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Alona [7]
3 years ago
12

SUPER EASY PROBLEM PLEASE HELP!

Mathematics
1 answer:
zzz [600]3 years ago
8 0

9514 1404 393

Answer:

  110%

Step-by-step explanation:

The whole left square is shaded, so that is 100%.

The right square is divided into 10 equal parts, so each one is 100%/10 = 10% of the whole. One of those is shaded.

The whole amount shaded is ...

  100% + 10% = 110%

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Alekssandra [29.7K]

Answer:

-1/2

It is a negative slope going down 1 over 2

7 0
3 years ago
Read 2 more answers
TELE
Aleonysh [2.5K]

Answer:

$23271.49

Step-by-step explanation:

The function C(t) = C(1 + r)^{t} models the rise in the cost of a product that has a cost of C today, subject to an average yearly inflation rate of r for t years.

Now, if the average annual rate of inflation over the next 8 years is assumed to be 2.5% then we have to find the inflation-adjusted cost of a $19100 motorcycle after 8 years.

Therefore, the cost will be C(8) = 19100(1 + \frac{2.5}{100} )^{8} = 23271.49 dollars. (Answer)

8 0
3 years ago
The force of gravity on Mars is different than on Earth. The function of the same situation on Mars would be represented by the
sweet-ann [11.9K]

Answer:

If thrown up with the same speed, the ball will go highest in Mars, and also it would take the ball longest to reach the maximum and as well to return to the ground.

Step-by-step explanation:

Keep in mind that the gravity on Mars; surface is less (about just 38%) of the acceleration of gravity on Earth's surface. Then when we use the kinematic formulas:

v=v_0+a\,*\,t\\y-y_0=v_0\,* t + \frac{1}{2} a\,\,t^2

the acceleration (which by the way is a negative number since acts opposite the initial velocity and displacement when we throw an object up on either planet.

Therefore, throwing the ball straight up makes the time for when the object stops going up and starts coming down (at the maximum height the object gets) the following:

v=v_0+a\,*\,t\\0=v_0-g\,*\,t\\t=\frac{v_0}{t}

When we use this to replace the 't" in the displacement formula, we et:

y-y_0=v_0\,* t + \frac{1}{2} a\,\,t^2\\y-y_0=v_0\,(\frac{v_0}{g} )-\frac{g}{2} \,(\frac{v_0}{g} )^2\\y-y_0=\frac{1}{2} \frac{v_0^2}{g}

This tells us that the smaller the value of "g", the highest the ball will go (g is in the denominator so a small value makes the quotient larger)

And we can also answer the question about time, since given the same initial velocity v_0 , the smaller the value of "g", the larger the value for the time to reach the maximum, and similarly to reach the ground when coming back down, since the acceleration is smaller (will take longer in Mars to cover the same distance)

3 0
3 years ago
Pls help me with this !!!!! it is urgent
OLEGan [10]
The answer is 1A definitely
4 0
3 years ago
12 yd<br> 4 yd<br> 7 yd<br> 7yd<br> 4yd<br> 4 yd<br> 4 yd<br> 4 yd<br> what is the area
N76 [4]
I think the answer to it is 2,688
6 0
3 years ago
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