2 years ago

If f(x)=x^2-3 and

1 answer:

4 0

Since it is f of g you have to start by plugging in g(x) into f(x)

(4-3x)^2-3

Then just solve the equation

(4-3x)^2-3
(4-3x)(4-3x)-3 Square the binomial
8-12x-12x+9x^2-3 Combine like terms
5-24x+9x^2 Final answer

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Three dozens of juice boxes cost $4.80, how much do 60 juice boxes cost

Leya [2.2K]

Answer:

Step-by-step explanation:

36 juice = $4.80

60 juice = ?

36/6 = 4.80/6

6 juice = 0.8

0.8x10=8

5 0

3 years ago

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A textbook store sold a combined total of 290 math and history textbooks in a week. The number of math textbooks sold was 64 mor

11Alexandr11 [23.1K]

Answer:

113

Step-by-step explanation:

290 - 64 = 226

226 ÷ 2 = 113

4 0

2 years ago

Help pleaseeeeeeeeee

harkovskaia [24]

97+x=27+64
x=27+64-97
x=-6

(Hopefully this is correct)

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2 years ago

What is 3/7 × 5/6 thanks

aleksley [76]

3/7*5/6

cross out 3 and 6

divide by 3

3/3= 1

6/3= 2

1/7*5/2

mutiply the denominators together

7*2= 14

mutiply the numerators together

1*5= 5

Answer:

5/14

3 0

2 years ago

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Find the product of z1 and z2, where z1 = 7(cos 40° + i sin 40°) and z2 = 6(cos 145° + i sin 145°).

Oduvanchick [21]

For two complex numbers $z_1=re^{i\theta}=r(\cos\theta+i\sin\theta)$ and $z_2=se^{i\varphi}=s(\cos\varphi+i\sin\varphi)$ , the product is

$z_1z_2=rse^{i(\theta+\varphi)}=rs(\cos(\theta+\varphi)+i\sin(\theta+\varphi))$

That is, you multiply the moduli and add the arguments. You have $z_1=7e^{i40^\circ}$ and $z_2=6e^{i145^\circ}$ , so the product is

$z_1z_2=7\times6(\cos(40^\circ+145^\circ)+i\sin(40^\circ+145^\circ)=42(\cos185^\circ+i\sin185^\circ)=42e^{i185^\circ}$

3 0

2 years ago

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