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3 years ago

Does (7, 8) make the equation y = x + 5 true?

Mathematics

2 answers:

Bumek [7]3 years ago

6 0

Answer:

nop0e

Step-by-step explanation:

NNADVOKAT [17]3 years ago

3 0

Answer:

Step-by-step explanation:

no the correct point would be (7,12)

7x1=7

7+5=12

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Answer:

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Which are the roots of the quadratic function f(q) = q^2 - 125? Select two options.

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3 years ago

Read 2 more answers

In each of the following equations, what is

timama [110]

The graph will cross at the coordinates (-2, 9)

<h3>How to solve equations?</h3>

y = 3x + 15

y = 3 - 3x

y = 3x + 15

Hence,

when x = 2

y = 3(2) + 15 = 21

when x = 0

y = 3(0) + 15 = 15

y = 3 - 3x

when x = 2

y = 3 - 3(2)

y = 3 - 6

y = -3

when x = 0

y = 3 - 3(0)

y = 3

Therefore, let's check if the equation will cross.

y = 3x + 15

y = 3 - 3x

using substitution,

3 - 3x = 3x + 15

3 - 15 = 3x + 3x

- 12 = 6x

x = -12 / 6

x = -2

y = 3 - 3(-2)

y = 3 + 6

y = 9

Therefore, the graph will cross at the coordinates (-2, 9)

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2 years ago

Type the correct answer in each box. A circle is centered at the point (5, -4) and passes through the point (-3, 2). The equatio

Galina-37 [17]

Answer:

$(x+ \boxed{-5})^2+(y+\boxed4)^2=\boxed{100}$

Step-by-step explanation:

Given:

Center of circle is at (5, -4).

A point on the circle is $(x_1,y_1)=(-3, 2)$

Equation of a circle with center $(h,k)$ and radius 'r' is given as:

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

Here, $(h,k)=(5,-4)$

Radius of a circle is equal to the distance of point on the circle from the center of the circle and is given using the distance formula for square of the distance as:

$r^2=(h-x_1)^2+(k-y_1)^2$

Using distance formula for the points (5, -4) and (-3, 2), we get

$r^2=(5-(-3))^2+(-4-2)^2\\r^2=(5+3)^2+(-6)^2\\r^2=8^2+6^2\\r^2=64+36=100$

Therefore, the equation of the circle is:

$(x-5)^2+(y-(-4))^2=100\\(x-5)^2+(y+4)^2=100$

Now, rewriting it in the form asked in the question, we get

$(x+ \boxed{-5})^2+(y+\boxed4)^2=\boxed{100}$

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