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bagirrra123 [75]
3 years ago
12

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
Read 2 more answers
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