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3 years ago

9

Laila is setting up a conical tent in the backyard. The height of the tallest part of the tent is 7 feet. The inside of the tent

has a volume of 56 cubic feet. How many square feet are covered by the base of the tent?

2 answers:

svp [43]3 years ago

8 0

Answer: 24

Step-by-step explanation:

I got it right on edge

postnew [5]3 years ago

7 0

Answer:

56*7= 392^2

Step-by-step explanation:

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Help me <br>heeeeeeeeeeeeeeeeeeeeelllllllllpppp

borishaifa [10]

Answer:

The diameter would also double

Step-by-step explanation:

3 0

4 years ago

PLEASE answer these math questions

zysi [14]

1. To find the square root, you have to square root them both.

Square root of 324 is 18, so x=-18, 1818

2. Divide 49 by 9. 49/9=5.44 then find the square root of 5.44 which is 2.3, x=-2.3, 2.3.

3. To do this one it is one of either 2 or 4 because it is not a perfect square, it can break down to square root of 16 and square root of 6, which then makes it +-4 square root 6.

4. This one is simple the square root of 34 is 5.8

5. Subtract each by 5 to get 6y^2=215 then divide each by 6. Y^2=34.8 then square root. Which makes y=+-5.9. Hope this helped.

3 0

3 years ago

Use the substitution x = et to transform the given Cauchy-Euler equation to a differential equation with constant coefficients.

Anika [276]

Answer:

$\boxed{\sf \ \ \ ax^2+bx^{-10} \ \ \ }$

Step-by-step explanation:

Hello,

let's follow the advise and proceed with the substitution

first estimate y'(x) and y''(x) in function of y'(t), y''(t) and t

$x(t)=e^t\\\dfrac{dx}{dt}=e^t\\y'(t)=\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}=e^ty'(x)y'(x)=e^{-t}y'(t)\\y''(x)=\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}(e^{-t}\dfrac{dy}{dt})=-e^{-t}\dfrac{dt}{dx}\dfrac{dy}{dt}+e^{-t}\dfrac{d}{dx}(\dfrac{dy}{dt})\\=-e^{-t}e^{-t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}\dfrac{dt}{dx}=-e^{-2t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}e^{-t}\\=e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt})$

Now we can substitute in the equation

$x^2y''(x)+9xy'(x)-20y(x)=0\\ e^{2t}[ \ e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}) \ ] + 9e^t [ \ e^{-t}\dfrac{dy}{dt} \ ] -20y=0\\ \dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}+ 9\dfrac{dy}{dt}-20y=0\\ \dfrac{d^2y}{dt^2}+ 8\dfrac{dy}{dt}-20y=0\\$

so the new equation is

$y''(t)+ 8y'(t)-20y(t)=0$

the auxiliary equation is

$x^2+8x-20=0\\ x^2-2x+10x-20=0\\x(x-2)+10(x-2)=0\\(x+10)(x-2)=0\\ x=-10\text{ or }x=2$

so the solutions of the new equation are

$y(t)=ae^{2t}+be^{-10t}$

with a and b real

as

$x(t)=e^t\\ t(x)=ln(x)$

$y(x)=ae^{2ln(x)}+be^{-10ln(x)}=ax^2+bx^{-10}$

hope this helps

do not hesitate if you have any questions

8 0

3 years ago

What are the x- and y- coordinates of point E, which partitions the directed line segment from A to B into a ratio of 1:2?

Solnce55 [7]

<u><em>The coordinates of the points A and B are not given. We'll assume: A(3,1) B(12,-5)</em></u>

Answer:

<em>The coordinates of E are (6,-1)</em>

Step-by-step explanation:

<u>Partition of a Segment</u>

The length of segment directed from A(xa,ya) to B(xb,yb) can be decomposed in its x,y coordinates:

$x_{AB}=x_b-x_a$

$y_{AB}=y_b-y_a$

If a point E is to be in the segment and partition it into a ratio 1:2, then

\displaystyle \frac{x_{AE}}{{x_{EB}}=\frac{y_{AE}}{{y_{EB}}=\frac{1}{2}

But

$x_{AE}=x_e-x_a$

$y_{AE}=y_e-y_a$

$x_{EB}=x_b-x_e$

$y_{EB}=y_b-y_e$

Then we set the equation:

$\frac{x_{AE}}{{x_{EB}}=\frac{1}{2}$

$2(x_e-x_a)=x_b-x_e$

Operating and rearranging

$3x_e=x_b+2x_a$

Solving

$\displaystyle x_e=\frac{x_b+2x_a}{3}=\frac{12+6}{3}$

$x_e=6$

Similarily

$\displaystyle y_e=\frac{y_b+2y_a}{3}=\frac{-5+2}{3}$

$y_e=-1$

The coordinates of E are (6,-1)

8 0

4 years ago

Read 2 more answers

939515 rounded to nearest ten

tino4ka555 [31]

Answer: 939520

Step-by-step explanation: your ones place has a five in it, therefore you round up your tens place

8 0

3 years ago

Read 2 more answers