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3 years ago

Jennifer is allowed to watch no more than

Mathematics

1 answer:

Vedmedyk [2.9K]3 years ago

5 0

Answer:

Step-by-step explanation:

10-4=6

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Find the value of X in each triangle (pls I need help this is a test grade)

jok3333 [9.3K]

Answer:

x=125°

Step-by-step explanation:

100° + 25° =125°

180°- 125° = 55°

and since adjacent angles on a line equal 180°

180° - 55° = 125°

x= 125°

8 0

3 years ago

A Blimp is 400 feet in the air. There is an angle of depression of 40 degrees to a football stadium. How far is the Blimp from t

VMariaS [17]

Answer:

Step-by-step explanation:

The diagram of the stadium and the blimp is shown in the attached photo. A right angle triangle, ABC is formed. Angle A is alternate to the angle of depression. This mean that they are equal. So angle A =40 degrees.

The distance of the the Blimp from the stadium is x feet which is the hypotenuse of the triangle.

Applying trigonometric ratio,

Sin 40 = 400/x

xsin40 = 400

x = 400/sin40 = 400 / 0.6428

x = 622.3 feet

5 0

3 years ago

Suppose the radius of a circle is 2. What is its circumference?

Scilla [17]

Answer:

12.57

Step-by-step explanation:

The formula to solve the circumference of a circle is:

2 x PI x R (radius)

=> 2 x PI x 2

=> 4 PI or 12.57

4 0

3 years ago

If f(x)=x2+10sinx, show that there is a number c such that f(c)=1000.

gayaneshka [121]

F(x) is continuous for all x.

Pick a point and show that f(x) is either negative or positive. Pick another point and show that f(x) is negative, if positive, or positive, if negative.

At x = 30, f(30) - 1000 = 900 + 10sin(30) - 1000 ≤ 0
Now, show at another point f(x) - 1000 is positive, and hence, there would be root between 30 and such point.

Let's pick 40.
At x = 40, f(40) - 1000 = 1600 + 10sin(40) - 1000 ≥ 0

Since f(x) - 1000 is continuous, there lies a root between 30 and 40, and hence, 30 ≤ c ≤ 40

7 0

3 years ago

Find thd <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{dy}{dx}" a

NARA [144]

$x^3y^2+\sin(x\ln y)+e^{xy}=0$

Differentiate both sides, treating $y$ as a function of $x$ . Let's take it one term at a time.

Power, product and chain rules:

$\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}$

$=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}$

$=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}$

$=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)$

$=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}$

$=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)$

$=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}$

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

$3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0$

Isolate the derivative, and solve for it:

$\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)$

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}$

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by $y$ to get rid of that fraction in the denominator.

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}$

3 0

3 years ago