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3 years ago

Use the substitution method to solve the system of equations. y=7x +8 y=x +20

Mathematics

1 answer:

DanielleElmas [232]3 years ago

3 0

Answer:

Step-by-step explanation:

$\left\{\begin{array}{ccc}y=7x+8&(1)\\y=x+20&(2)\end{array}\right\qquad\text{substitute (1) to (2):}\\\\7x+8=x+20\qquad\text{subtract 8 from both sides}\\7x=x+12\qquad\text{subtract x from both sides}\\6x=12\qquad\text{divide both sides by 6}\\x=2\\\\\text{put the value of x to (2):}\\\\y=2+22\\y=22$

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What is the total of 4/7 +3/7

lilavasa [31]

Answer:

7/7 also known as a whole (1)

Step-by-step explanation:

All you have to do is add 4 and 3 then keep the denominator the same (7)

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2 years ago

(4n-3n^3)-(3n^3+4n) answer

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3 years ago

raph the equation with a diameter that has endpoints at (-3, 4) and (5, -2). Label the center and at least four points on the ci

andreyandreev [35.5K]

Answer:

Equation:

${x}^{2} + {y}^{2} + 2x - 2y - 35= 0$

The point (0,-5), (0,7), (5,0) and (-7,0)also lie on this circle.

Step-by-step explanation:

We want to find the equation of a circle with a diamterhat hs endpoints at (-3, 4) and (5, -2).

The center of this circle is the midpoint of (-3, 4) and (5, -2).

We use the midpoint formula:

$( \frac{x_1+x_2}{2}, \frac{y_1+y_2,}{2} )$

Plug in the points to get:

$( \frac{ - 3+5}{2}, \frac{ - 2+4}{2} )$

$( \frac{ -2}{2}, \frac{ 2}{2} )$

$( - 1, 1)$

We find the radius of the circle using the center (-1,1) and the point (5,-2) on the circle using the distance formula:

$r = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

$r = \sqrt{ {(5 - - 1)}^{2} + {( - 2- - 1)}^{2} }$

$r = \sqrt{ {(6)}^{2} + {( - 1)}^{2} }$

$r = \sqrt{ 36+ 1 } = \sqrt{37}$

The equation of the circle is given by:

$(x-h)^2 + (y-k)^2 = {r}^{2}$

Where (h,k)=(-1,1) and r=√37 is the radius

We plug in the values to get:

$(x- - 1)^2 + (y-1)^2 = {( \sqrt{37}) }^{2}$

$(x + 1)^2 + (y - 1)^2 = 37$

We expand to get:

${x}^{2} + 2x + 1 + {y}^{2} - 2y + 1 = 37$

${x}^{2} + {y}^{2} + 2x - 2y +2 - 37= 0$

${x}^{2} + {y}^{2} + 2x - 2y - 35= 0$

We want to find at least four points on this circle.

We can choose any point for x and solve for y or vice-versa

When y=0,

${x}^{2} + {0}^{2} + 2x - 2(0) - 35= 0$

${x}^{2} +2x - 35= 0$

$(x - 5)(x + 7) = 0$

$x = 5 \: or \: x = - 7$

The point (5,0) and (-7,0) lies on the circle.

When x=0

${0}^{2} + {y}^{2} + 2(0) - 2y - 35= 0$

${y}^{2} - 2y - 35= 0$

$(y - 7)(y + 5) = 0$

$y = 7 \: or \: y = - 5$

The point (0,-5) and (0,7) lie on this circle.

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2 years ago

[!]Urgent[!] What is another way to summarize the outcome of this proof?

Marina CMI [18]

Answer:

B. Vertical angles are congruent

Step-by-step explanation:

The instructions are to prove that ∠ANR and ∠VNE are congruent, and ∠ANR and ∠VNE are vertical angles.

I hope this helps :)

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2 years ago

Is (-3, 1) a solution to this system of inequalities?

Maksim231197 [3]

Answer:

Step-by-step explanation:

5(-3)= -15

5(1)= 5

-15+5=-10

-10 is greater than -19 not less than or equal too.

Hope this helps!

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2 years ago

Use the substitution method to solve the system of equations. y=7x +8 y=x +20​

Use the substitution method to solve the system of equations. y=7x +8 y=x +20