To put it in the slope-intercept formula, we should isolate the y as follows:
17x + 15y = 5
15y = -17x + 5
y = (-17/15) x + (5/15)
y = $\frac{-17}{15} x$ + $\frac{1}{3}$
where:
m is the slope = -17/15
c is the y-intercept = 1/3

Hope this helps :)

7 0

3 years ago

WILL GIVE BRAINLEST ! <br> hAVE A GOOD DAy :)!

gayaneshka [121]

Answer:

6. B

7. C

8. B

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Solve the simultaneous equation<br> 2x+4y=1<br> 3x+5y=7b

Lyrx [107]

Step-by-step explanation:

3x+5y=7b equation 1
2x+4y=1

2x=-4y+1

x=(-4y+1)/2
x=(-4y+1)/2 in equation 1

3×(-4y+1)/2+5y=7b

(-12y+3)/2+5y=7b

(-12y+3+10y)/2=7b

(-12y+3)/2=7b

-12y+3=14b

-12y=14b-3

y=3-14b/12
y=3-14b/12 in x=(-4y+1)/2

x=(-4×{3-14b/12}+1)/2

x=(-3+14b)/3+1/2

x=14b/3×1/2

x=7b/3

8 0

2 years ago

9 is a member of A {1,3,8} intersecting D {3,9}

Jet001 [13]

The intersection of the two sets is the list of members that appear in both sets. That is {3}.

Your statement is FALSE.

5 0

3 years ago

A school is planning to construct two rectangular play areas in the playground. The length of play area A must be 1 foot longer

Anna11 [10]

Answer:

А.The system has two solutions, but only one is viable because the other results in a negative width.

Step-by-step explanation:

Given

Let:

$L_A \to$ length of play area A

$W_A \to$ width of play area A

$L_B \to$ length of play area B

$W_B \to$ width of play area B

$x \to$ Area of A

$y \to$ Area of B

From the question, we have the following:

$L_A = 1 + 4W_A$

$W_B = 2 + W_A$

$L_B = 2 + 3W_B$

$x = y$

The area of A is:

$x = L_A * W_A$

This gives:

$x = (1 + 4W_A) * W_A$

Open bracket

$x = W_A + 4W_A^2$

The area of B is:

$y = L_B * W_B$

$y = (2 + 3W_B) * ( 2 + W_A)$

Substitute: $W_B = 2 + W_A$

$y = (2 + 3(2 + W_A)) * ( 2 + W_A)$

Open brackets

$y = (2 + 6 + 3W_A) * ( 2 + W_A)$

$y = (8 + 3W_A) * ( 2 + W_A)$

Expand

$y = 16 + 8W_A + 6W_A + 3W_A^2$

$y = 16 + 14W_A + 3W_A^2$

We have that:

$x = y$

This gives:

$W_A + 4W_A^2 = 16 + 14W_A + 3W_A^2$

Collect like terms

$4W_A^2 - 3W_A^2 + W_A -14W_A - 16 =0$

$W_A^2 -13W_A - 16 =0$

Using quadratic calculator, we have:

$W_A = -14.1$ or $W = 1.13$ --- approximated

But the width can not be negative; So:

$W = 1.19$

7 0

3 years ago