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SVEN [57.7K]
3 years ago
8

Solve the system of linear equations by graphing. y = –5/2 x – 7 x + 2y = 4 What is the solution to the system of linear equatio

ns?
A(–4.5, 4.25)
B(–1.7, –2.8)
C(0, –7)
D(3, 0.5)

Mathematics
2 answers:
Andrei [34K]3 years ago
6 0

Answer:

(–4.5, 4.25)

Step-by-step explanation:

Anestetic [448]3 years ago
4 0

Keywords

system; linear equations; graphing; solution

we have

y=-\frac{5}{2}x-7 -------> equation A

x+2y=4

x=4-2y ----> equation B

The <u>system</u> of <u>linear equations</u> is composed of equation A and equation B

Substitute the equation B in equation A

y=-\frac{5}{2}[4-2y]-7

y=-10+5y-7

4y=17

y=17/4=4.25

Find the value of x

x=4-2[17/4]=-4.5

The<u> solution</u> is the point (-4.5,4.25)

Using a <u>graphing</u> tool

see the attached figure

therefore

the answer is the option A

(-4.5,4.25)


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Given that Triangle ABC ~ DEF, solve for X.​
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Answer:

x = 7

Step-by-step explanation:

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The ratio of AB to ED is given as 5/30 = 1/6

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3 years ago
Write the value on the number line next to letter
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I dont see it though
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Shishir bought 4000 orange at 70 paisa each. But 400 of them were rotten. He sold 2000 oranges at 90 paisa each.If he plans to m
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Answer:

He needs to sell the rest of the oranges at 75 paisa each.

Step-by-step explanation:

Consider the given information that, Shishir bought 4000 orange at 70 paisa each.

Note: 1 rupees = 100 paisa

Thus, 70 paisa = 70/100 rupees = 0.70 rupees

Therefore, the cost price of 4000 oranges is:

4000×0.70 rupees = 2800 rupees

The selling price of 2000 oranges is:

2000×0.90 rupees = 1800 rupees

The number of oranges now Shishir have:

4000 - 2000 - 400 = 1600

He wants to make a profit of RS 200. Thus the selling price of 4000 oranges should be:

2800 rupees + 200 rupees = 3000 rupees

He earned 1800 rupees by selling 2000 oranges at 90 paisa. So, the remaining amount that he needs to make with 1600 oranges is:

3000 rupees - 1800 rupees = 1200 rupees

Therefore, the cost of one orange is:

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4 0
3 years ago
Algebra 1 question 4.
Brrunno [24]

Greetings!

To start this problem, let's first assign a variable for the missing, consecutive odd numbers. Since they are consecutive and odd, we add two.

<u>Proof:</u> <em>3-1=2, 5-3=2</em>

The first, consecutive, odd number: x

The second, consecutive, odd number: x+2

The third, consecutive, odd number: x+4

The fourth, consecutive, odd number: x+6

The sum of the values are equal to 3 times the sum of the first two numbers, of which this is equal to 35 less than the fourth number. Let's create an equation to simplify this:

3((x)+(x+2))=(x+6)-35

Complete the operations inside the parenthesis:

3(2x+2)=(x+6)-35

Distribute the parenthesis (utilizing the distributive property)

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(6x+6)=(x+6)-35

Simplify both sides:

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Add -6 and -x to both sides of the equation:

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5x=-35

Divide both sides of the equation by 5:

\frac{5x}{5}=\frac{-35}{5}

x=-7

If x is equal to -7:

x+2=-5

x+4=-3

x+6=-1

The four numbers are:

\boxed{-7,-5,-3,-1}

I hope this helps!

-Benjamin

4 0
3 years ago
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