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3 years ago

The length of a rectangle is 3 yd longer than its width. if the perimeter of the rectangle is 50 yd , find its length and width.

1 answer:

7 0

Length = w+3
P = 2L +2W
50 = 2(w+3) + 2w
50 = 2w +6 +2w
50 = 4w +6
50-6 =4w-6
44 = 4w
44/4 = 4w/4
w = 11
l = 11+3
l = 14

50 = 2L+2W
50 = 2(14) + 2(11)
50 = 28 +22
50=50

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Evaluate the following and show steps. f(x) = 15x – 12 and g(x) = -15x2 + 14x - 10 find g(f(7))

diamong [38]

When we have a function such as h(x) = 2x, and we want to find the value of h at a given x value, we plug in the given number for x. For example, h(3) = 2*3 = 6.

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4 years ago

Find the area of the surface. The part of the sphere x2 + y2 + z2 = a2 that lies within the cylinder x2 + y2 = ax and above the

sineoko [7]

Answer:

The area of the sphere in the cylinder and which locate above the xy plane is $\mathbf{ a^2 ( \pi -2)}$

Step-by-step explanation:

The surface area of the sphere is:

$\int \int \limits _ D \sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 ) } \ dA$

and the cylinder $x^2 + y^2 =ax$ can be written as:

$r^2 = arcos \theta$

$r = a cos \theta$

where;

D = domain of integration which spans between $\{(r, \theta)| - \dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2}, 0 \leq r \leq acos \theta\}$

and;

the part of the sphere:

$x^2 + y^2 + z^2 = a^2$

making z the subject of the formula, then :

$z = \sqrt{a^2 - (x^2 +y^2)}$

Thus,

$\dfrac{\partial z}{\partial x} = \dfrac{-2x}{2 \sqrt{a^2 - (x^2+y^2)}}$

$\dfrac{\partial z}{\partial x} = \dfrac{-x}{ \sqrt{a^2 - (x^2+y^2)}}$

Similarly;

$\dfrac{\partial z}{\partial y} = \dfrac{-2y}{2 \sqrt{a^2 - (x^2+y^2)}}$

$\dfrac{\partial z}{\partial y} = \dfrac{-y}{ \sqrt{a^2 - (x^2+y^2)}}$

So;

$\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\begin {pmatrix} \dfrac{-x}{\sqrt{a^2 -(x^2+y^2)}} \end {pmatrix}^2 + \begin {pmatrix} \dfrac{-y}{\sqrt{a^2 - (x^2+y^2)}} \end {pmatrix}^2+1}$ $\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\dfrac{x^2+y^2}{a^2 -(x^2+y^2)}+1}$

$\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\dfrac{x^2+y^2+a^2 -(x^2+y^2)}{a^2 -(x^2+y^2)}}$

$\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\dfrac{a^2}{a^2 -(x^2+y^2)}}$

$\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = {\dfrac{a}{\sqrt{a^2 -(x^2+y^2)}}$

From cylindrical coordinates; we have:

$\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = {\dfrac{a}{\sqrt{a^2 -r^2}}$

dA = rdrdθ

By applying the symmetry in the x-axis, the area of the surface will be:

$A = \int \int _D \sqrt{ (\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2+1} \ dA$

$A = \int^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}} \int ^{a cos \theta}_{0} \dfrac{a}{\sqrt{a^2 -r^2 }} \ rdrd \theta$

$A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \sqrt{a^2 -r^2} \end {bmatrix}^{a cos \theta}_0 \ d \theta$

$A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \sqrt{a^2 - a^2cos^2 \theta} + a \sqrt{a^2 -0}} \end {bmatrix} d \theta$ $A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \ sin \theta +a^2 } \end {bmatrix} d \theta$

$A = 2a^2 [ cos \theta + \theta ]^{\dfrac{\pi}{2} }_{0}$

$A = 2a^2 [ cos \dfrac{\pi}{2}+ \dfrac{\pi}{2} - cos (0)- (0)]$

$A = 2a^2 [0 + \dfrac{\pi}{2}-1+0]$

$A = a^2 \pi - 2a^2$

$\mathbf{A = a^2 ( \pi -2)}$

Therefore, the area of the sphere in the cylinder and which locate above the xy plane is $\mathbf{ a^2 ( \pi -2)}$

6 0

3 years ago

Three equivalent ratios are shown on the graph

kvasek [131]

To plot more the points, we move up one and right two.

<h3>How to plot more the points?</h3>

The ratios are given as:

(1, 0.5), (2, 1), and (5, 2.5)

Calculate the slope (m) using:

$m = \frac{y_2 -y_1}{x_2 -x_1}$

So, we have:

$m = \frac{1 - 0.5}{2 - 1}$

Evaluate

$m = \frac{0.5}{1}$

This gives

$m = \frac{1}{2}$

The above means

Vertical : Horizontal = 1 : 2

This means a vertical movement of 1, followed by an horizontal movement of 1

Hence, to plot more the points, we move up one and right two.

Read more about equivalent ratios at:

brainly.com/question/13513438

#SPJ1

Complete question

Three equivalent ratios are shown on the graph. On a coordinate plane, points (1, 0.5), (2, 1), and (5, 2.5) are plotted. Starting on one of the plotted points, how can you plot more equivalent ratio points?

5 0

2 years ago