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Sveta_85 [38]
2 years ago
13

At Best Buy they have a 43 tv that sells for 865 and is on sale for 30% off how much would they pay for the tv after the discoun

t
Mathematics
1 answer:
otez555 [7]2 years ago
4 0

Answer:

$605.50 after the discount

Step-by-step explanation:

Take the original price, 865, and multiply it by 70% of the discount still on.

865 x 0.7 = $605.50 after the discount

If this answer is correct, please make me Brainliest!

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Which expression is equivalent to 9x – 2x2 + 4x – x?
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-2x^2+12x

Step-by-step explanation:

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2 years ago
Help me with A and B please
g100num [7]

Part A.

Ashwin had a 4-cm cube.

volume of cube = side^3

volume = (4 cm)^3 = 64 cm^3

Ashwin has 64 small cubes.

We need to find the volumes of all prisms in Part A. Only prisms with at most 64 cm^3 volume can be the answer.

A. v = 2 * 4 * 7 = 28

B. v = 4 * 5 * 6 = 120

C. v = 5 * 5 * 4 = 100

D. v = 4 * 7 * 5 = 140

E. v = 3 * 5 * 4 = 60

Part A. answer: A, E

Part B.

Dora had a 5-cm cube.

volume of cube = side^3

volume = (5 cm)^3 = 125 cm^3

Dora has 125 small cubes.

We need to find the volumes of all prisms in Part B. Only prisms with at most 125 cm^3 volume can be the answer.

A. v = 3 * 3 * 8 = 72

B. v = 3 * 4 * 5 = 60

C. v = 6 * 6 * 4 = 144

D. v = 5 * 8 * 4 = 160

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6 0
3 years ago
Please help me with this please and thank you please actually help me
fgiga [73]

Answer:

-5, -8, -11

Step-by-step explanation:

it every -3

5 0
2 years ago
Read 2 more answers
Simplify the expression. tan(sin^−1 x)
Blizzard [7]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2799412

_______________


Let  \mathsf{\theta=sin^{-1}(x)\qquad\qquad-\dfrac{\pi}{2}\ \textless \ \theta\ \textless \ \dfrac{\pi}{2}.}

(that is the range of the inverse sine function).


So,

\mathsf{sin\,\theta=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\theta=x\qquad\quad(i)}


Square both sides:

\mathsf{sin^2\,\theta=x^2\qquad\qquad(but~sin^2\,\theta=1-cos^2\,\theta)}\\\\ \mathsf{1-cos^2\,\theta=x^2}\\\\ \mathsf{1-x^2=cos^2\,\theta}\\\\ \mathsf{cos^2\,\theta=1-x^2}


Since \mathsf{-\,\dfrac{\pi}{2}\ \textless \ \theta\ \textless \ \dfrac{\pi}{2},} then \mathsf{cos\,\theta} is positive. So take the positive square root and you get

\mathsf{cos\,\theta=\sqrt{1-x^2}\qquad\quad(ii)}


Then,

\mathsf{tan\,\theta=\dfrac{sin\,\theta}{cos\,\theta}}\\\\\\ \mathsf{tan\,\theta=\dfrac{x}{\sqrt{1-x^2}}}\\\\\\\\ \therefore~~\mathsf{tan\!\left[sin^{-1}(x)\right]=\dfrac{x}{\sqrt{1-x^2}}\qquad\qquad -1\ \textless \ x\ \textless \ 1.}


I hope this helps. =)


Tags:  <em>inverse trigonometric function sin tan arcsin trigonometry</em>

3 0
3 years ago
Read 2 more answers
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