All categories

2 years ago

Question 2 on plato geometry 2 70 points

Mathematics

2 answers:

Verizon [17]2 years ago

7 0

Answer:

m∠CFB m∠CAB m∠EAD m ∠CAB + m∠EAD

45° 19.93° 70.07° 90°

50° 29.91° 70.09° 100°

60° 49.92° 70.08° 120°

75° 79.92° 70.08° 150°

m∠CFB BF CF DF EF BF + DF CF + EF BF × DF CF × EF

45° 2.69 2.74 9.09 8.91 11.78 11.65 24.45 24.41

50° 3.7 3.63 8.07 8.23 11.77 11.86 29.86 29.87

60° 5.35 4.72 6.42 7.28 11.77 12.00 34.35 34.36

75° 7.3 5 4.47 6.52 11.77 11.52 32.63 32.6

Step-by-step explanation:

edmentum

Lemur [1.5K]2 years ago

3 0

Answer:

PLATO

Step-by-step explanation:

You might be interested in

At south side middle school 20% of the students play an instrument . There are 305 students . How many students play an instrume

bulgar [2K]

That's 20% of 305.
You could set up the problem as a ratio.
20 x
--- = ------
100 305

100x = 6100
x = 61

3 0

3 years ago

Read 2 more answers

A friend asks to borrow $250 for 100 days anD aGREES tO pAY 36.5%

aksik [14]

95% percent that friend is not going to pay you back

7 0

3 years ago

Increase £110 by 10%

UNO [17]

Answer:

121

Step-by-step explanation:

Find 10% of 110: $\frac{10}{100}$ × $\frac{110}{1} = \frac{11}{1} = 11$
Increase means add: 110 + 11 = 121

8 0

3 years ago

Read 2 more answers

Find the EXACT value of sin(A−B) if cos A = 3/5 where A is in Quadrant IV and cos B = 12/13 where B is in Quadrant IV. Assume al

MissTica

$\bf \textit{Sum and Difference Identities} \\\\ sin(\alpha - \beta)=sin(\alpha)cos(\beta)- cos(\alpha)sin(\beta)$

well, for both angles A and B we're on the IV Quadrant, meaning, the sine is negative, the cosine is positive, likewise, the opposite side is negative and the adjacent side for the angle is positive.

$\bf cos(A)=\cfrac{\stackrel{adjacent}{3}}{\underset{hypotenuse}{5}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{5^2-3^2}}\implies b = \pm 4 \\\\\\ \stackrel{IV~Quadrant}{b = -4}\qquad \qquad sin(A)=\cfrac{\stackrel{opposite}{-4}}{\underset{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill\\\\ cos(B)=\cfrac{\stackrel{adjacent}{12}}{\underset{hypotenuse}{13}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{13^2-12^2}}\implies b = \pm 5$

$\bf \stackrel{IV~Quadrant}{b = -5}\qquad \qquad sin(B)=\cfrac{\stackrel{opposite}{-5}}{\underset{hypotenuse}{13}} \\\\[-0.35em] ~\dotfill\\\\ sin(A-B)=\cfrac{-4}{5}\cdot \cfrac{12}{13}-\left( \cfrac{3}{5}\cdot \cfrac{-5}{13} \right)\implies sin(A-B)=\cfrac{-48}{65} - \left( \cfrac{-15}{65} \right) \\\\\\ sin(A-B)=\cfrac{-48}{65} + \cfrac{15}{65}\implies sin(A-B)=\cfrac{-33}{65}$

4 0

2 years ago

HURRY WILL GIVE BRAINLIEST

chubhunter [2.5K]

If you separate the triangles, you'll notice that both are literally the same triangle because all angles are the same.

8 0

2 years ago

Read 2 more answers