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Hunter-Best [27]
3 years ago
8

Find an equation of the line that passes through the given point and has the indicated slope m.

Mathematics
1 answer:
REY [17]3 years ago
4 0

Answer:

y = \frac{4}{5} x + 2

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here m = \frac{4}{5} , thus

y = \frac{4}{5} x + c ← is the partial equation

To find c substitute (- 5, - 2) into the partial equation

- 2 = - 4 + c ⇒ c = - 2 + 4 = 2

y = \frac{4}{5} x + 2 ← equation of line in slope- intercept form

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kherson [118]

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3 years ago
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Which three expressions are equivalent to 10x +11
BARSIC [14]

Answer:

B, C, and E

Step-by-step explanation:

<u>Simplify all answer choices:</u>

A.

5(2x+10)+1=

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B.

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C.

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D.

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E.

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Answer choices B, C, and E simplify to 10x+11

8 0
3 years ago
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AleksAgata [21]

Answer:

Step-by-step explanation:

Let C be the number of correct answers

Let B be the number of questions not answered

Let W be the wrong answers

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5 0
3 years ago
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3 years ago
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inysia [295]

Answer:

(x+3)^2+(y-2)^2=9

Step-by-step explanation:

<u>Equation of a circle</u>

(x-a)^2+(y-b)^2=r^2

where:

  • (a, b) is the center
  • r is the radius

From inspection of the diagram, the center of the circle <em>appears</em> to be at point (-3, 2), although this is not very clear.  Therefore, a = -3  and  b = 2.

Substitute these values into the general form of the equation of a circle:

\implies (x-(-3))^2+(y-2)^2=r^2

\implies (x+3)^2+(y-2)^2=r^2

Again, from inspection of the diagram, the <u>maximum vertical point</u> of the circle appears to be at y = 5.  Therefore, to calculate the radius, subtract the y-value of the center point from the y-value of the maximum vertical point:

⇒ radius (r) = 5 - 2 = 3

Substitute the found value of r into the equation:

\implies (x+3)^2+(y-2)^2=3^2

Therefore, the final equation of the given circle is:

\implies (x+3)^2+(y-2)^2=9

7 0
1 year ago
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