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3 years ago

WHO NEED POINTS< COME AND GET POINTS!!!!!

Mathematics

2 answers:

KiRa [710]3 years ago

6 0

Answer:

thank you by a ton

Step-by-step explanation:

Vaselesa [24]3 years ago

3 0

Answer:

Wow Tysm!

Step-by-step explanation:

TYSM

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For the figures that have rotational symmetry, list the angles of rotation less than 360 degrees. For figures without rotational

elixir [45]

Answer:
This shape has no rotational symmetry.

Explanation:
Shapes with rotational symmetry have the property of looking the same after being rotated a degree smaller than 360.

Although this shape in the image does have symmetry, it does not have rotational symmetry because there is no way for it to look the exact same after being rotated.

I hope this helps!

6 0

3 years ago

An airplane flew with the wind for 2.5 hours and returned the same distance against the wind in 3.5 hours. If the cruising speed

KengaRu [80]

Answer:

60 miles per hour.

Step-by-step explanation:

Let r represent speed of wind blowing in miles per hour.

We have been given that the cruising speed of the plane was a constant 360 mph in air. The speed of the plane is the direction of wind would be $360+r$ .

The speed of the plane is the opposite direction of wind would be $360-r$ .

$\text{Distance}=\text{Speed}\times \text{Time}$

Distance covered in the direction of wind would be $2.5(360+r)$ .

Distance covered in the opposite direction of wind would be $3.5(360-r)$ .

Since both distances are same, so we will get:

$2.5(360+r)=3.5(360-r)$

$900+2.5r=1260-3.5r$

$900-900+2.5r+3.5r=1260-900-3.5r+3.5r$

$6r=360$

$\frac{6r}{6}=\frac{360}{6}$

$r=60$

Therefore, the wind is blowing at a rate of 60 miles per hour.

6 0

3 years ago

My mother is 12 years more than twice my age.after 8 years she will be 20 years less than three times my age...find my age and m

Mnenie [13.5K]

You are correct for the first part, you would be 16 and your Mom would be 44. After 8 years, you would be 24 and your mother would be 52. You solve the second part by adding 8 to 16 which equals 24. Then multiply 24 times 3 which equals 72. Lastly, subtract 20 from 72 to find your mother's new age. Which would be 52.

Hope this helps !

4 0

3 years ago

What fraction of 6 weeks is 6 days

adelina 88 [10]

Answer:

$\frac{1}{7}$

Step-by-step explanation:

6 weeks is 42 days.

6 days/42 days = 1/7

6 0

2 years ago

Find the exact area of the surface obtained by rotating the curve about the x-axis. y = 1 + ex , 0 ≤ x ≤ 9

tekilochka [14]

The surface area is given by

$\displaystyle2\pi\int_0^9(1+e^x)\sqrt{1+e^{2x}}\,\mathrm dx$

since $y=1+e^x\implies y'=e^x$ . To compute the integral, first let

$u=e^x\implies x=\ln u$

so that $\mathrm dx=\frac{\mathrm du}u$ , and the integral becomes

$\displaystyle2\pi\int_1^{e^9}\frac{(1+u)\sqrt{1+u^2}}u\,\mathrm du$

$=\displaystyle2\pi\int_1^{e^9}\left(\frac{\sqrt{1+u^2}}u+\sqrt{1+u^2}\right)\,\mathrm du$

Next, let

$u=\tan t\implies t=\tan^{-1}u$

so that $\mathrm du=\sec^2t\,\mathrm dt$ . Then

$1+u^2=1+\tan^2t=\sec^2t\implies\sqrt{1+u^2}=\sec t$

so the integral becomes

$\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\left(\frac{\sec t}{\tan t}+\sec t\right)\sec^2t\,\mathrm dt$

$=\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\left(\frac{\sec^3t}{\tan t}+\sec^3 t\right)\,\mathrm dt$

Rewrite the integrand with

$\dfrac{\sec^3t}{\tan t}=\dfrac{\sec t\tan t\sec^2t}{\sec^2t-1}$

so that integrating the first term boils down to

$\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\frac{\sec t\tan t\sec^2t}{\sec^2t-1}\,\mathrm dt=2\pi\int_{\sqrt2}^{\sqrt{1+e^{18}}}\frac{s^2}{s^2-1}\,\mathrm ds$

where we substitute $s=\sec t\implies\mathrm ds=\sec t\tan t\,\mathrm dt$ . Since

$\dfrac{s^2}{s^2-1}=1+\dfrac12\left(\dfrac1{s-1}-\dfrac1{s+1}\right)$

the first term in this integral contributes

$\displaystyle2\pi\int_{\sqrt2}^{\sqrt{1+e^{18}}}\left(1+\frac12\left(\frac1{s-1}-\frac1{s+1}\right)\right)\,\mathrm ds=2\pi\left(s+\frac12\ln\left|\frac{s-1}{s+1}\right|\right)\bigg|_{\sqrt2}^{\sqrt{1+e^{18}}}$

$=2\pi\sqrt{1+e^{18}}+\pi\ln\dfrac{\sqrt{1+e^{18}}-1}{1+\sqrt{1+e^{18}}}$

The second term of the integral contributes

$\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\sec^3t\,\mathrm dt$

The antiderivative of $\sec^3t$ is well-known (enough that I won't derive it here myself):

$\displaystyle\int\sec^3t\,\mathrm dt=\frac12\sec t\tan t+\frac12\ln|\sec t+\tan t|+C$

so this latter integral's contribution is

$\pi\left(\sec t\tan t+\ln|\sec t+\tan t|\right)\bigg|_{\pi/4}^{\tan^{-1}(e^9)}=\pi\left(e^9\sqrt{1+e^{18}}+\ln(e^9+\sqrt{1+e^{18}})-\sqrt2-\ln(1+\sqrt2)\right)$

Then the surface area is

$2\pi\sqrt{1+e^{18}}+\pi\ln\dfrac{\sqrt{1+e^{18}}-1}{1+\sqrt{1+e^{18}}}+\pi\left(e^9\sqrt{1+e^{18}}+\ln(e^9+\sqrt{1+e^{18}})-\sqrt2-\ln(1+\sqrt2)\right)$

$=\boxed{\left((2+e^9)\sqrt{1+e^{18}}-\sqrt2+\ln\dfrac{(e^9+\sqrt{1+e^{18}})(\sqrt{1+e^{18}}-1)}{(1+\sqrt2)(1+\sqrt{1+e^{18}})}\right)\pi}$

4 0

4 years ago