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ivolga24 [154]
2 years ago
6

PLEASE HELP ASAP Due in 1 day!!!

Mathematics
1 answer:
Helen [10]2 years ago
7 0

Step-by-step explanation:

Q1 . (f+g)(x) = f(x) + g(x)

=4x-4 +2x^2 -3x

= 2x^2 + x -4

Q2. (f-g)(x) = f(x) - g(x)

= 2x^2−2 - (4x+1)

= 2x^2 -2 -4x -1

= 2x^2 - 4x -3

Q3. h(x)=3x−3 and g(x)=x^2+3

(h.g)(x) = h(x) × g(x)

= (3x-3) × (x^2 + 3)

=3x^3 -3x^2 + 9x -9

Q4.f(x)=x+4 and g(x)=x+6

(f/g)(x) = f(x) ÷ g(x)

= x+4 / x+6

the domain restriction is x>-6

x<-6

x doesn't equal (-6)

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All the edges of a cube are expanding at a rate of 4 in. per second. How fast is the volume changing when each edge is 10 in. lo
Greeley [361]

Given:

All the edges of a cube are expanding at a rate of 4 in. per second.

To find:

The rate of change in volume when each edge is 10 in. long.

Solution:

Let a be the edge of the cube.

According to the question, we get

\dfrac{da}{dt}=4\text{ in./sec}

a=10\text{ in.}

We know that, volume of a cube is

V=a^3

Differentiate with respect to t.

\dfrac{dV}{dt}=\dfrac{d}{dt}a^3

\dfrac{dV}{dt}=(3a^2)\times \dfrac{da}{dt}

Putting the given values, we get

\dfrac{dV}{dt}=(3(10)^2)\times 4

\dfrac{dV}{dt}=3(100)\times 4

\dfrac{dV}{dt}=300\times 4

\dfrac{dV}{dt}=1200\text{ in}^3\text{/sec}

Therefore, the rate of change in volume 1200 cubic inches per second.

5 0
3 years ago
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Step-by-step explanation:

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3 0
3 years ago
Help me please, i need help with all the questions. My teacher did not explain very well
belka [17]
The questions that are 1-6 above:

Use the term
P- Parenthesis
E- Exponents
M-Multiplying
D-Dividing
A- Addition
S-Subtraction
Do the equations from that order doing Parenthesis first, Exponents second, etc.
If there's none of one of the operations, skip to the next one. Remember to solve the equation starting from parenthesis and then going down the order.
Sorry I can't help you with the bottom 1-6 Good luck
8 0
2 years ago
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