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3 years ago

Find the coordinates of point P along the direct line segment AB so that AP to PB is the given ratio.

Mathematics

1 answer:

Grace [21]3 years ago

4 0

I need answers ASAP please let me know when you guys find an answer

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5. Thunderstorms are more common in some areas than in others. The map shows the average number of thunderstorms per year in dif

natta225 [31]

Answer:

where is the map?

Step-by-step explanation:

4 0

2 years ago

What is the equation of the line ?

Ronch [10]

Look at the picture.

The point-slope form:

$y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (-3, -3) and (2, 0). Substitute:

$m=\dfrac{0-(-3)}{2-(-3)}=\dfrac{3}{5}\\\\y-0=\dfrac{3}{5}(x-2)\qquad|\text{use distributive property}\\\\y=\dfrac{3}{5}x-\dfrac{6}{5}\qquad|\cdot5\\\\5y=3x-6\qquad|-5y\\\\0=3x-5y-6\qquad|+6\\\\3x-5y=6$

Answer:

point-slope form: $y=\dfrac{3}{5}(x-2)$

slope-intercept form: $y=\dfrac{3}{5}x-\dfrac{6}{5}$

standard form: $3x-5y=6$

7 0

3 years ago

V²=u²+2as make u as the subject <br><br>

yKpoI14uk [10]

Answer:

u as the subject of the given formula is, $u = \sqrt{v^{2}-2as }$

Step-by-step explanation:

Given;

v² = u²+2as

To make u the subject of the formula, the following steps are taken;

v² = u²+2as

v² - 2as = u²

take the square - root of both sides of the equation;

$\sqrt{u^{2} } = \sqrt{v^{2}-2as } \\\\u = \sqrt{v^{2}-2as }$

Thus, u as the subject of the given formula is, $u = \sqrt{v^{2}-2as }$

5 0

3 years ago

1. The length of a rectangle is 5 mm longer than its width. Its perimeter is more than 30 mm. Let w equal the width of the recta

VLD [36.1K]

The of the rectangle is w > 5mm

4 0

3 years ago

Find the x-intercept and y-intercept of the line.

castortr0y [4]

The y intercept is when x = 0 so it is:-

8(0) + y = 8
0 + y = 8

y = 6 <------- y-interdept

x intercept is when y = 0 so:-
8x + 0 = 8

x = 1 <-------- x intercept

8 0

3 years ago

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