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viva [34]
3 years ago
12

Triangle ABC and DEF are right triangles, as shown. Triangle ABC is similar to triangle DEF. which ratios are equal to sin C?

Mathematics
1 answer:
Dafna11 [192]3 years ago
4 0

Answer:

First option and Fourth option.

Step-by-step explanation:

To solve this exercise you need to use the following Trigonometric Identity:

sin\alpha=\frac{opposite}{hypotenuse}

In this case you know that:

\alpha =C

Since triangle ABC is similar to triangle DEF:

\angle C=\angle F

Let's begin with the triangle ABC.

You can identify that:

opposite=AB\\hypotenuse=AC

Then, substituing values, you get:

sinC=\frac{AB}{AC}

In triangle DEF, you know that:

opposite=DE\\hypotenuse=DF

So, substituing values, you get:

sinF=sinC=\frac{DE}{DF}

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because there isnt any full circles. in order to have a square you need 4 corners .they all only have 1 inside that circle.

7 0
3 years ago
A candy shop collected the following information regarding purchases from 100 of its customers.
charle [14.2K]

Answer:

at least 25

Step-by-step explanation:

as not everybody answered the suvey, more people probable purchased only candy bars. so ar least 25.

3 0
3 years ago
Read 2 more answers
Plz help last one ty!!!!!!!
azamat

My apologies on answering late...

Same situation as the previous problem, but this time, all you need to do is state the degree of the angle instead of just providing the angle itself.


ΔABC ≅ ΔDEF

Now, we can see that ∠C ≅ ∠F. Using this information, we can find ∠C on the first triangle ( which is 75° ).

Since ∠C ≅ ∠F,

m∠F is 75°.


Hope I caught your question in time!

Have a good one! If you need anymore help, let me know.

4 0
3 years ago
Consider the function f(x)=2|x|-3. What is f(-3)?
olganol [36]

f(-3)=2|-3|-3=6-3=3\\f(-3)=3

Hope I helped and have a great day!

P.S. brainliest would really help im one away from virtuoso

8 0
3 years ago
Help please!<br><br> | 2x^2 + 5x + 3 | &gt; 0<br> solve the inequality
vovikov84 [41]

Factorize the quadratic.

2x^2 + 5x + 3 = (2x + 3) (x + 1)

We have |ab| = |a||b|, so

|2x^2 + 5x + 3| = |2x + 3| |x + 1| > 0

Now, both |2x+3|\ge0 and |x+1|\ge0 (since the absolute value of any number cannot be negative), so we just need to worry about when the left side is exactly zero. This happens for

(2x + 3) (x + 1) = 0 \implies 2x+3 = 0\text{ or }x+1 = 0 \\\\ \implies x = -\dfrac32 \text{ or } x = -1

So the solution to the inequality is the set

\left\{x \in \Bbb R \mid x\neq-\dfrac32 \text{ and } x\neq-1\right\}

3 0
1 year ago
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