50% of 180 dancers said they prefer hip hop class over tap class. How many dancers prefer hip hop class?

Help me with my math ASAP!!!!

almond37 [142]

Answer:

$f$ = 78

$g$ = 3.5

Step-by-step explanation:

First we need to find $f$ .

We can use the equation $72 + (f + 30) = 180$ to solve for $f$ .

We can then change that equation to $72 + f + 30 = 180$ , since the Commutative Property of Addition says that you can have any addition in any order.

Now, we can solve the equation.

$102 + f = 180\\(102 - 102) + f = 180 - 102\\f = 78$

Now that we solved $f$ , we can now solve for $g$ .

Since 25 equals $8g -3$ , we can solve the equation $8g - 3 = 25$ .

Here is how you solve it:

$8g - 3 = 25\\8g + (-3 + 3) = 25 + 3\\8g = 28\\g = \frac{28}{8}$

Since $\frac{28}{8}$ equals 3.5, which is the simplest form, that is the answer.

Hope this helps, and please mark me brainliest! :)

3 0

3 years ago

19 - k = 4. I need help

Marysya12 [62]

Answer:

15

Step-by-step explanation:

first you have to subtract 4 from 19 to get what k is. that should be your answer. good luck!

5 0

2 years ago

-6 + x over 4=-5 what is the value of x

My name is Ann [436]

Answer:x=26

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E

Ipatiy [6.2K]

Answer:

(A) $y=ke^{2t}$ with $k\in\mathbb{R}$ .

(B) $y=ke^{2t}/2-1/2$ with $k\in\mathbb{R}$

(C) $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$

(D) $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ ,

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then $0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}$ . With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form $e^{2t}$ with $t$ real.

(B) Proceeding and the previous item, we obtain $1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2$ . Which is not a vector space with the usual operations (this is because $-1/2$ ), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set $y=e^{mt} \Rightarrow y''=m^2e^{mt}$ and we obtain the characteristic equation $0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$ , and as in the first items the set of solutions form a vector space.

(D) Using C, let be $y=me^{3t}$ we obtain that it must satisfies $3^2m-4m=1\Rightarrow m=1/5$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ , and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).

4 0

3 years ago

Find the value of the question mark

zvonat [6]

The value of the question mark I 8.
Step by step solution:

Values:

Brown frosted donut: 3

Orange frosted donut: 1/3

White frosted donut: 1

Hope this helps!! :)

6 0

2 years ago