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Liono4ka [1.6K]
3 years ago
8

In tan A = 7/9 the measure of

Mathematics
1 answer:
TiliK225 [7]3 years ago
8 0
If the tangent of angle 'A' is 7/9 , then
angle 'A' is either 37.87° or 217.87°.
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E/3=7. please can some one help
RUDIKE [14]

Answer:

e=21

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
If 3:y=18:24 find y.
VashaNatasha [74]

y = 4

Explanation:\begin{gathered} 3\text{ : y = 18 : 24} \\ 3\colon y\text{ = }\frac{3}{y} \\ 18\colon24\text{ = }\frac{18}{24} \end{gathered}\begin{gathered} \frac{3}{y}\text{ = }\frac{18}{24} \\ \text{cross mult}iply\colon \\ 3(24)\text{ = y(18)} \\ \text{divide both sides by 18:} \\ \frac{3(24)}{18}\text{ = }\frac{18y}{18} \end{gathered}\begin{gathered} y\text{ = }\frac{24}{6} \\ y\text{ = 4} \end{gathered}

7 0
1 year ago
Evaluate your expression for m =12
Feliz [49]

There's no way for me to do that, because my expression
is totally blank, and doesn't involve ' m ' in any way.

But if you'll come back and give us <u>your</u> expression, I'll
evaluate it for m=12, and I'll also show you how.


3 0
3 years ago
If XY=18, YZ=14, and XZ=20, find the radius of each circle.
Pachacha [2.7K]
Let`s assume that points M, N and P are the touching points of those 3 circles:Then:Y M + M Z = 14,Z N + N X = 20X P + P Y = 18And also: M Z = ZN, Y M = P Y and N X = X P.Now we have a system of 3 equations ( Y M, M Z and X P  are the radii of each circle ):Y M + M Z = 14M Z + X P = 20X P + Y M = 18 Y M - M Z = - 14+X P  + Y M = 18 X P - M Z  = 4Y M - M Z = - 14+M Z + X P = 20 X P - Y M = 6  /* ( - 1 )X P  - M Z = 4 X P + Y M = - 6  X P - M Z = 4    Y M - M Z = - 2           Y M + M Z = 14            2 Y M = 12  =>  Y M = 6M Z - 6 = 2  =>   M Z = 8X P + 6 = 18 
 X P = 12
Radii of the circles are: 12, 8 and 6. 
8 0
3 years ago
The differential equation in Example 3 of Section 2.1 is a well-known population model. Suppose the DE is changed to dP dt = P(a
LuckyWell [14K]

Answer:

Decreases

Step-by-step explanation:

We need to determine the integral of the DE;

dP/dt=P(aP-b)

dP=P(aP-b)dt

1/(dP^2-bP)dP=dt

We can solve this by integration by parts on the left side. We expand the fraction 1/P²:

1/(d-b/P)\cdot{P^2} dP

let

u=d-b/P

du/dP=b/P^2

dP=\int\limits {P^2/b} \, du

P=lnu/b

Substitute u in:

P=ln(d-b/P)/b

Therefore the equation is:

ln(d-b/P)/b=t

We simplify:

d-b/P=e^b^t

P=b/(d-e^b^t)

As t increases to infinity P will decrease

6 0
3 years ago
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