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rusak2 [61]
3 years ago
12

Abed says he has written a system of two linear equations that has an infinite number of solutions. One of the equations of the

system is y = 3x – 1. Which could be the other equation?
y = 3x + 2
3x – y = 2
3x – y = 1
3x + y = 1
Mathematics
1 answer:
LenKa [72]3 years ago
3 0

Answer:

3x - y = 1

Step-by-step explanation:

This will be the same equation written in a different way'

The original equation is:

y = 3x - 1

0 = 3x - y - 1

3x - y = 1.

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The side of a square is 3^2 inches. Using the area formula A = s2, determine the area of the square.
earnstyle [38]

The area of the square would be; 512 square inches.

<h3>What is the area of a square?</h3>

The area of a square can be calculated as the square of the sides of a given figure.

The area of a square = side x side

We are given that:

A side of a square = (8)³/² inches.

To find the area of a square we use the formula as;

Area = s²

Now Substitute with the side length in the rule given to get the area as follows:

Area=(8)³/²)²

Area = 8³

Area = 512 square inches

Learn more about the area;

brainly.com/question/1658516

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7 0
11 months ago
What is the first step you will use to solve this equation:<br><br> 4x +2 = 7x - 13
Liono4ka [1.6K]
I would subtract 4x from 7x to get x on one side.
6 0
3 years ago
Odd numbers are always divisible by 2
Stells [14]
Odd numbers are not divisible by 2. Only even numbers are visible by 2. Such as, 4,6,8.

Multiples of 10 do always end in 0.

& i'm not so sure about the last one.

4 0
3 years ago
Is true or false 18= 18.00​
larisa86 [58]

Answer:

it is true.  any whole number has a 0 following the decimal

4 0
3 years ago
If 180° &lt; α &lt; 270°, cos⁡ α = −817, 270° &lt; β &lt; 360°, and sin⁡ β = −45, what is cos⁡ (α + β)?
eduard

Answer:

cos(\alpha+\beta)=-\frac{84}{85}

Step-by-step explanation:

we know that

cos(\alpha+\beta)=cos(\alpha)*cos(\beta)-sin(\alpha)*sin(\beta)

Remember the identity

cos^{2} (x)+sin^2(x)=1

step 1

Find the value of sin(\alpha)

we have that

The angle alpha lie on the III Quadrant

so

The values of sine and cosine are negative

cos(\alpha)=-\frac{8}{17}

Find the value of sine

cos^{2} (\alpha)+sin^2(\alpha)=1

substitute

(-\frac{8}{17})^{2}+sin^2(\alpha)=1

sin^2(\alpha)=1-\frac{64}{289}

sin^2(\alpha)=\frac{225}{289}

sin(\alpha)=-\frac{15}{17}

step 2

Find the value of cos(\beta)

we have that

The angle beta lie on the IV Quadrant

so

The value of the cosine is positive and the value of the sine is negative

sin(\beta)=-\frac{4}{5}

Find the value of cosine

cos^{2} (\beta)+sin^2(\beta)=1

substitute

(-\frac{4}{5})^{2}+cos^2(\beta)=1

cos^2(\beta)=1-\frac{16}{25}

cos^2(\beta)=\frac{9}{25}

cos(\beta)=\frac{3}{5}

step 3

Find cos⁡ (α + β)

cos(\alpha+\beta)=cos(\alpha)*cos(\beta)-sin(\alpha)*sin(\beta)

we have

cos(\alpha)=-\frac{8}{17}

sin(\alpha)=-\frac{15}{17}

sin(\beta)=-\frac{4}{5}

cos(\beta)=\frac{3}{5}

substitute

cos(\alpha+\beta)=-\frac{8}{17}*\frac{3}{5}-(-\frac{15}{17})*(-\frac{4}{5})

cos(\alpha+\beta)=-\frac{24}{85}-\frac{60}{85}

cos(\alpha+\beta)=-\frac{84}{85}

4 0
3 years ago
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