Write an equation for the line parallel to the given line that contains C. C(3,6); y= -2x+7

Mathematics

1 answer:

Anarel [89]2 years ago

6 0

Answer:

y=-2x+12

Step-by-step explanation:

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The function d(t) =300,000t represents the distance (in kilometers) that light travels in t seconds

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We were given the distance, d as a function of time, t.

$d(t)=300,000t$ .

To find $d(15)$ , we put $t=15$ , in to

$d(t)=300,000t$

This implies that,

$d(15)=300,000(15)$

$d(15)=4,500,000$

This means that in 15 seconds light covers a distance of 4,500,000 kilometers

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3 years ago

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Which description is true about pre-image ABC and image A′B′C′ ?

scoray [572]

Answer:

Option (C)

Step-by-step explanation:

From the graph attached,

In ΔABC and ΔA'B'C',

∠A ≅ ∠A'

∠B ≅ ∠B'

Therefore, ΔABC ~ ΔA'B'C'

Corresponding sides of these triangles will be proportional.

$\frac{\text{AB}}{\text{A'B'}}=\frac{\text{BC}}{\text{B'C'}}=\frac{\text{AC}}{\text{A'C'}}=\frac{2}{6}$

Therefore, ratio of the sides, AC : A'C' = 1 : 3 shows that image triangle A'B'C' is a dilated form of pre-image ABC with a scale factor of 3.

Option (C) will be the correct option.

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3 years ago

A ball rolls down an inclined plane such that the distance (in centimeters) that it rolls in t seconds is given by s(t) = 2t^3 +

IRISSAK [1]

The ball's velocity will be represented by the derivative of its distance function:

$s'(t)=6t^2+6t$

Now find the times $t$ for which this is equal to 30, i.e. solve

$6t^2+6t=30\implies t^2+t-5=0$

This has two solutions, $t=-\dfrac{1\pm\sqrt{21}}2$ , but only one is positive and falls in the interval $[0,3]$ . So the velocity reaches 30 cm/s when $t=-\dfrac{1-\sqrt{21}}2\approx1.79$ .