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3 years ago

X/7=21 what is the answer

Mathematics

2 answers:

Tems11 [23]3 years ago

5 0

here is the answer x = -147

Hope its the answer

igomit [66]3 years ago

4 0

You have to multiply -21 and 7, so you get x=-147

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Find an equation for y (in terms of β and α). Then find tan y.

dmitriy555 [2]

This can be expanded through the tangent angle addition formula:

tan(α+β)=tanα+tanβ1−tan -α + tanβ

Thus,tan(x+y)=tanx+tany1−tan x tany

Hope it helps you

7 0

2 years ago

Read 2 more answers

4(PX+1)=64 what is the value of x in terms of p and what is the value of x when p is -5 ?

Oksana_A [137]

Answer:

x = -3

Step-by-step explanation:

$4(PX+1)=64\\\\p= -5$

Substitute -5 for p in the given equation and solve

$4(-5(x)+1)=64\\\\\mathrm{Divide\:both\:sides\:by\:}4\\\frac{4\left(-5x+1\right)}{4}=\frac{64}{4}\\\\Simplify\\-5x+1=16\\\\\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}\\-5x+1-1=16-1\\\\Simplify\\-5x =15\\\\\mathrm{Divide\:both\:sides\:by\:}-5\\\frac{-5x}{-5}=\frac{15}{-5}\\\\Simplify\\x = -3$

4 0

2 years ago

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

1 year ago

In the country of United States of Heightlandia, the height measurements of ten-year-old children are approximately normally dis

aliya0001 [1]

(A)

P(X < 61.25) = P((X - 55.4)/4.1 < (61.25 - 55.4)/4.1)

… ≈ P(Z ≤ 0.1427)

… ≈ 0.5567

(B)

P(X > 46.5) = P((X - 55.4)/4.1 > (46.5 - 55.4)/4.1)

… ≈ P(Z > -2.1707)

… ≈ 1 - P(Z ≤ -2.1707)

… ≈ 0.9850

5 0

3 years ago

BRAINLIEST GOING TO THE FIRST ANSWER

Lesechka [4]

Answer:

w + 2.3 > 18

Step-by-step explanation:

MARK ME BRAINLIEST!!!!!!!!!

5 0

3 years ago

Read 2 more answers