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Orlov [11]
3 years ago
9

Find length and width .the lenght is 10 times as long as the width.P=308

Mathematics
1 answer:
patriot [66]3 years ago
8 0

<span>Width(height): x
length of rectangle: 10x 
perimeter: width +width +length +length = 308 inches
or ------- 2width+2length=308
2x+2(10x)=308
2x+20x=308
22x=308</span>

<span>
x=14 the width is 14 inches 
length:
10x
10*14
140 inches 
ANSWER: 
Length:140 inches
Width: 14 inches</span>

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The area of the circle is 64pie cm. What is the radius in cm?
Vinil7 [7]

Answer: 8 centimeters

Step-by-step explanation: To find the radius of the circle, remember that the formula for the area of a circle is πr² and since we're given that the area of our circle is 64π, we can set up the equation 64π = πr².

To solve for <em>r</em>, we first divide both sides of the equation by π.

On the left side, the π's cancel and we're left with 64 and on the right side, the π's cancel and we're left with r².

So we have 64 = r².

Next, we take the square root of both sides to get 8 = r.

So the radius of our circle is 8 centimeters.

8 0
3 years ago
Read 2 more answers
Find the value of each variable. write your answers in simplest radical form
erica [24]

Answer:

y=3.364

x=4

Step-by-step explanation:

hope this helps

7 0
2 years ago
2. The path of a high diver is given by y = + (2x2 – 9x – 56), where y is the height of the diver
timofeeve [1]

Answer:

The diver will be 8 feet from the end of the board when he hits the water.

Step-by-step explanation:

The diver hits the water when y = 0.

To find the distance, we have to find the values of x when y = 0.

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = (x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this problem, we have that:

y = 2x^{2} - 9x - 56

2x^{2} - 9x - 56 = 0

So

a = 2, b = -9, c = -56

Then

\bigtriangleup = b^{2} - 4ac = (-9)^{2} - 4*2(-56) = 529

x_{1} = \frac{-(-9) + \sqrt{529}}{2*2} = 8

x_{2} = \frac{-(-9) - \sqrt{529}}{2*2} = -3.5

It is a horizontal distance, so the answer is a positive value.

The diver will be 8 feet from the end of the board when he hits the water.

7 0
3 years ago
X2 = 121 Please HELP!!!
Allushta [10]
The answer will be
x=root 121
x=11
3 0
2 years ago
Read 2 more answers
Consider the following region R and the vector field F. a. Compute the​ two-dimensional curl of the vector field. b. Evaluate bo
Shalnov [3]

Looks like we're given

\vec F(x,y)=\langle-x,-y\rangle

which in three dimensions could be expressed as

\vec F(x,y)=\langle-x,-y,0\rangle

and this has curl

\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle

which confirms the two-dimensional curl is 0.

It also looks like the region R is the disk x^2+y^2\le5. Green's theorem says the integral of \vec F along the boundary of R is equal to the integral of the two-dimensional curl of \vec F over the interior of R:

\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of R by

\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle

\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt

with 0\le t\le2\pi. Then

\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt

=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0

7 0
3 years ago
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