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2 years ago

13

A car dealership sold 850 cars last year. This year the dealership sold 715 cars. What is the percent decrease, to the nearest p

ercent, in the number of cars sold from last year to this year?????

2 answers:

Bumek [7]2 years ago

3 0

The answer is 16%

Hope this helps :)

Nitella [24]2 years ago

3 0

Answer:

15.8824% decrease

Step-by-step explanation:

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Please help! i might be wrong so please correct me. thanks for the help i will reward points!

olya-2409 [2.1K]

Hold on let me solve this problem meet em in the comments for the awnser

4 0

3 years ago

Find the x-coordinates where f '(x) = 0 for f(x) = 2x - sin(2x) in the interval [0, 2π].

Serggg [28]

Answer:

The x-coordinates where f '(x) = 0 for f(x) = 2x - sin(2x) in the interval [0, 2π] are: x=0, π, and 2π.

Step-by-step explanation:

f(x)=2x-sin(2x)

f'(x)=[f(x)]'

f'(x)=[2x-sin(2x)]'

f'(x)=(2x)'-[sin(2x)]'

f'(x)=2-cos(2x) (2x)'

f'(x)=2-cos(2x) (2)

f'(x)=2-2 cos(2x)

f'(x)=0, then:

2-2 cos(2x)=0

Solving for x. First solving for cos(2x): Subtracting 2 from both sides of the equation:

2-2 cos(2x)-2=0-2

-2 cos(2x)=-2

Dividing both sides of the equation by -2:

-2 cos(2x) / (-2)=-2 / (-2)

cos(2x)=1

If cos(2x)=1, then:

2x=2nπ

Dividing both sides of the equation by 2:

2x/2=2nπ/2

x=nπ

If n=-1→x=-1π→x=-π, and -π is not in the interval [0,2π], then x=-π is not in the solution.

If n=0→x=0π→x=0, and 0 is in the interval [0,2π], then x=0 is in the solution.

If n=1→x=1π→x=π, and π is in the interval [0,2π], then x=π is in the solution.

If n=2→x=2π, and 2π is in the interval [0,2π], then x=2π is in the solution.

If n=3→x=3π, and 3π is not in the interval [0,2π], then x=2π is not in the solution.

Solution: x=0, π, and 2π.

5 0

2 years ago

Drag each capacity to match it to an equivalent capacity.

weqwewe [10]

Answer:

2 cups ⇔ 16 fluid ounces
96 ounces ⇔ 6 pints
12 pints ⇔ 24 cups
4 quarts ⇔ 1 gallon

<em>Also see attached</em>

3 0

2 years ago

Read 2 more answers

Use a linear approximation (or differentials) to estimate the given number. (Round your answer to five decimal places.) 3 217

Soloha48 [4]

Answer:

$f(216) \approx 6.0093$

Step-by-step explanation:

Given

$\sqrt[3]{217}$

Required

Solve

Linear approximated as:

$f(x + \triangle x) \approx f(x) +\triangle x \cdot f'(x)$

Take:

$x = 216; \triangle x= 1$

So:

$f(x) = \sqrt[3]{x}$

Substitute 216 for x

$f(x) = \sqrt[3]{216}$

$f(x) = 6$

So, we have:

$f(x + \triangle x) \approx f(x) +\triangle x \cdot f'(x)$

$f(215 + 1) \approx 6 +1 \cdot f'(x)$

$f(216) \approx 6 +1 \cdot f'(x)$

To calculate f'(x);

We have:

$f(x) = \sqrt[3]{x}$

Rewrite as:

$f(x) = x^\frac{1}{3}$

Differentiate

$f'(x) = \frac{1}{3}x^{\frac{1}{3} - 1}$

Split

$f'(x) = \frac{1}{3} \cdot \frac{x^\frac{1}{3}}{x}$

$f'(x) = \frac{x^\frac{1}{3}}{3x}$

Substitute 216 for x

$f'(216) = \frac{216^\frac{1}{3}}{3*216}$

$f'(216) = \frac{6}{648}$

$f'(216) = \frac{3}{324}$

So:

$f(216) \approx 6 +1 \cdot f'(x)$

$f(216) \approx 6 +1 \cdot \frac{3}{324}$

$f(216) \approx 6 + \frac{3}{324}$

$f(216) \approx 6 + 0.0093$

$f(216) \approx 6.0093$

6 0

3 years ago

Check out if I'm right<br> Plz explain..if u have a different answer

Ronch [10]

Answer:

A C

Step-by-step explanation:

A

A is true.The numerator does have 3 terms. The constant term (the 3 at the end) is still a term and counts as a term.

B

A constant term does not count as a coefficient so b is not right, just as you have indicated by not underlining it.

C

C is true. In fact, 2 is the leading coefficent of the denominator.

D

Either A is true or D is, but I don't think they both are. You need a modifier to claim that D is true. If the statement said the numerator has <em>at least </em>2 terms then both A and D would be true. Without the at least, you have to pick one and the one I choose is A

E

This is the tough one. Very sticky. You would think it is true, but it isn't. A constant term is not a coefficient. Coefficients only count when they are with "x"s.

8 0

2 years ago

Read 2 more answers