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Veronika [31]
3 years ago
8

A gallon of paint covers 1 1/3 walls. How many gallons does it take to cover 1 wall? How about 3 walls?

Mathematics
1 answer:
Alekssandra [29.7K]3 years ago
6 0

Answer:

1 wall is 0.75 gallons. 3 walls is 2.25 gallons.

Step-by-step explanation:

if 1 gallon covers 1 1/3 wall, divide 1/1.3333 = 0.75

multiply 0.75(1 wall) by 3.... 0.75×3=2.25

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In Problems 23–30, use the given zero to find the remaining zeros of each function
Talja [164]

Answer:

x =  2i, x = -2i and x = 4 are the roots of given polynomial.

Step-by-step explanation:

We are given the following expression in the question:

f(x) = x^3 - 4x^2+ 4x - 16

One of the zeroes of the above polynomial is 2i, that is :

f(x) = x^3 - 4x^2+ 4x - 16\\f(2i) = (2i)^3 - 4(2i)^2+ 4(2i) - 16\\= -8i+ 16+8i-16 = 0

Thus, we can write

(x-2i)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16

Now, we check if -2i is a root of the given polynomial:

f(x) = x^3 - 4x^2+ 4x - 16\\f(-2i) = (-2i)^3 - 4(-2i)^2+ 4(-2i) - 16\\= 8i+ 16-8i-16 = 0

Thus, we can write

(x+2i)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16

Therefore,

(x-2i)(x+2i)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16\\(x^2 + 4)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16

Dividing the given polynomial:

\displaystyle\frac{x^3 - 4x^2 + 4x - 16}{x^2+4} = x -4

Thus,

(x-4)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16

X = 4 is a root of the given polynomial.

f(x) = x^3 - 4x^2+ 4x - 16\\f(4) = (4)^3 - 4(4)^2+ 4(4) - 16\\= 64-64+16-16 = 0

Thus, 2i, -2i and 4 are the roots of given polynomial.

4 0
3 years ago
Please help this is on a quiz
AleksAgata [21]

Answer:

4,5,8,11 are domain

Step-by-step explanation:

Since they are on the x side they are domain

6 0
3 years ago
For the third week of February, Betty Robinson worked 46 hours. Betty earns $17.50 an hour. Her employer pays overtime for all h
anygoal [31]

Answer:

1. 700

2. $157.50

3. $857.50

Step-by-step explanation:

Regular pay is calculated by taking 40 hours x $17.50 to get $700.

Overtime is calculated by taking 6 hours x 17.50 x 1.50 to get $157.50

Gross pay would be adding these two amounts together.

6 0
3 years ago
Estimate the probability that an Earthbound meteorite either disintegrates above or impacts with a body of water. Round to the n
Ad libitum [116K]

Answer:

0.71 is the answer

Step-by-step explanation:

Answered on edu

5 0
3 years ago
1. (5pts) Find the derivatives of the function using the definition of derivative.
andreyandreev [35.5K]

2.8.1

f(x) = \dfrac4{\sqrt{3-x}}

By definition of the derivative,

f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}

We have

f(x+h) = \dfrac4{\sqrt{3-(x+h)}}

and

f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}

Combine these fractions into one with a common denominator:

f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}

3.1.1.

f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3

Differentiate one term at a time:

• power rule

\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4

\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}

\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}

The last two terms are constant, so their derivatives are both zero.

So you end up with

f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}

8 0
2 years ago
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