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3 years ago

Which equation has a solution of -10? -28+32+x=-40 -28x+32x=-40 -28−32+x=6 -28x+23x=2

Mathematics

2 answers:

Leno4ka [110]3 years ago

8 0

I think it is the 3rd one but I might be wrong...

Damm [24]3 years ago

5 0

The other person is right! It is the third one...

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Flying against the wind, a jet travels 2190 miles in 3 hours. Flying with the wind, the same jet travels 9520 miles in 8 hours.

dezoksy [38]

Answer:

Velocity of jet in still air is 970 miles per hour and velocity of wind is 210 miles per hour.

Step-by-step explanation:

Jet's velocity against wind is

3040/4 = 760 miles per hour and flying with wind it is

8260/7 = 1180 miles per hour.

Let the velocity of jet in still air be x miles per hour and velocity of wind be y miles per hour.

As such its velocity against wind is x − y and with wind is x + y and therefore

x − y = 760 and x + y = 1180

Adding the two 2 x = 1940 and x = 970 and y = 1180 − 970 = 210

Hence velocity of jet in still air is 970 miles per hour and velocity of wind is

210 miles per hour.

Actual Answer:

https://socratic.org/questions/flying-against-the-wind-a-jet-travels-3040-miles-in-4-hours-flying-with-the-wind

7 0

3 years ago

What expression is equivalent to 5 (x+6)-2x+9.

myrzilka [38]

5x+30- 2x+9

3x+39 is the expression.

7 0

3 years ago

You are competing in a race. The table shows the times from last year's race. You want your time to be last year's median time w

olasank [31]

Answer:

28 ≤ x ≤40

Step-by-step explanation:

Arrange the data from the smallest to the largest

26, 27, 28,32, 32, 33, 35, 36, 38, 40,42,48

Identify the median value

26, 27, 28,32, 32,<u> 33, 35</u>, 36, 38, 40,42,48

Getting the median

(33+35)/2 = 68/2 = 34 minutes

You want your time to be last year's median time with an absolute deviation of at most 6 minutes

This mean ±6 minutes, hence this will be 34 ± 6

34+6=40 and 34-6=28

The inequality will be 28 ≤ x ≤40

7 0

3 years ago

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??anyone wanna help plz give me a answer not a link and ty

pishuonlain [190]

Answer:

light blue

bbbbbbbbccccc i said and brainly wont just let me type the answer if im wrong sorry

8 0

3 years ago

Need help on this calculus

N76 [4]

$\displaystyle\int\sin^3t\cos^3t\,\mathrm dt$

One thing you could do is to expand either a factor of $\sin^2t$ or $\cos^2t$ , then expand the integrand. I'll do the first.

You have

$\sin^2t=1-\cos^2t$

which means the integral is equivalent to

$\displaystyle\int\sin t(1-\cos^2t)\cos^3t\,\mathrm dt$

Substitute $u=\cos t$ , so that $\mathrm du=-\sin t\,\mathrm dt$ . This makes it so that the integral above can be rewritten in terms of $u$ as

$\displaystyle-\int(1-u^2)u^3\,\mathrm du=\int(u^5-u^3)\,\mathrm du$

Now just use the power rule:

$\displaystyle\int(u^5-u^3)\,\mathrm du=\dfrac16u^6-\dfrac14u^4+C$

Back-substitute to get the antiderivative back in terms of $t$ :

$\dfrac16\cos^6t-\dfrac14\cos^4t+C$

4 0

3 years ago

Read 2 more answers