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3 years ago

What similarity statement can you write relating the three triangles in the diagram?

Mathematics

1 answer:

GarryVolchara [31]3 years ago

8 0

Answer:

They are both right-angled triangles.

Step-by-step explanation:

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Write two linear expressions with a sum of -5x+4

murzikaleks [220]

-10x+8 and 5x-4 is a possible combination. Those are both linear equations when added together will give you -5x+4

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3 years ago

Suppose you are choosing between checking account A, which charges a $7.50 monthly fee, $1 per check, and $2 per ATM visit; and

Kisachek [45]

I'd go with Checking account B because its cheaper .

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3 years ago

A rectangle is removed from a right triangle to create the shaded region shown below. Find the area of the shaded region.

irina [24]

Answer:

I can't tell what the triangle is

Step-by-step explanation:

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3 years ago

If sin theta = (4)/(7), theta in quadrant II, find the exact value of (a) cos theta (b) sin (theta + (pi) / (6) ) (c) cos (the

EleoNora [17]

Answer:

a) $\cos(\theta) = \frac{\sqrt[]{33}}{7}$

b) $\sin(\theta + \frac{\pi}{6})\frac{-3\sqrt[]{11}+4}{14}$

c) $\cos(\theta-\pi)=\frac{\sqrt[]{33}}{7}$

d) $\tan(\theta + \frac{\pi}{4}) = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}$

Step-by-step explanation:

We will use the following trigonometric identities

$\sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)$

$\cos(\alpha+\beta) = \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$ $\tan(\alpha+\beta) = \frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$ .

Recall that given a right triangle, the sin(theta) is defined by opposite side/hypotenuse. Since we know that the angle is in quadrant 2, we know that x should be a negative number. We will use pythagoras theorem to find out the value of x. We have that

$x^2+4^2 = 7 ^2$

which implies that $x=-\sqrt[]{49-16} = -\sqrt[]{33}$ . Recall that cos(theta) is defined by adjacent side/hypotenuse. So, we know that the hypotenuse is 7, then

$\cos(\theta) = \frac{-\sqrt[]{33}}{7}$

b)Recall that $\sin(\frac{\pi}{6}) =\frac{1}{2} , \cos(\frac{\pi}{6}) = \frac{\sqrt[]{3}}{2}$ , then using the identity from above, we have that

$\sin(\theta + \frac{\pi}{6}) = \sin(\theta)\cos(\frac{\pi}{6})+\cos(\alpha)\sin(\frac{\pi}{6}) = \frac{4}{7}\frac{1}{2}-\frac{\sqrt[]{33}}{7}\frac{\sqrt[]{3}}{2} = \frac{-3\sqrt[]{11}+4}{14}$

c) Recall that $\sin(\pi)=0, \cos(\pi)=-1$ . Then,

$\cos(\theta-\pi)=\cos(\theta)\cos(\pi)+\sin(\theta)\sin(\pi) = \frac{-\sqrt[]{33}}{7}\cdot(-1) + 0 = \frac{\sqrt[]{33}}{7}$

d) Recall that $\tan(\frac{\pi}{4}) = 1$ and $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}=\frac{-4}{\sqrt[]{33}}$ . Then

$\tan(\theta+\frac{\pi}{4}) = \frac{\tan(\theta)+\tan(\frac{\pi}{4})}{1-\tan(\theta)\tan(\frac{\pi}{4})} = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}$

5 0

3 years ago

At the beginning of the day, it was -14 degrees. Throughout the day, the temperature increased 16 degrees. What was the temperat

konstantin123 [22]

Answer:

The temperature at the end of the day is 2 degrees

Step-by-step explanation:

Here, we are interested in calculating the temperature at the end of the day.

In the question, we are told that at the beginning of the day, temperature was -14 degrees.

But throughout the day, the temperature increased by 16 degrees;

Thus, the temperature at the end of the day = -14 degrees + 16 degrees = 2 degrees

7 0

3 years ago