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Drupady [299]
3 years ago
14

Find (f x g)(2)? Thanks

Mathematics
1 answer:
omeli [17]3 years ago
7 0

Let me break this up.

Note: (f o g)(2) = f(g(2)).

Note: g(2) = (2, 2) means when x is 2 y is 2.

In other words, g(2) is 2.

We know that f(g(2)) = f(2).

You can see that f(2) = -3. This is given in the point (2, -3).

So, f(g(2)) = -3.

Done.

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One more thanfour times a number is thirteen what is the number
brilliants [131]

Answer:

3

Step-by-step explanation:

subtract 1 from 13 divide the number by 4

7 0
3 years ago
If p varies directly as q and q= 70 when p= 10, find q when p = 12
stellarik [79]

\qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad \stackrel{\textit{constant of variation}}{y=\stackrel{\downarrow }{k}x~\hfill } \\\\ \textit{\underline{x} varies directly with }\underline{z^5}\qquad \qquad \stackrel{\textit{constant of variation}}{x=\stackrel{\downarrow }{k}z^5~\hfill } \\\\[-0.35em] ~\dotfill

\stackrel{\textit{"p" varies directly as "q"}}{p~~ = ~~kq}\qquad \textit{we also know that} \begin{cases} q=70\\ p=10 \end{cases}\implies 10=k70 \\\\\\ \cfrac{10}{70}=k\implies \cfrac{1}{7}=k\hspace{15em}\boxed{p=\cfrac{1}{7}k} \\\\\\ \textit{when p = 12, what is "q"?}\qquad 12=\cfrac{1}{7}q\implies 12=\cfrac{q}{7}\implies 84=q

5 0
1 year ago
What is the scientific notation of 0.000000000012
cestrela7 [59]

Answer:

1.2x10^-11

You never start with a zero so you move the decimal 11 spots the right which makes it negative

6 0
3 years ago
A 10.0 kg and a 2.0 kg cart approach each other on a horizontal frictionless air track. Their
Flura [38]

the  final speed in m/s of the 10.0 kg is 2.53 m/s .

<u>Step-by-step explanation:</u>

Here we have , A 10.0 kg and a 2.0 kg cart approach each other on a horizontal friction less air track. Their  total kinetic energy before collision is 96 ). Assume their collision is elastic. We need to find What is the  final speed in m/s of the 10.0 kg mass if that of the 2.0 kg mass is 8.0 m/s . Let's find out:

We know that in an elastic collision :

⇒ Total kinetic energy before collision  = Total kinetic energy after collision

⇒ 96 = \frac{1}{2}M_1(v_1)^2 + \frac{1}{2}M_2(v_2)^2

⇒ 96 = \frac{1}{2}(10)(v_1)^2 + \frac{1}{2}(2)(8)^2

⇒ 96=5(v_1)^2 + 64

⇒ 5(v_1)^2  =32

⇒ (v_1)^2  =6.4

⇒ v_1  =2.53 m/s

Therefore , the  final speed in m/s of the 10.0 kg is 2.53 m/s .

8 0
3 years ago
I’ll give brainliest
Nimfa-mama [501]

Answer:

Step-by-step explanation:

Remark

Area of an entire circle = pi * r^2

Area of a sector = angle / 360 * pi * r^2

Givens

angle = 140

radius = 8

Solution

Area = (140/360) * 3.14 * 8^2

Area =0.3889 * 3.14 * 64

Area = 78.15

4 0
3 years ago
Read 2 more answers
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