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3 years ago

Solve systems of equations for x + y, please!! -2x - 7y = -13 3x + 6y = 15

Mathematics

1 answer:

Andrei [34K]3 years ago

6 0

Answer:

y=-4

Step-by-step explanation:

i just did it

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Express as fraction in simplest form

Arte-miy333 [17]

As a percent: 2.5%

As a decimal: .025

As a fraction: 25/1000

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2 years ago

Quadrilateral DEFG - quadrilateral KLMN. Find the given side length.

Delvig [45]

Answer:

Step-by-step explanation:

By the diagram FG = MN.

3y + 3 = 5y - 25 Add 25 to both sides

3y + 3 + 25 = 5y Subtract 3y from both sides

28 = 5y - 3y Combine

28 = 2y Divide by 2

28/2 = 2y/2

y = 15

15 is not your answer. It is just the value of y.

3y + 3 = 3*15 + 3 = 45 + 3 = 48

3 0

2 years ago

Tanya and three friends have saved $75 to help protect the pandas. If they each gave the same amount, how much did each person c

Ber [7]

Answer:

Step-by-step explanation:

If it is Tanya and 3 friends, there are 4 people in total. If they all contributed the saem amount, you would do 75/4= 18.75 dollars each

8 0

3 years ago

A tennis court is 24 m long and 11 m wide. What is the perimeter of a tennis court? A. 78 m B. 70 m C.264 m D.35 m

scoundrel [369]

The answer is B..........................................................

6 0

3 years ago

Read 2 more answers

The following set of coordinates most specifically represents which figure? (−5, 6), (−1, 8), (3, 6), (−1, 4) (6 points) Paralle

Vladimir [108]

Answer:

Rhombus

Step-by-step explanation:

The given points are A(−5, 6), B(−1, 8), C(3, 6), D(−1, 4).

We use the distance formula to find the length of AB.

$|AB|=\sqrt{(-1--5)^2+(8-6)^2}$

$|AB|=\sqrt{16+4}$

$|AB|=\sqrt{20}$

The length of AD is

$|AD|=\sqrt{(-1--5)^2+(6-4)^2}$

$|AD|=\sqrt{16+4}$

$|AD|=\sqrt{20}$

The length of BC is:

$|BC|=\sqrt{(-1-3)^2+(8-6)^2}$

$|BC|=\sqrt{16+4}$

$|BC|=\sqrt{20}$

The length of CD is

$|CD|=\sqrt{(-1-3)^2+(6-4)^2}$

$|CD|=\sqrt{16+4}$

$|CD|=\sqrt{20}$

Since all sides are congruent the quadrilateral could be a rhombus or a square.

Slope of AB $=\frac{8-6}{-1--5}=\frac{1}{2}$

Slope of BC $=\frac{8-6}{-1-3}=-\frac{1}{2}$

Since the slopes of the adjacent sides are not negative reciprocals of each other, the quadrilateral cannot be a square. It is a rhombus

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3 years ago

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