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svetlana [45]
3 years ago
12

30 points if you answer this question and please show work !!

Mathematics
1 answer:
worty [1.4K]3 years ago
5 0
I don't know what is between f(x) and g(x), so (for now) there are two posible answers:

1. f(x)g(x) = f(x)\cdot g(x) - function multiplication:

f(x)g(x)=(2x-3)\cdot(3x+8)=2x\cdot3x+2x\cdot8-3\cdot3x-3\cdot8=\\\\=6x^2+16x-9x-24=\boxed{6x^2+7x-24}

2. f(x)g(x)=f(x)\circ g(x) - function composition:

f(x)g(x) = f(x)\circ g(x) = f\big(g(x)\big)=2(3x+8)-3=2\cdot3x+2\cdot8-3=\\\\=6x+16-3=\boxed{6x+13}


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Find the measure of the angle formed by the hands of a clock when it is 7:00 (the angle formed between 7 to 12, clockwise).
aev [14]
The answer you are looking for is an obtuse so it would be more than 90 degrees. may i have brainlest
6 0
2 years ago
1. h(2) + g(2) *<br> h(t) = 3-5<br> g(t) = 2t - 5
kap26 [50]

Answer:

h(2)+g(2) = -3

Step-by-step explanation:

h(2) +g(2)\\ \\h(t) = 3-5, g(t) = 2t -5

        Replace the variable  (t)  with (

2)  in the expression.

h (2) = 3 - 5

        Replace the variable  (t)  with (

2)  in the expression.

g(2) = 2(2) -5

        Replace the function designators in h(2) +g(2) with the actual functions.

h(t) = 3 - 5 +2 (2) ← plug h(2) into 2(t)

         Remove parentheses.

3 - 5 + 2(2)

         Multiply 2 by 2

3 - 5 + 4 - 5

         Subtract  5  from  3

.

-2 + 4 - 5

          Add -2 and 4

2 - 5

          Subtract 5 from 2

-3

7 0
3 years ago
Find the difference -11 - -5, 25 - (-3) ,-7 - (-2), 4 - 15, 6 - (-9)
Leviafan [203]

Answer:

-11-(-5)= -6

25-(-3)= 28

-7-(-2)= -5

4-15= -11

6-(-9)= 15

7 0
3 years ago
Area A, equals the product of b and h divided by 2, slove for h
gizmo_the_mogwai [7]
A = bh/2
2a = bh
2a/b = h

h = 2a/b
7 0
3 years ago
Calculate the total area of the shaded region.
LUCKY_DIMON [66]

so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

\stackrel{f(x)}{2x^3-x^2-5x}~~ = ~~\stackrel{g(x)}{-x^2+3x}\implies 2x^3-5x=3x\implies 2x^3-8x=0 \\\\\\ 2x(x^2-4)=0\implies x^2=4\implies x=\pm\sqrt{4}\implies x= \begin{cases} 0\\ \pm 2 \end{cases}

so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill

\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}

7 0
2 years ago
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