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3 years ago

3. Tyler said this net cannot be a net for

Mathematics

1 answer:

Keith_Richards [23]3 years ago

3 0

Agree..............

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WOULD YOU RATHER use a coupon for $20 off 10

tatiyna

Answer:

$20 off 10

Step-by-step explanation

it really depends on how much there is in your order, but if its more than 10 i would use $20 off 10 because its less stuff so the prices would go lower.. if that makes any sense

6 0

3 years ago

Find dy/dx of the function y = √x sec*-1 (√x)

ELEN [110]

Hi there!

$\large\boxed{\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \frac{1}{2|\sqrt{x}|\sqrt{{x} - 1}}}$

$y = \sqrt{x} * sec^{-1}(-\sqrt{x}})$

Use the chain rule and multiplication rules to solve:

g(x) * f(x) = f'(x)g(x) + g'(x)f(x)

g(f(x)) = g'(f(x)) * 'f(x))

Thus:

f(x) = √x

g(x) = sec⁻¹ (√x)

$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \sqrt{x} * \frac{1}{\sqrt{x}\sqrt{\sqrt{x}^{2} - 1}} * \frac{1}{2\sqrt{x}}$

Simplify:

$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \sqrt{x} * \frac{1}{2|x|\sqrt{{x} - 1}}$

$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \frac{1}{2|\sqrt{x}|\sqrt{{x} - 1}}$

5 0

3 years ago

Read 2 more answers

Find the area of the rectangle below.

Sloan [31]

(3x-2)*(4x-7)
12x^2 - 21x - 8x + 14
(12x^2 - 29x + 14) square units
Final answer: C

7 0

3 years ago

How do you find ratio

ser-zykov [4K]

<span>Identify the known ratio and the unknown ratio.Set up the proportion.Cross-multiply and solve.Check answer by plugging result into unknown ratio. Look here for the formula: </span>http://prntscr.com/e0qtq3

6 0

3 years ago

the area of a rectangle is 13 square inches. the length of the rectangle is 3.25 inches. what is the width of the rectangle? wha

IrinaK [193]

since the area of the rectangle is 13 square inches and the length is 3.25 inches subtract 3.25 from 13

$13 - 3.25 = 9.75$

which makes your width 9.75

to find the perimeter you add

9.75 + 3.25 + 9.75 + 3.25= 26.5

The perimeter of the rectangle is 26.5

4 0

3 years ago