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Klio2033 [76]
3 years ago
11

If f(x) = 3x + 2 and g(x) = x2 + 1, which expression is equivalent to (F.g) (x)

Mathematics
1 answer:
Misha Larkins [42]3 years ago
3 0

For those who may still need it in the future, the correct answer is actually: D: 3(x^2 + 1) +2

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4 1/6 pounds of rice is divided into 5 equal portions. What is the weight in pounds of each portion?
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5/6 or 0.83 lbs is the anwser
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Zach spent 2/3 hour reading on Friday and 1 1/3 hours reading on Saturday. how much more time did he read on Saturday than on Fr
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11/3 - 2/3 = 9/3 = 3 so he spent 3 more hours reading on saturday than on friday


Hope this helps!
6 0
3 years ago
Orthogonalizing vectors. Suppose that a and b are any n-vectors. Show that we can always find a scalar γ so that (a − γb) ⊥ b, a
jeka57 [31]

Answer:

\\ \gamma= \frac{a\cdot b}{b\cdot b}

Step-by-step explanation:

The question to be solved is the following :

Suppose that a and b are any n-vectors. Show that we can always find a scalar γ so that (a − γb) ⊥ b, and that γ is unique if b \neq 0. Recall that given two vectors a,b  a⊥ b if and only if a\cdot b =0 where \cdot is the dot product defined in \mathbb{R}^n. Suposse that b\neq 0. We want to find γ such that (a-\gamma b)\cdot b=0. Given that the dot product can be distributed and that it is linear, the following equation is obtained

(a-\gamma b)\cdot b = 0 = a\cdot b - (\gamma b)\cdot b= a\cdot b - \gamma b\cdot b

Recall that a\cdot b, b\cdot b are both real numbers, so by solving the value of γ, we get that

\gamma= \frac{a\cdot b}{b\cdot b}

By construction, this γ is unique if b\neq 0, since if there was a \gamma_2 such that (a-\gamma_2b)\cdot b = 0, then

\gamma_2 = \frac{a\cdot b}{b\cdot b}= \gamma

6 0
3 years ago
Samuel must choose a 4-digit PIN number. The first 2 digits must be numbers cannot repeat. The 3 and 4th digits are letters that
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7 0
3 years ago
the measurements of a photo and it's frame are shown in the diagram. Write a polynomial that represents the width of the photo.
suter [353]

Answer:

The width of the photo is 4w^2+6w+4.

Step-by-step explanation:

From the given figure it is notices that the total width of the frame is

6w^2+8

The photo is covered by a frame border and the width of the border is

w^2-3w+2

To find the width of the photo we have to subtract the width of upper frame border and lower frame border from the total width of frame.

Width of the photo is

\text{Width of the photo}=\text{Width of the frame}-2(\text{Width of the frame border})

\text{Width of the photo}=6w^2+8-2(w^2-3w+2)

\text{Width of the photo}=6w^2+8-2w^2+6w-4

\text{Width of the photo}=4w^2+6w+4

Therefore the width of the photo is 4w^2+6w+4.

4 0
3 years ago
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