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3 years ago

What is the value of the expression –12 – 3 + 9?

Mathematics

2 answers:

klasskru [66]3 years ago

5 0

-12-3+9 answer is B. -6

wariber [46]3 years ago

4 0

B is the correct answer

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Quadrilateral ABCD is inscribed in this circle,

neonofarm [45]

Where’s the picture? i can’t answer it without seeing the measurements

3 0

2 years ago

1. Three apples, a, and two peaches, p, sell for $1.25.

Simora [160]

Answer:

$0.25

Step-by-step explanation:

You can use systems of equations to solve this.

3a+2p = 1.25

5a+p = 1.5

Multiply the second equation by 2.

3a+2p = 1.25

10a+2p = 3

Subtract the bottom from the top

-7a= -1.75

Divide by -7

a= 0.25

Each apple is $0.25 (in this case, the peach also costs $0.25)

5 0

2 years ago

the third term of an arithmetic sequence is 24 and the fifth is 32 if the first term is a1. which is an equation nth term of the

const2013 [10]

An = a1 + (n - 1)(d)
Where a1 is the first term and d is the common difference.
First find d, the common difference.
24, ____, 32
a3 a4 a5
Subtract 32-24 = 8
Subtract a5 - a3 = 2
Divide 8/2 = 4
d = 4
Use d and one of the values they give us to find a1.
a3 = 24
24 = a1 + (3 - 1)(4)
24 = a1 + 2(4)
24 = a1 + 8
Subtract 8 from both sides
16 = a1
an = 16 + (n - 1)(4)
Can also be written
an = 16 + 4n - 4
an = 4n + 12

8 0

2 years ago

Read 2 more answers

Can someone solve this?

Vlada [557]

Answer:

1) B and C

2) 0 < 5x - 10 < 50

10 < 5x < 60

2 < x < 12

3) B

5 0

2 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

2 years ago