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3 years ago

10

A textbook that Janelle needs for her psychology class costs $119 plus 7% sales tax. Which of the following equations correctly

calculates the final cost, C, of the textbook after taxes?A.119(0.7)=C B.119(1.07) = C C.119(1.7) = C D. 119(0.07 = C

1 answer:

Pani-rosa [81]3 years ago

3 0

119(1.07) = C <== ur equation

there is another equation that would also work...
119 + 0.07(119) = C

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The hypotenuse and one of the legs of a right triangle form an angle that has a cosine of 2 √2 .

Tems11 [23]

Quick answer I don't think this has an answer.

If you take the cos-1(2 sqrt(2)) your calculator should have a fit. Let's check that out. Mine certainly does. So there is something wrong with the question. If there is something to add in please do it and I will it least put an answer in the comments. As it stands, nothing will work.

If you put your calculator in radians, you will get an answer but it will not be anything resembling the choices you've listed.

If you meant sqrt(2) / 2 that would give 45o. Put it in your calculator like this 2 ^ 0.5 divided by 2 = 0.707
Cos - 1 (0.707) = 45

7 0

4 years ago

Read 2 more answers

Evaluate each trigonometric equation for θ.

liberstina [14]

Answer:

sinФ = 8/17

cosФ = 15/17

tan Ф = 8/15

csc Ф = 17/8

sec Ф = 17/15

cot Ф = 15/8

Step-by-step explanation:

Let us revise the trigonometry functions

Sin(x) = opposite/hypotenuse

Cos(x) = adjacent/hypoteouse

Tan(x) = opposite/adjacent

Csc(x) = hypotenuse/opposit

Sec(x) = hypotenues/adjacent

Cot(x) = adjacent/opposite

In the given figure

The opposite side to angle Ф = 8

The adjacent side to angle Ф = 15

Find the hypotenuse using Pythagoras' theorem

$Hypotenuse = \sqrt{8^{2}+15^{2} }$

$Hypotenuse = \sqrt{64+225}$

$Hypotenuse = \sqrt{289}$

$Hypotenuse = 17$

Let us use the rules above to find the trigonometry functions

sinФ = 8/17

cosФ = 15/17

tan Ф = 8/15

csc Ф = 17/8

sec Ф = 17/15

cot Ф = 15/8

4 0

3 years ago

15% of Alina’s workout consisted of cardio exercise. 25% of Dane’s workout consisted of cardio exercise. Which statement must be

Arisa [49]

The correct answer is B. If Alina’s workout was 80 minutes, and Dane’s workout was 48 minutes, they spent the same amount of time on cardio exercise.

To figure this out, first find 15% of 80 and then 25% of 48.

To find 15% of 80, you can first find 10% of 80 which is 8(just move the decimal one place to the left). The remaining 5% will be half of 8 since 8 is 10%. Add both the numbers up and you will get 12. (8+4=12)

To find 25% of 48, you can divide 48 by 4. Since percent means 'out of 100', 25% is 25 of 100. 100 divided by 25 is equal to 4. So, 48 divided by 4 is equal to 12.

So, the final answer is B. They both equal to 12.

Hope it helps :)

6 0

3 years ago

the outdoor temperature was 8 degrees at midnight the. temperature declined 5 degrees during each. of the next 3 hours what was

LenKa [72]

-7 degrees would be the answer

6 0

3 years ago

Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai

marta [7]

$f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1$

Lets check with every option

(a) [-4,-3]

We plug in -4 for x and -3 for x

$f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55$

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3 for x and -2 for x

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.

(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0 for x and 1 for x

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.

(f) [1,2]

We plug in 1 for x and 2 for x

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

$f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]

8 0

3 years ago

Read 2 more answers