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2 years ago

Two parallel lines, m and n, are cut by a transversal, t, as shown in the figure below.

Mathematics

1 answer:

Misha Larkins [42]2 years ago

6 0

Answer:

Step-by-step explanation:

∠2 and ∠7 are alternate interior angles

alternate interior angles are congruent meaning ∠2 ≅ ∠7

we have ∠2 = 2x + 7 and ∠7 = 3x - 13

hence, 2x + 7 = 3x - 13

==> add 13 to both sides

2x + 20 = 3x

==> subtract 2x from both sides

20 = x

now to find the measure of ∠7 we plug in the value of x

∠7 = 3x - 13

==> plug in x = 20

∠7 = 3(20) - 13

==> multiply 3 and 20

∠7 = 60 - 13

==> subtract 13 from 60

∠7 = 47

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Rewrite the equation of the line 5y = -1/4x - 6 in standard form

netineya [11]

Answer:

Step-by-step explanation:

The equation of a line in standard form is

Ax + By = C

Given

5y = - $\frac{1}{4}$ x - 6 ( multiply through by 4 to clear the fraction )

20y = - x - 24 ( add x to both sides )

x + 20y = - 24 → D

7 0

3 years ago

jason voorhes was buying hocky masks. 1 hockey mask cost 10 dollars how much does 10 masks it cost in all?

Furkat [3]

The total cost would be $100

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3 years ago

Read 2 more answers

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

2 years ago

Miguel volunteers at his local food pantry and takes note of how much money is donated each day during a 10-day

ryzh [129]

Answer:

Step-by-step explanation:

The answer is c

5 0

3 years ago

The two-way table shows the distribution of gender to favorite film genre for the senior class at Mt. Rose High School.

Studentka2010 [4]

Answer:

The second statement is correct

Step-by-step explanation:

Hello!

The table shows the information of the favorite film genre of the students of the class regarding their gender.

You have to prove which statement is correct:

1)The probability of randomly selecting a student who has a favorite genre of drama and is also female is about 17 percent.

If you chose a student at random, you need to calculate the probability of its favorite genre being "Drama" (D) and the student being female (F), symbolically: P(D∩F)

To do so you have to divide the number of observed students that are female and like drama by the total number of students:

P(D∩F)= $\frac{24}{240}= \frac{1}{10} =0.10$

This means that the probability of choosing a student at random and it being a female that likes drama is 10%.

This statement is incorrect.

2) Event F for female and event D for drama are independent events.

Two events are independent when the occurrence of one of them doesn't affect the probability of occurrence of the other one.

So if F and D are independent then:

P(F)= P(F|D)

-or-

P(D)=P(D|F)

The probability of the event "Female" is equal to $P(F)= \frac{Total females in the class}{n} = \frac{144}{240} = \frac{3}{5}= 0.6$

The probability of the event "Drama" is:

$P(D)= \frac{Total students that like "Drama"}{n}= \frac{40}{240}= \frac{1}{6}= 0.166$

$P(F|D)= \frac{P(FnD)}{P(D)}= \frac{\frac{1}{10} }{\frac{1}{6} }= \frac{3}{5} = 0.6$

As you can see P(F)= 0.6 and P(F|D)= 0.6 so both events are independent.

This statement is correct.

3) The probability of randomly selecting a male student, given that his favorite genre is horror, is 16/40

This is a conditional probability, you already know that the student likes horror movies (H), and out of that group you want to know the probability of the student being male (M):

$P(M|H)= \frac{number of male students that like horror movies}{total students that like horror movies}= \frac{16}{38}= \frac{8}{19} = 0.42$