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LenaWriter [7]
2 years ago
13

Solve for x. Round to the nearest tenth of a degree, if necessary.

Mathematics
1 answer:
brilliants [131]2 years ago
8 0

Answer:

78

Step-by-step explanation:

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The hypotenuse of the right triangle is 5y inches long. The lengths of the legs are x + 8 and x + 3 inches. If the perimeter of
Zolol [24]

Answer: 35 inches.

Step-by-step explanation:

We know that:

hypotenuse = 5*y in

cathetus 1 = (x + 8) in

cathetus 2 = (x + 3) in

The perimeter of the triangle is 76 inches, then:

5*y + (x + 8) + (x + 3) = 76

5*y + 2*x + 13 = 76

We also know that the length of the hypotenuse minus the length of the shorter leg is 17 in.

The shorter leg is x + 3, then:

5*y - (x + 3) = 17

Then we have the equations:

5*y + 2*x + 11 = 76

5*y - (x + 3) = 17

With only these two we can solve the system, first we need to isolate one of the variables in one of the equations, i will isolate x in the second equation.

x = 5*y  - 3 - 17 = 5*y - 20

x = 5*y - 20

Now we can replace this in the other equation, we get:

5*y + 2*x + 11 = 76

5*y + 2*(5*y - 20) + 11 = 76

15*y - 40 + 13 = 76

15*y - 29 = 76

15*y = 76 + 29 = 105

and remember that the hypotenuse is equal to 5*y, then we want to get:

3*(5*y) = 105

5*y = 105/3 = 35

5*y = 35

Then te length of the hypotenuse is 35 inches.

6 0
3 years ago
Find all the missing angles: a, b, c, d.
morpeh [17]

Answer:

a= 180-60-55=65

we know that the angle of a flat(?) ground would be 180, so we can take away the 60 and 55 to find a.

b=180-90-65=25

the sum of interior angle of a triangle would be 180, now we know what a is, also the right angle is 90, now we can take away the 90 and 65 to find b.  

c=25

c is 25, because two angles on both sides of a X is the same

6 0
3 years ago
Evaluate the integral by changing to polar coordinatesye^x dA, where R is in the first quadrant enclosed by the circle x^2+y^2=2
34kurt
\displaystyle\iint_Rye^x\,\mathrm dA=\int_{\theta=0}^{\theta=\pi/2}\int_{r=0}^{r=5}r^2\sin\theta e^{r\cos\theta}\,\mathrm dr\,\mathrm d\theta

which follows from the usual change of coordinates via

\begin{cases}\mathbf x(r,\theta)=r\cos\theta\\\mathbf y(r,\theta)=r\sin\theta\end{cases}

and Jacobian determinant

|\det J|=\left|\begin{vmatrix}\mathbf x_r&\mathbf x_\theta\\\mathbf y_r&\mathbf y_\theta\end{vmatrix}\right|=|r|

Swap the order, so that the integral is

\displaystyle\int_{r=0}^{r=5}\int_{\theta=0}^{\theta=\pi/2}r^2\sin\theta e^{r\cos\theta}\,\mathrm d\theta\,\mathrm dr

and now let \sigma=r\cos\theta, so that \mathrm d\sigma=-r\sin\theta. Now, you have

\displaystyle\int_{r=0}^{r=5}\int_{\sigma=0}^{\sigma=r}re^\sigma\,\mathrm d\sigma\,\mathrm dr=\int_{r=0}^{r=5}r(e^r-1)\,\mathrm dr=4e^5-\dfrac{23}2
7 0
3 years ago
EASY BRAINLIEST PLEASE HELP!!
antoniya [11.8K]

ANSWER:

We know that angle fromed by same arc are congruent. so

∠ACB ~ ∠ADB

∠CAD ~ ∠CBD

Therefore, ΔCAF ~ ΔDBF by AA similarity. So, By CPCT

arcAC = arcBD.

4 0
2 years ago
Read 2 more answers
I need help with the second one
labwork [276]

the answer you're choosing is correct

8 0
3 years ago
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