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3 years ago

Y = 2x x + y = −6 the answer should be like this (x,y) or ( , )

Mathematics

1 answer:

melisa1 [442]3 years ago

6 0

Answer:

(-2, -4)

Step-by-step explanation:

Given:

The system of equations to solve is given as:

$y=2x\\\\x+y=-6$

In order to solve this system of linear equations, we use substitution method.

In substitution method, we substitute the value of any one variable in the other equation.

So, we substitute the value of 'y' from first equation in the second equation.

This gives,

$x+2x=-6\\\\3x=-6$

Dividing both sides by 3, we get:

$\frac{3x}{3}=\frac{-6}{3}\\\\x=-2$

Now, $y=2x=2\times -2=-4$

Therefore, the solution is (-2, -4)

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Solve each proportion. 2/8=5/r+8

brilliants [131]

Answer:

r=-20/31

Step-by-step explanation:

1)divide the numbers

2) subtract 8 from both sides of the equation

6 0

2 years ago

READ WHOLE QUESTION!:There are 50 campers at day camp and 10 counselors.Write the ratio campers to counselors as a fraction.Expl

Dovator [93]

Given that there are 50 campers at day camp and 10 counselors.

The ratio of campers to counselors as a fraction is given by $\frac{50}{10}$ .

Using equivalent fractions we have $\frac{50\div10}{10\div10} = \frac{5}{1}$ .

Thus, the related rate and unit rate to the ratio of campers to counselors is given by $\frac{5}{1}$ .

The unit rate indicates that for every counselor, there are 5 campers.

6 0

3 years ago

A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient

ankoles [38]

Answer:

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

Solution:

We need to show that the gradient of the curve at A is 1

Here given that ,

$y=(x-a) \sqrt{(x-b)}$ --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

$0=(b+1-a) \sqrt{(b+1-b)}$

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

$y^{\prime}=u^{\prime} y+y^{\prime} u$

so, we get

${u}^{\prime}=1$

$v^{\prime}=\frac{1}{2} \sqrt{(x-b)}$

$y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}$

By putting value of point A and putting value of eq 2 we get

$y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}$

$y^{\prime}=\frac{d y}{d x}=1$

Hence proved that the gradient of the curve at A is 1.

7 0

3 years ago

Please help thanks so much.

Alik [6]

You can see if they are similar by looking at the sides.

If each of the sides have the same proportion then they are similar.

24/3 = 8
32/4 = 8
112/14 = 8

Yes they are similar and their ratio is 1:8

Hope this helps :)

3 0

3 years ago

Read 2 more answers

A. 37 B. 38 C. 39 D. 40

Dvinal [7]

Hello ^w^

I believe the answer you are looking for is A

I found this answer by adding up the lengths of all the sides.

S -> T = 8

S -> R = 7

Q -> R = 6

P -> Q = 6

P -> T = 10

10 + 6 + 6 + 7 + 8 -> 37

A -> 37

If I am incorrect, please inform me.

Have a good day!

4 0

3 years ago