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2 years ago

6

So what is 27+22? Is it A 8338 B 8282 C 8228 D 8234

2 answers:

ivann1987 [24]2 years ago

6 0

Answer:

You put the add sign, so it would be 49. I don't believe that is what you meant to put though.

Explanation:

Y_Kistochka [10]2 years ago

4 0

i gotchu, 27+22 so you take the 8 and multiply that by the number of times you blink every 2 minutes and then divide that by 0×1 and carry the 3 and that gives you -7

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A child, hunting for his favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The ang

olga55 [171]

Answer:

9.22 s

Explanation:

One-quarter of a turn away is 1/4 of 2π, or π/2 which is approximately 1.57 rad

Let t (seconds) be the time it takes for the child to catch up with the horse. We would have the following equation of motion for the child and the horse:

For the child: $s_c = \omega_ct = 0.233t$

For the horse: $s_h = s_0 + a_ht^2/2 = 1.57 + 0.0136t^2/2 = 1.57 + 0.0068t^2$

For the child to catch up with the horse, they must cover the same angular distance within the same time t:

$s_c = s_h$

$0.233t = 1.57 + 0.0068t^2$

$0.0068t^2 - 0.233t + 1.57 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{0.233\pm \sqrt{(-0.233)^2 - 4*(0.0068)*(1.57)}}{2*(0.0068)}$

$t= \frac{0.233\pm0.11}{0.0136}$

t = 25.05 or t = 9.22

Since we are looking for the shortest time we will pick t = 9.22 s

6 0

3 years ago

Does a mirror work because of refraction

Semenov [28]

No. A mirror works because of reflection.

3 0

3 years ago

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WallyGPX accelerates from 0 m/s to 8 m/s in 3 seconds. What is his acceleration? Is this acceleration higher than that of a car

olga nikolaevna [1]

My Phone is +2348181686682

4 0

3 years ago

Two airplanes leave an airport at the same time.The velocity of the first airplane is 700 m/h at a heading of 31.3 the velocity

MArishka [77]

Here in order to find out the distance between two planes after 3 hours can be calculated by the concept of relative velocity

$v_{12} = v_1 - v_2$

here

speed of first plane is 700 mi/h at 31.3 degree

$v_1 = 700 cos31.3\hat i + 700 sin31.3\hat j$

$v_1 = 598.12\hat i + 363.7\hat j$

speed of second plane is 570 mi/h at 134 degree

$v_2 = 570 cos134 \hat i + 570 sin134 \hat j$

$v_2 = -396\hat i + 410\hat j$

now the relative velocity is given as

$v_{12} = (598.12 + 396)\hat i + (363.7 - 410)\hat j$

$v_{12} =994.12\hat i -46.3 \hat j$

now the distance between them is given as

$d = v* t$

$d = (994.12 \hat i - 46.3 \hat j)* 3$

$d = 2982.36\hat i - 138.9\hat j$

so the magnitude of the distance is given as

$d = \sqrt{2982.36^2 + 138.9^2}$

$d = 2985.6$ miles

so the distance between them is 2985.6 miles

5 0

2 years ago

Read 2 more answers

algol13

the answer should be C.

4 0

3 years ago

Read 2 more answers