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3 years ago

How do you do this to this geometry question?

Mathematics

1 answer:

olga55 [171]3 years ago

8 0

B is the correct answer. BDC bisects at angle B so first angle of BDC is 58/2= 29. Second angle is 119 because it is half of 122 = 61 and also the 58 which gets added so 119. 180-119-29=32

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Find the requested information based on the given facts. Base: $54,300: Total Sales: $950,000; Commission: 2.75% Determine the t

Dafna11 [192]

Answer:

$871068.25

Step-by-step explanation:

Given the base of starting the business is $54300 and the total sale after the business is $950000.

Therefore, the profit of the business is $(950000-54300) = $895700.

Now, the commission has to be deducted from the profit at the rate of 2.75% of the profit.

Hence, total profit of the business is given by

895700 [1 - $\frac{2.75}{100}$ ] = $871068.25 (Answer)

6 0

4 years ago

Can someone explain to me on how to do this?

sp2606 [1]

Answer:$1304.89

Step-by-step explanation:

P(1 + r/n)^r*t

500 ( 1 + 0.08/52) ∧ 52*12

500 (1.00154)^624

500 ( 2.61)

1304.88523

7 0

3 years ago

Find the 3rd term in the expansion of (a+b)^8 in simplest form.

Art [367]

Answer:

The fifth term is the middle term of nine, with coefficient given by all the ways of choosing 4 items out of 8 , namely the ways of choosing 4 a 's out of 8 binomial factors.

Step-by-step explanation:

4 0

3 years ago

29 kg of onions cost $145.how much would 12 kg cost?

Romashka-Z-Leto [24]

I am pretty sure the answer is 60 because 145 divided by 29 is 5 you multiply by 12 you get 60

7 0

4 years ago

Intergal of 8sin^4(x)dx

Romashka-Z-Leto [24]

$\int\limits8sin^4(x)dx =$

Take the constant out $\int\limits a*f(x)dx=a* \int\limitsf(x)dx$

$= 8 \int\limits sin^4(x)dx$

$\int\limits sin^4(x)dx = - \frac{cos(x)sin^3(x)}{4} = \frac{3}{4} \int\limits sin^2(x)dx$

$= 8( -\frac{cos(x)sin^3(x)}{4}+ \frac{3}{4} \int\limits sin^2(x)dx)
$

Use the following identify : $sin^2(x) = \frac{1-cos(2x)}{2}$

$= 8(- \frac{cos(x)sin^3(x)}{4} + \frac{3}{4} \frac{1}{2}( \int\limits 1 - cos(2x)dx)$

$\int\limits 1dx = x$

$\int\limits cos(2x)dx = \frac{1}{2} sin(2x)$

Apply the sum rule :

$=8(- \frac{cos(x)sin^3(x)}{4} + \frac{3}{4} \frac{1}{2} (x- \frac{1}{2} sin(2x))$

Simplify :

$8( \frac{3}{8} (x- \frac{1}{2} sin(2x)) - \frac{1}{4} sin^3(x)cos(x))$

Therefore add a constant to the solution:

$= 8 ( \frac{3}{8} (x- \frac{1}{2} sin(2x)) - \frac{1}{4} sin^3(x)cos(x))+C$

hope this helps!

3 0

3 years ago