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2 years ago

Can anyone help me with this?? i dont know how to get letter h.

Mathematics

1 answer:

MAVERICK [17]2 years ago

4 0

This can be solved by using the 30 60 90 triangle. The 30-60 side is equal to 2x, here it is 2x=18. The 60-90 side is equal to x. So this side is 18/2=9. The 30-90 side is by the way equal to the root of x, so here the root of 18

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Nine times the difference of a number and 4 is the same as the product of 2 and the same

nevsk [136]

Answer:

x = 8

Step-by-step explanation:

9(x - 4) = 2(x + 10)

9x - 36 = 2x + 20

7x = 56

x = 8

Please mark as brainliest :)

6 0

3 years ago

What is the solution to the equation?

ki77a [65]

Answer:

Step-by-step explanation:

$3 {(x + 9)}^{ \frac{3}{4} } = 24 \\ \\ {(x + 9)}^{ \frac{3}{4} } = \frac{24}{3} \\ \\ {(x + 9)}^{ \frac{3}{4} } = 8 \\ \\ {(x + 9)}^{ \frac{3}{4} } = {2}^{3} \\ \\ (x + 9) = {2}^{3 \times \frac{4}{3} } \\ \\ x + 9 = {2}^{4} \\ \\ x + 9 = 16 \\ \\ x = 16 - 9 \\ \\ x = 7$

7 0

3 years ago

BC is a secant intersecting the circle at point X: The figure shows a circle with points A and B on it and point C outside it. S

yuradex [85]

Answer:

Is their a picture that goes along with this?

8 0

3 years ago

Read 2 more answers

Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )

DedPeter [7]

(a) First find the intersections of $y=e^{2x-x^2}$ and $y=2$ :

$2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}$

So the area of $R$ is given by

$\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx$

If you're not familiar with the error function $\mathrm{erf}(x)$ , then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line $y=1$ with $y=e^{2x-x^2}$ .

$1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2$

So the area of $S$ is given by

$\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx$
$\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx$

which is approximately 1.546.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve $y=e^{2x-x^2}$ and the line $y=1$ , or $e^{2x-x^2}-1$ . The area of any such circle is $\pi$ times the square of its radius. Since the curve intersects the axis of revolution at $x=0$ and $x=2$ , the volume would be given by

$\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx$

5 0

3 years ago

3a+2(4a-b)<br> please simplify this equation with processing

Alexeev081 [22]

$\bf \colorbox{black} {\bf {\textcolor{blue}{topic= }{ \bf { \textcolor{orange}{mathematics= }{ \bf{ \textcolor{cyan}{answer = }}}}}}}$

3a+2(4a-b)=

3a+8a-2b=

$\bf \textcolor{green}{ \underbrace{11a - 2b}}$

<h2>xLasersx✨</h2>

8 0

3 years ago