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3 years ago

Table of values 3x+2y=24

1 answer:

4 0

3
x
+
2
y
>
24
3
x
+
2
y
>
24
Solve for
y
y
.
Tap for more steps...
y
>
−
3
x
2
+
12
y
>
-
3
x
2
+
12
Use the slope-intercept form to find the slope and y-intercept.
Tap for more steps...
Slope:
−
3
2
-
3
2
Y-Intercept:
12
12
Graph a dashed line, then shade the area above the boundary line since
y
y
is greater than
−
3
x
2
+
12
-
3
x
2
+
12
.
y
>
−
3
x
2
+
12
y
>
-
3
x
2
+
12

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Find the hypotenuse of a right triangle if side a=3 and side b=4

Slav-nsk [51]

Answer:

5 units

Step-by-step explanation:

By Pythagoras theorem:

$(hypotenuse)^2 = a^2+ b^2\\\therefore hypotenuse = \sqrt{a^2+ b^2} \\\therefore hypotenuse = \sqrt{3^2+ 4^2} \\\therefore hypotenuse = \sqrt{9+ 16} \\\therefore hypotenuse = \sqrt{25} \\\therefore hypotenuse = 5\: units$

4 0

4 years ago

A rectangular swimming pool is twice as long as it is wide. A small concrete walkway surrounds the pool. The walkway is a 2 feet

Ksivusya [100]

Answer:

The width and the length of the pool are 12 ft and 24 ft respectively.

Step-by-step explanation:

The length (L) of the rectangular swimming pool is twice its wide (W):

$L_{1} = 2W_{1}$

Also, the area of the walkway of 2 feet wide is 448:

$W_{2} = 2 ft$

$A_{T} = W_{2}*L_{2} = 448 ft^{2}$

Where 1 is for the swimming pool (lower rectangle) and 2 is for the walkway more the pool (bigger rectangle).

The total area is related to the pool area and the walkway area as follows:

$A_{T} = A_{1} + A_{w}$ (1)

The area of the pool is given by:

$A_{1} = L_{1}*W_{1}$

$A_{1} = (2W_{1})*W_{1} = 2W_{1}^{2}$ (2)

And the area of the walkway is:

$A_{w} = 2(L_{2}*2 + W_{1}*2) = 4L_{2} + 4W_{1}$ (3)

Where the length of the bigger rectangle is related to the lower rectangle as follows:

$L_{2} = 4 + L_{1} = 4 + 2W_{1}$ (4)

By entering equations (4), (3), and (2) into equation (1) we have:

$A_{T} = A_{1} + A_{w}$

$A_{T} = 2W_{1}^{2} + 4L_{2} + 4W_{1}$

$448 = 2W_{1}^{2} + 4(4 + 2W_{1}) + 4W_{1}$

$224 = W_{1}^{2} + 8 + 4W_{1} + 2W_{1}$

$224 = W_{1}^{2} + 8 + 6W_{1}$

By solving the above quadratic equation we have:

W₁ = 12 ft

Hence, the width of the pool is 12 feet, and the length is:

$L_{1} = 2W_{1} = 2*12 ft = 24 ft$

Therefore, the width and the length of the pool are 12 ft and 24 ft respectively.

I hope it helps you!

8 0

3 years ago

A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 31 ft/s. (a) A

julsineya [31]

Answer:

a) -13.9 ft/s

b) 13.9 ft/s

Step-by-step explanation:

a) The rate of his distance from the second base when he is halfway to first base can be found by differentiating the following Pythagorean theorem equation respect t:

$D^{2} = (90 - x)^{2} + 90^{2}$ (1)

$\frac{d(D^{2})}{dt} = \frac{d(90 - x)^{2} + 90^{2})}{dt}$

$2D\frac{d(D)}{dt} = \frac{d((90 - x)^{2})}{dt}$

$D\frac{d(D)}{dt} = -(90 - x) \frac{dx}{dt}$ (2)

Since:

$D = \sqrt{(90 -x)^{2} + 90^{2}}$

When x = 45 (the batter is halfway to first base), D is:

$D = \sqrt{(90 - 45)^{2} + 90^{2}} = 100. 62$

Now, by introducing D = 100.62, x = 45 and dx/dt = 31 into equation (2) we have:

$100.62 \frac{d(D)}{dt} = -(90 - 45)*31$

$\frac{d(D)}{dt} = -\frac{(90 - 45)*31}{100.62} = -13.9 ft/s$

Hence, the rate of his distance from second base decreasing when he is halfway to first base is -13.9 ft/s.

b) The rate of his distance from third base increasing at the same moment is given by differentiating the folowing Pythagorean theorem equation respect t:

$D^{2} = 90^{2} + x^{2}$