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3 years ago

Segment BD is an altitude of triangle ABC. Find the area of the triangle.

Mathematics

2 answers:

inn [45]3 years ago

8 0

The answer is 7.5
Just count the squares and I did it so it is right.

hodyreva [135]3 years ago

3 0

Area = 1/2 * base * height
the height is BDand the base is AC
so its area = 1/2 * BD * AC

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The computer game Peter wants to buy will cost at least $50 and not more than $70. He earns $3 an hour running errands for his g

Lunna [17]

Answer:

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6 0

3 years ago

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Find the value of tan(sin^-1(1/2))

suter [353]

If you know that $\sin\dfrac\pi3=\dfrac12$ , then you know right away

$\tan\left(\sin^{-1}\dfrac12\right)=\tan\dfrac\pi3=\dfrac1{\sqrt}3=\dfrac{\sqrt3}3$

###

Otherwise, you can derive the same result. Let $\theta=\sin^{-1}\dfrac12$ , so that $\sin\theta=\dfrac12$ . $\sin^{-1}$ is bounded, so we know $-\dfrac\pi2\le\theta\le\dfrac\pi2$ . For these values of $\theta$ , we always have $\cos\theta\ge0$ .

So, recalling the Pythagorean theorem, we find

$\cos^2\theta+\sin^2\theta=1\implies\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\left(\dfrac12\right)^2}=\dfrac{\sqrt3}2$

Then

$\tan\theta=\tan\left(\sin^{-1}\dfrac12\right)=\dfrac{\sin\theta}{\cos\theta}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}=\dfrac{\sqrt3}3$

as expected.

5 0

3 years ago

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Need help what the awnser is.<br>b - 6 = 21

Alex787 [66]

27 hope this helps!!!

3 0

3 years ago

4/5 of 40

algol [13]

4/5 of 40 = 32
1/3 of 3 = 1
1/3 of 15 muffins = 10 muffins

3 0

3 years ago

Urgent algebra 2 help please

In-s [12.5K]

Answer:

y = 4x - 5

Step-by-step explanation:

y = mx + b

By putting above values in equation (i)

-1 = 4 (1) + b

-1 = 4 + b

-5 = b

b = -5

y = 4x - 5

5 0

1 year ago