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3 years ago

Draw a rectangle with an area of 42 square units

1 answer:

6 0

We know, Area = Length * Width

So, with area 42, dimensions could be: 1 * 42
2 * 21
3 * 14
6 * 7

Just draw the lines with that measures, two pair of two equal opposite lines.

Hope this helps!

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marcys breakfast table has a square table top with an area of 36 square feet. what is the diagonal length of the table top

MAXImum [283]

Answer:

√36 = 6

a^2 + b^2 = c^2

6^2 + 6^2 = c^2

36 + 36 = c^2

72 = c^2

√72 = c

2 36

2 18

2 9

3 3

6√2 = c

6√2 = (estimate rounded up, 8.49)

7 0

3 years ago

Answer to 5 and 6 please?

yKpoI14uk [10]

Answer:

5. ST 6. AT

Step-by-step explanation:

For a product to be divisible by a number, one of the factors must be that number. This can be achieved by one of the consecutive numbers being that number, or a multiple of that number.

5. 4 x 5 x 6 = 120 which is divisible by 5

1 x 2 x 3 = 6 which is not divisible by 5

6. 1 x 2 x 3 = 6 which is divisible by 6

2 x 3 x 4 = 24 which is divisible by 6

3 x 4 x 5 = 60 which is divisible by 6

4 x 5 x 6 = 120 which is divisible by 6

And now we will just repeat ourselves.

7 0

3 years ago

Please help ASAP will give brainilest to best answer

kykrilka [37]

Answer: 100

Step-by-step explanation:

1 cubic inch would just be 100 because each cube is 1 cubic inch. So 100(number of cubes)/1(cubic inch)=100 I hope I'm right :)

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7

salantis [7]

Answer:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}$

When we directly plug in x = 0, we see that we would have an indeterminate form:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}$

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}$

Plugging in x = 0 again, we would get:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}$

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}$

Substitute in x = 0 once more:

$\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}$

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

6 0

3 years ago

How do you write a function as an equation?

Ket [755]

No x=5 since 5 is the input. The input is always equal to x and the output is always equal to y. So x=5 and y=x+5

8 0

4 years ago