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3 years ago

What’s the measure of angle x?

Mathematics

2 answers:

Sonja [21]3 years ago

3 0

M<E = 39
m<JIH = 105
m<JIE + m<JIH = 180
m<JIE = 180 - 105 = 75

m<E + m<JIE + m<IJE = 180
39 + 75 + m<IJE = 180
m<IJE = 66

x = m<IJE (vertical angles)

x = 66

uranmaximum [27]3 years ago

3 0

I attached pictures of my work. Please let me know if there is something that you do not understand.

Answer:
x= 66°

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Is the distance jumped a constant or variable

Advocard [28]

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8 0

3 years ago

For the function y=-2sin(pi/12(x-2) what is the maximum value

pogonyaev

Answer:

Step-by-step explanation:

The max value of the sine function is 1.

If we multiply it by 2, we get 2 sin x, whose max. value is 2.

If we multiply the sine function by -2, the max value of this functin is still 2.

3 0

2 years ago

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

4(2x+8)=10x+2-2x+30 what is x?

Anastaziya [24]

Let's solve your equation step-by-step.

4(2x+8)=10x+2−2x+30

Step 1: Simplify both sides of the equation.

4(2x+8)=10x+2−2x+30

(4)(2x)+(4)(8)=10x+2+−2x+30(Distribute)

8x+32=10x+2+−2x+30

8x+32=(10x+−2x)+(2+30)(Combine Like Terms)

8x+32=8x+32

Step 2: Subtract 8x from both sides.

8x+32−8x=8x+32−8x

32=32

Step 3: Subtract 32 from both sides.

32−32=32−32

0=0

Answer:

All real numbers are solutions.

5 0

2 years ago

plz help whats the answer to this question !!! calculate the mass of a gold bar that is 18.00cm long, 9.21cm wide, and 4.45cm hi

erik [133]

We have to find the mass of the gold bar.

We have gold bar in the shape of a rectangular prism.

The length, width, and the height of the gold bar is 18.00 centimeters, 9.21 centimeters, and 4.45 centimeters respectively.

First of all we will find the volume of the gold bar which is given by the volume of rectangular prism:

Volume of the gold bar $= length \times width\times height$

Plugging the values in the equation we get,

Volume of the gold bar $=18.00 \times 9.21 \times 4.45= 737.721cm^3$

Now that we have the volume we can find the mass by using the formula,

$Mass= density \times volume$

The density of the gold is 19.32 grams per cubic centimeter. Plugging in the values of density and volume we get:

$Mass = 19.32\times 737.721=14252.769$ grams

So, the mass of the gold bar is 14252.769 grams

7 0

3 years ago