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2 years ago

3 times a number equals 45

Mathematics

2 answers:

marta [7]2 years ago

7 0

Answer:

MY ANSWER IS 15

Step-by-step explanation:

BECAUSE 3×15=45

Marysya12 [62]2 years ago

5 0

Answer:

Step-by-step explanation:

if you divide 45 by 3 you get 15 and then if you multiply 15 by 3 you get 45

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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

Is the equation true, false, or open? 9p+8=10p+7

stich3 [128]

The answer is A because it has a variable. HOPE THIS HELPS!!! <3

6 0

2 years ago

the sum of the three number is 92. the third is 3 times the first, the first number is 7 less than the second. what are the numb

vichka [17]

Let x represent the first number, y represent the second, and z represent the third.

x + y + z = 92
z = 3x
x = y - 7

Solve x = y - 7 for y.
x = y - 7
x + 7 = y

Fill in y and z so that we only have to worry about x for now.

x + y + z = 92
x + (x + 7) + (3x) = 92
x + x + 7 + 3x = 92
5x + 7 = 92
5x = 92 - 7
5x = 85
x = 85/5
x = 17

Solve for y.
y = x + 7
y = (17) + 7
y = 24

Solve for z.
z = 3x
z = 3(17)
z = 51

So x = 17, y = 24, and z = 51.

8 0

3 years ago

At Pizza Pi, 68% of the pizzas made last week had extra cheese. If 17 pizzas had extra cheese, how many pizzas in all were made

Andrei [34K]

Answer: 50

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

A rare form of malignant tumor occurs in 11 children in a million, so its probability is 0.000011. Four cases of this tumor occ

Shkiper50 [21]

Answer:

a) The mean number of cases is 0.14608 cases.

b) The probability that the number of cases is exactly 0 or 1 is 0.990.

c) The probability of more than one case is 0.010

d) No, because the probability of more than one case is very small

Step-by-step explanation:

We can model this problem with a Poisson distribution, with parameter:

$\lambda=r*t=0.000011*13,280=0.14608$

a) The mean amount of cases is equal to the parameter λ=0.14608.

b) The probability of having 0 or 1 cases is:

$P(k=0)=\frac{\lambda^0 e^{-\lambda}}{0!}=\frac{1*0.864}{1} =0.864\\\\ P(k=1)=\frac{\lambda^1 e^{-\lambda}}{0!}=\frac{0.14608*0.864}{1} =0.126\\\\P(k\leq1)=0.864+0.126=0.990$

c) The probability of more than one case is:

$P(k>1)=1-P(k\leq 1)=1-0.990=0.010$

d) The cluster of 4 cases can not be due to pure chance, as it is a very high proportion of cases according to the average rate. Just having more than one case has a probability of 1%.

7 0

3 years ago