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3 years ago

11

On the surface, it seems easy. Can you think of the integers for x, y, and z so that x³+y³+z³=8? Sure. One answer is x = 1, y =

-1, and z = 2. But what about the integers for x, y, and z so that x³+y³+z³=42? what do the integers add up to????? solve it anytime

1 answer:

rosijanka [135]3 years ago

7 0

Answer:

Well ,lets see∉ if you add theose numbers togetherФ it maybe it he answer is 4=67 *64 ---4 67x54 5xe4t4 6x

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40 points!!! Please Help!!! (MATH) Will give Brainliest!!!

Vedmedyk [2.9K]

Answer:

A =62x^2

Step-by-step explanation:

The formula for surface area of a rectangular prism is given by

A=2(wl+hl+hw)

Given a width of 2x and length of 3x and a height of 5x, substitute into the equation

A = 2(2x*3x+ 5x*3x+ 2x*5x)

A = 2(6x^2 + 15x^2 +10x^2)

Combine like terms

A = 2(31x^2)

A =62x^2

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3 years ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

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4 years ago

Hi can someone help me with question and how you got the answer? thanks :)

Triss [41]

Answer:

Convert 27 4/3 to an improper fraction.

Step-by-step explanation:

Exact Form:

85/3

Decimal Form:

28.¯3

Mixed Number Form:

28 1/3

5 0

3 years ago

Read 2 more answers

A school P is 16km due west of a school Q. what is the bearing of Q from P

Irina-Kira [14]

Answer:

16Km due east of school P

Step-by-step explanation:

Given

A school P is 16km due west of a school Q

Thus, we can say that distance PQ = 16 km.

________________________________

now we have to find the bearing of Q from P

As distance is same

distance PQ = distance QP

Thus,

Distance will remain same of 16 km.

For direction,

If Q is west of P, then P will be east of Q $P------------------>Q$

as shown in figure P is west of Q,

now from point P , Q is west P.

Thus,

Bearing of School Q from P is 16Km due east of school P

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4 years ago

Need help- Find the measure of the angle.<br><br> THANK YOU

snow_tiger [21]

Answer:

47

Step-by-step explanation:

Remember triangle always equal to 180

180=70+4x-5+6x-15

x=13

4(13)-5

47

Angle A is 47

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3 years ago