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spayn [35]
1 year ago
13

What is the area of a triangle with a base of 10 cm and a height of 12 cm?

Mathematics
2 answers:
STALIN [3.7K]1 year ago
7 0

Answer:

60 square cm

Step-by-step explanation:

Tpy6a [65]1 year ago
7 0
60 square cm or 120 square cm
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3. One weekend, 152 peopleattended an art exhibit. Thefollowing weekend, 298 peopleattended. To the nearest percent,what was the
castortr0y [4]

We can calculate the percent change dividing the change by the original amount of people attending the exhibit.

The increase is 298-152 = 146.

Then, we can calculate the percent change as:

\frac{298-152}{152}=\frac{146}{152}=0.9605=96\%

There was a 96% increase in the number of people attending the exhibit.

7 0
1 year ago
Please help I’m stuck on this question
yan [13]
It helps demonstrate it since it forms a right triangle in the middle with 2 base lengths and a hypotenuse. The areas of the squares can be plugged into the pythagorean theorem in order to find the answer
3 0
2 years ago
What is the quotient of 6 and -1/2
Grace [21]
Quotient means divide, meaning we'll divide 6 by -1/2.
6 \div - \frac{1}{2}
When dealing with fraction division, we flip the fraction in the second place and then change the problem to multiplication.
6 \cdot - \frac{2}{1}
Anything over 1 is just itself, so -2/1 becomes -2.
6 \cdot -2
Finally, we multiply 6 and -2.
6 \cdot -2 = -12
So the quotient of 6 and -1/2 is -12.

8 0
3 years ago
Read 2 more answers
I'll mark you brainlist
tamaranim1 [39]

Answer:

don't know but I'll give it a shot

Step-by-step explanation:

-6 < X < 6

-6 is smaller than or equal to X and X is smaller than or equal to 6

3 0
2 years ago
At time t = 0 a car has a velocity of 16 m/s. It slows down with an acceleration given by –0.50t, in m/s^2 for t in seconds. By
anygoal [31]

The car's velocity at time t is

v(t)=16\dfrac{\rm m}{\rm s}+\displaystyle\int_0^t\left(\left(-0.50\frac{\rm m}{\mathrm s^2}\right)u\right)\,\mathrm du=16\dfrac{\rm m}{\rm s}+\left(-0.25\dfrac{\rm m}{\mathrm s^2}\right)t^2

It comes to rest at

v(t)=0\implies16\dfrac{\rm m}{\rm s}=\left(0.25\dfrac{\rm m}{\mathrm s^2}\right)t^2\implies t=8.0\,\mathrm s

Its velocity over this period is positive, so that the total distance the car travels is

\displaystyle\int_0^{8.0}v(t)\,\mathrm dt=\left(16\dfrac{\rm m}{\rm s}\right)(8.0\,\mathrm s)+\frac13\left(-0.25\dfrac{\rm m}{\mathrm s^2}\right)(8.0\,\mathrm s)^3=\boxed{85\,\mathrm m}

so the answer is D.

4 0
2 years ago
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