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Vlad [161]
4 years ago
8

i Dont get how u make a the subject and get a on its own can u please help me with all the questions even if u only know the ans

wers too a few of them

Mathematics
1 answer:
FrozenT [24]4 years ago
7 0
It just means to get a by itself on one side of the equal sign. For the first equation subtract 7p from both side. The equation would then look like 3a=8-7p. After, you divide 3 from both sides and the answer for the first part would be a=8-7p/3 (8-7p will all be divided by 3). If you are having problems with the other equation I'll be happy to help.
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ollegr [7]

Answer:

A counter clockwise

Step-by-step explanation:

4 0
3 years ago
Find the slope of the line that passes through the given points (7, 3) and (13, 8)​
k0ka [10]

Answer:

The slope is 5/8.

Step-by-step explanation:

Use the slope formula:

m = y2 - y1 / x2 - x1

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m = 8 - 3 / 13 - 7

m = 5/8

4 0
3 years ago
What is the area of 1/2x7(4+2)
solong [7]
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8 0
3 years ago
If y = 3 when x = 2. find x when y = 5.
ahrayia [7]
Y = 3   x = 2
if y = 1  x = ?

x = 3/2
x = 1.5
x = 5*1.5 ( y = 5  x = ?)
x = 7.5       
(IF THE ANSWER IS WRONG)
may be this can be that answer

x = 2/3
x = 0.67
x = 0.67*5
x = 3.35

check the answer either first or second method is correct

8 0
3 years ago
State the vertical asymptote of the rational function. f(x) =((x-9)(x+7))/(x^2-4)
Lubov Fominskaja [6]
D:x^2-4\not=0\\
D:x^2\not=4\\
D:x\not=-2 \wedge x\not =2\\\\
\displaystyle
\lim_{x\to-2^-}\dfrac{(x-9)(x+7)}{x^2-4}=\\
\dfrac{(-2-9)(-2+7)}{(-2^-)^2-4}=\dfrac{-11\cdot5}{4^+-4}=\dfrac{-55}{0^+}=-55\cdot\infty=-\infty\\
\dfrac{(-2-9)(-2+7)}{(-2^+)^2-4}=\dfrac{-11\cdot5}{4^--4}=\dfrac{-55}{0^-}=-55\cdot(-\infty)=\infty


\displaystyle
\lim_{x\to2^-}\dfrac{(x-9)(x+7)}{x^2-4}=\\
\dfrac{(2-9)(2+7)}{(2^-)^2-4}=\dfrac{-7\cdot9}{4^--4}=\dfrac{-63}{0^-}=-63\cdot(-\infty)=\infty\\
\dfrac{(2-9)(2+7)}{(2^+)^2-4}=\dfrac{-7\cdot9}{4^+-4}=\dfrac{-63}{0^+}=-63\cdot\infty=-\infty\\

So, the vertical asymptotes are x=\pm 2
5 0
4 years ago
Read 2 more answers
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