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3 years ago

Corgurent or supplementary

Mathematics

1 answer:

nataly862011 [7]3 years ago

5 0

Answer:

Congruent:

B, E, F, G

The rest are supplementary.

Step-by-step explanation:

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Please help me solve for x

kkurt [141]

$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

The shown pair of angles are Co interior angle pair, therefore the sum of those angles is 180°

$\qquad \sf \dashrightarrow \: 130 \degree + 7x + 1 \degree = 180 \degree$

$\qquad \sf \dashrightarrow \: 131 \degree + 7x = 180 \degree$

$\qquad \sf \dashrightarrow \: 7x= 180 \degree - 131 \degree$

$\qquad \sf \dashrightarrow \: 7x= 49 \degree$

$\qquad \sf \dashrightarrow \: x=49 \degree \div 7$

$\qquad \sf \dashrightarrow \: x=7 \degree$

The required value of x is 7°

3 0

2 years ago

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The number of fans who attended a softball game increased from 1,200 for the first game to 1,500 for the second game. How many f

erma4kov [3.2K]

1875 fans will attend

6 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

NEED HELP ASAP

CaHeK987 [17]

Answer:

The y-intercept is 6

Step-by-step explanation:

The y-intercept of a function is where the graph of an equation touches or crosses the vertical axis. This is usually the value of y when x = 0.

For this particular function, when x = 0 ⇒

f(0) = 0^2 + 4(0) + 6

= 0 + 0 + 6

= 6

Therefore, the y-intercept of this function is 6

8 0

2 years ago

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Pick out the prime numbers from the following numbers 16. 13. 19.27.54.51.2, 73. 87.89

Marrrta [24]

Answer:

13, 19, 73, 89

Step-by-step explanation:

Prime numbers only have two factors, 1 and itself.

6 0

3 years ago

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