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iVinArrow [24]
3 years ago
11

Could any of y'all help me?

Mathematics
1 answer:
Scorpion4ik [409]3 years ago
5 0
Start using calculator
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Please help me!!!!! How do you get 67.3 and 112.7 from .9228
antoniya [11.8K]

Answer and Step-by-step explanation:

You are correct in that we need to use the Law of Sines: \frac{c}{sinC} =\frac{b}{sinB}=\frac{a}{sinA}.

Here, when we use the Law of Sines, we have: \frac{28}{sin(63)}=\frac{29}{sinB}.

Cross multiply:

(sinB) * 28 = (sin63) * 29

28sinB ≈ 25.839

sinB ≈ 0.9228

Now, in order to solve for B, we need to use inverse sin (sin^{-1}):

sin^{-1}(sinB)=sin^{-1}(0.9228)

The sines on the left cancel out, and we're left with:

B ≈ 67.3 degrees

Now, one thing to keep in mind when doing Law of Sines is that there is potentially more than one answer possible for the degree measure. The other degree measure can be found by subtracting this one from 180:

180 - 67.3 = 112.7 degrees.

Hope this helps!

7 0
3 years ago
Read 2 more answers
Its -20 but how? pls helppp. ill give brainliest
MissTica

w-10=2(w+5)

w-10=2w+10

w-2w=10+10

-w=20

w=-20

4 0
2 years ago
Read 2 more answers
What is the distance between the following points?
likoan [24]

Answer:

11.

Step-by-step explanation:

9 right, and 2 up

7 0
2 years ago
X = mb + y <br><br> Is this written out in slope intercept form?
Anna [14]

Answer:

no

Step-by-step explanation:

y=mx+b

hope this helps

8 0
3 years ago
Read 2 more answers
100 POINTS AND BRAINLIEST
zysi [14]

Answer:

8 square units and \frac{40}{3} square units

Step-by-step explanation:

The area of the triangle ABC is 24 square units.

1. Triangles ABC and FBG are similar with scale factor \frac{1}{3}, then

\dfrac{A_{\triangle FBG}}{A_{\triangle ABC}}=\dfrac{1}{9}\Rightarrow A_{\triangle FBG}=\dfrac{1}{9}\cdot 24=\dfrac{8}{3}\ un^2.

2. Triangles ABC and DBE are similar with scale factor \frac{2}{3}, then

\dfrac{A_{\triangle DBE}}{A_{\triangle ABC}}=\dfrac{4}{9}\Rightarrow A_{\triangle DBE}=\dfrac{4}{9}\cdot 24=\dfrac{32}{3}\ un^2.

3. Thus, the area of the quadrilateral DFGE is

A_{DFGE}=A_{\triangle DBE}-A_{\triangle FBG}=\dfrac{32}{3}-\dfrac{8}{3}=8\ un^2.

and the area of the quadrilateral ADEC is

A_{ADEC}=A_{\triangle ABC}-A_{\triangle DBE}=24-\dfrac{32}{3}=\dfrac{40}{3}\ un^2.

4 0
2 years ago
Read 2 more answers
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