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SashulF [63]
3 years ago
7

Write an equation that is perpendicular to y = 2x + 5 and passes through (3 , -2)

Mathematics
1 answer:
Maslowich3 years ago
3 0
To find the perpendicular line, we need to find the negative inverse slope.

y = -1/2x - 2

This line also passes through the point since both have the y intercept / value of -2.

Hope this helps!
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Given f (x) = x +1 and g(x) = x², what is (gºf)(x)?
Gelneren [198K]

Answer:

Solution given:

f(x) = x +1 and g(x) = x²,

now

(gºf)(x) =g(fx)=g(x+1)=(x+1)²

<u>B) (gºf)(</u><u>x</u><u>) = (x + 1)^2</u><u />

7 0
2 years ago
help pleaseeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
krek1111 [17]

Answer:

-31>v

Step-by-step explanation:

look above there

hope it helps

8 0
2 years ago
Read 2 more answers
Can someone help me Please
alisha [4.7K]
Jane:  Think:  the hypotenuse of her ramp is 14 inches and the angle opposite the rise is 30 degrees.  Thus, cos 30 deg = (adj side) / (hyp) = (adj side) / (14 in).

Solving for (adj side), we get   (adj side) = (14 in)(sqrt(3)/2) = 7 sqrt(3) inches.
3 0
3 years ago
Cna i get some help with dis plz
Tomtit [17]

Let b_1,b_2,\ldots,b_{20} be the 20 marks of the boys, and g_1,g_2,\ldots,g_{10} be the 10 marks of the girls.

We know that the global mean was 70, meaning that

\dfrac{b_1+b_2+\ldots+b_{20}+g_1+g_2+\ldots+g_{10}}{30}=70

Multiplying both sides by 30 we deduce that the sum of the scores of the whole classroom is

b_1+b_2+\ldots+b_{20}+g_1+g_2+\ldots+g_{10}=2100

By the same logic, we work with the marks of the boys alone: we know the average:

\dfrac{b_1+b_2+\ldots+b_{20}}{20}=62

And we deduce the sum of the marks for the boys:

b_1+b_2+\ldots+b_{20}=1240

Which implies that the sum of the marks of the girls is 2100-1240=860

And finally, the mean for the girls alone is

\dfrac{860}{10}=86

6 0
3 years ago
How do I get it (it’s worth 3 marks)
Stella [2.4K]

Answer:

58.5

Step-by-step explanation:

6.5*9=58.5

3 0
3 years ago
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