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3 years ago

10

Help me this equation show work plzz

1 answer:

vlabodo [156]3 years ago

5 0

X=20,000 is the anwser

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The average of 7 numbers is 60. The average of the first three numbers is 50, while the average of the last three is 70. What mu

Svetradugi [14.3K]

From the last two facts, we get

$\dfrac{x_1+x_2+x_3}3=50\implies\dfrac{x_1+x_2+x_3}7=\dfrac{150}7$

$\dfrac{x_5+x_6+x_7}3=70\implies\dfrac{x_5+x_6+x_7}7=30$

Then

$\dfrac{x_1+x_2+\cdots+x_6+x_7}7=60\implies\dfrac{x_4}7=60-\dfrac{150}7-30\implies\boxed{x_4=60}$

4 0

4 years ago

Can someone help me with 36 and 38?

lukranit [14]

Answer: the height for 38. is 7

i don't know 36 so sorry.

Step-by-step explanation:

3 0

4 years ago

shawn is typing a paper for class. he can type 1 1/7 pages in 1/3 of an hour. how many pages can shawn type in one hour?

SCORPION-xisa [38]

He could type 3 3/7 in one hour.I think this because since it is 1/3 you would have to multiply it by 3.So you multiply 1/3 by 3,It would give one hour.When you multiply 1 1/7 by 3 it would give you 3 3/7.

4 0

3 years ago

A steady wind blows a kite due west. The kite's height above ground from horizontal position x = 0 to x = 80 ft is given by y =

GaryK [48]

Answer:

The distance travelled by the kite is 122.8 ft ( approx )

Step-by-step explanation:

Here, the given function,

$y=150-\frac{1}{40}(x-50)^2$

Differentiating with respect to x,

$y'=-\frac{1}{20}(x-50)$

∵ arc length of a curve is,

$L=\int_{a}^{b} \sqrt{1+y'^2}dx$

Where, y shows the height of the curve for a ≤ x ≤ b,

Thus, the arc length of the given curve is,

$L=\int_{0}^{80} \sqrt{1+(-\frac{1}{20}(x-50)^2}dx$

Put $-\frac{1}{20}(x-50)=tan\theta$

$\implies -dx=-20 sec^2\theta d\theta$

$\implies L=-20\int_{0}^{80} \sqrt{1+tan^2\theta}sec^2\theta d\theta$

$=-20\int_{0}^{80} (sec \theta ) sec^2\theta d\theta$

$=-20\int_{0}^{80} (sec \theta ) sec^2\theta d\theta$

By integration by parts,

$=|-\frac{20}{2}(sec \theta tan\theta +ln|sec\theta +tan\theta |) |^{x=80}_{x=0}$

If x = 80, $tan \theta = -\frac{1}{20}(30-50)=\frac{3}{2}$

$sec \theta = \frac{\sqrt{13}}{2}$

$\implies \theta = \frac{1}{20}(0-50)=\frac{5}{2}$

$sec \theta = \frac{\sqrt{29}}{2}$

Thus, the length of the curve is,

$=-10(\frac{\sqrt{13}}{2}(-\frac{3}{2}) +ln|\frac{13}{2}-\frac{3}{2}|) + 10(\frac{5\sqrt{29}}{4} + ln |\frac{29}{2} + \frac{5}{2} |)$

$\approx 122.8\text{ feet}$

8 0

3 years ago

Help!!! Please help me I need help on my homework!

Serjik [45]

Answer:

16

Step-by-step explanation:

the answer is 16 4 times 4.

6 0

3 years ago