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3 years ago

10PTS pls help me on this math problem and pls explain how you did it.

Mathematics

2 answers:

erastovalidia [21]3 years ago

4 0

Answer:

y = 8

Step-by-step explanation:

Line1 with points (4, 1) and (0, 0):

slope = (y₂ - y₁) / (x₂ - x₁)

= (0 - 1) / (0 - 4)

= 1/4

Perpendicular line is opposite inverse so original slope 1/4 becomes -4/1 or simply -4.

Line2 with points (1, 0) and (-1, y):

slope = (y₂ - y₁) / (x₂ - x₁)

= (y - 0) / (-1 - 1)

= y / -2

y must equal to 8 for the slope to be -4 as stated above.

ivann1987 [24]3 years ago

4 0

Answer:

Step-by-step explanation:

First, you need to find the slope of the first line by using point slope formula (y2-y1) = m(x2-x1)

0 - 1 = m(0-4)

-1 = -4m

--- ----

- 4 -4

1/4 = m (slope)

Because the first line is perpendicular to the second line, the slope of the second line is the inverse. 1/4 ----> - 4/1

y - 0 = -4(-1-1)

y - 0 = -4(-2)

y = 8

These are the line equations:

y = -4x + 7

y = 1/4x

I think this is right but if anyone sees something wrong or needs more clarification please comment!

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So, the distance from focus to vertex is p = -5/2 = -2.5. This places the focus 2.5 units below the vertex. Then the vertex is at (h, k) = (0, -1) +(0, 2.5) = (0, 1.5).

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diamong [38]

Yo sup??

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Hope this helps.

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Anna71 [15]

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Solution:

Given that,

$(8a^{-3})^{\frac{-2}{3}$

We have to write in simplest form

Use the following law of exponent

$(a^m)^n = a^{mn}$

Using this, simplify the given expression

$(8a^{-3})^{\frac{-2}{3}} = 8^{\frac{-2}{3}} \times a^{ -3 \times \frac{-2}{3}}\\\\Simplifying\ we\ get\\\\(8a^{-3})^{\frac{-2}{3}} = 8^{\frac{-2}{3}} \times a^2\\\\We\ know\ that\ 8 = 2^3\\\\Therefore\\\\(8a^{-3})^{\frac{-2}{3}} =2^3^{\frac{-2}{3}} \times a^2\\\\(8a^{-3})^{\frac{-2}{3}} =2^{-2} \times a^2\\\\(8a^{-3})^{\frac{-2}{3}} = \frac{a^2}{4}$

Thus the given expression is simplified

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3 years ago