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2 years ago

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Graph a parabola whose xxx-intercepts are at x=-3x=−3x, equals, minus, 3 and x=5x=5x, equals, 5 and whose minimum value is y=-4y

=−4y, equals, minus, 4.

1 answer:

DENIUS [597]2 years ago

7 0

Answer:

Vertex: (1, -4)

intercept: (-3, 0)

Step-by-step explanation:

lets be real, you don't care about an explanation you just want the answer.

BUT... the vertex is the Y line and the Intercept is the X line

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Y = 5.5x is a linear equation (i.e. the highest power of the variables is 1). Therefore, the line will be a straight line with a slope of 5.5

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Mr. Peterson needs topsoil for his garden. His rectangular garden is 78 in long and 48 in wide. A bag of topsoil covers an area

nata0808 [166]

Answer:

3744 inches squared, and 8 bags

Step-by-step explanation:

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˜”*°•.˜”*°• Area: •°*”˜.•°*”˜

Well, the formula for area is L*W

L=78 inches

W=48 inches

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2 years ago

Water is poured into a conical paper cup at the rate of 3/2 in3/sec (similar to Example 4 in Section 3.7). If the cup is 6 inche

aliya0001 [1]

Answer:

The water level rising when the water is 4 inches deep is $\frac{3}{8\times \pi} inch/s$ .

Step-by-step explanation:

Rate of water pouring out in the cone = R= $\frac{3}{2} inch^3/s$

Height of the cup = h = 6 inches

Radius of the cup = r = 3 inches

$\frac{r}{h}=\frac{3 inch}{6 inch}=\frac{1}{2}$

r = h/2

Volume of the cone = $V=\frac{1}{3}\pi r^2h$

$V=\frac{1}{3}\pi r^2h$

$\frac{dV}{dt}=\frac{d(\frac{1}{3}\pi r^2h)}{dt}$

$\frac{dV}{dt}=\frac{d(\frac{1}{3}\pi (\frac{h}{2})^2h)}{dt}$

$\frac{dV}{dt}=\frac{1}{3\times 4}\pi \times \frac{d(h^3)}{dt}$

$\frac{dV}{dt}=\frac{1\pi }{12}\times 3h^2\times \frac{dh}{dt}$

$\frac{3}{2} inch^3/s=\frac{1\pi }{12}\times 3h^2\times \frac{dh}{dt}$

h = 4 inches

$\frac{3}{2} inch^3/s=\frac{1\pi }{12}\times 3\times (4inches )^2\times \frac{dh}{dt}$

$\frac{3}{2} inch^3/s=\pi\times 4\times \frac{dh}{dt} inches^2$

$\frac{dh}{dt}=\frac{3}{8\times \pi} inch/s$

The water level rising when the water is 4 inches deep is $\frac{3}{8\times \pi} inch/s$ .

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Irina18 [472]

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Answer:

2:1 that is 2 inches in real life is equal to 1 inch in the drawing.

Step-by-step explanation:

We are given the following in the question:

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We find the ratio of the real length of pool and the measurement of the length of the pool in the drawing.

$\text{Scale} = \dfrac{10\text{ inches}}{5\text{ inches}} = 2:1$

Thus, we can say Paco's new drawing follows a scale of 2:1 that is 2 inches in real life is equal to 1 inch in the drawing.

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