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3 years ago

Y=6x^2, y=-4.5x^2,y=-x^2 Widest to narrowest

Mathematics

1 answer:

Ilya [14]3 years ago

8 0

Widest to narrowest. y=-x^2, y=-4.5x^2, y=6x^2

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Find the total area of a cylinder whose radius is 2 cm and height is 4 cm.

kolezko [41]

Answer:

78.95cm^2

Step-by-step explanation:

The equation for a cylinder is rπ^2 · h

2π²·4≈78.95

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3 years ago

Sales of a new line of athletic footwear are crucial to the success of a newly formed company, Fleet Shoes. Fleet wishes to esti

Murrr4er [49]

Answer:

The number of weeks required = 216 weeks

Step-by-step explanation:

Given that:

Margin of Error E = 200

Confidence interval = 95% = 0.95

Level of SIgnificance = 1 - C.I

= 1 - 0.95

= 0.05

Standard deviation = 1500

The Critical value for Z :

$Z_{\alpha/2} =Z_{0.05/2} \\ \\ = Z_{0.025} = 1.96$

The number of weeks( i.e the sample size (n) ) required is :

$n = (\dfrac{Z_{\alpha/2} \times \sigma}{E})^2$

$n = (\dfrac{1.96 \times 1500}{200})^2$

$n = (14.7)^2$

n = 216.09

n ≅ 216 weeks

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3 years ago

Write 20 billion 1 million and thirty thousand and one as a number

qaws [65]

20 001 030 001

The best way to answer these types of questions is to space your work out.

4 0

3 years ago

Read 2 more answers

The quadrilateral shown is a parallelogram. If AC = 28, what is the measure of AE?

Vera_Pavlovna [14]

Answer:

Step-by-step explanation:

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3 years ago

The focal lengths of the objective lens and the eyepiece of a microscope are 0.50 cm and 2.0 cm, respectively, and their separat

VladimirAG [237]

Answer:

- 103.7

Step-by-step explanation:

Given:

Focal length of the eyepiece, f = 2.0 cm

Focal length of the objective lens, f' = 0.50 cm

Separation for minimum eyestrain = 6,0 cm

Image distance, v = - 25 cm

Now, from the lens formula,

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

here, u is the object distance

on substituting the respective values, we get

$\frac{1}{2}=\frac{1}{u}+\frac{1}{-25}$

u = 1.852 cm

also, the separation is adjusted for minimum eyestrain,

therefore, image distance for the objective lens, v' = 6 - 1.852 = 4.148 cm

Now, for the objective lens

using the lens formula, we get

$\frac{1}{0.5}=\frac{1}{u'}+\frac{1}{4.418}$

Here, u' is the distance between the physical object and objective lens

u' = 0.568 cm

Thus,

Magnification, m = $-\frac{vv'}{ff'}$

m = $-\frac{25\times4.418}{0.5\times2}$

m = - 103.7

4 0

3 years ago