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3 years ago

I need help on this

Mathematics

1 answer:

Sever21 [200]3 years ago

3 0

Answer:

The heat from the sun is making the sand hot and heat from the food on plate is making the plate hot

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Joe asked the children in his class which flavours of ice-cream they like.

Leya [2.2K]

Answer:

a. 19

b. 14

Step-by-step explanation:

From the venn diagram, we see that:

9 children like only Vanilla

7 like vanilla and chocolate

12 like only chocolate, and

2 like neither chocolate nor vanilla

Thus:

a. Number of children that liked Chocolate ice-cream = those that like chocolate only + those that like both chocolate and vanilla = 12 + 7 = 19

19 children like chocolate ice-cream.

b. Number of children who do not like Vanilla ice-cream = those that like chocolate only + those that do not like neither chocolate nor vanilla = 12 + 2 = 14

14 children do not like vanilla ice-cream.

5 0

3 years ago

The price of tickets for the dance was 1 ticket for $6 or 2 tickets for $8 she looked inside the cash box and found $200 and tic

Aleks04 [339]

We can say that exchanging one couple's ticket for an individual's ticket would increase the money in the cash box from 200 to 202 and it would result in an even number of couples tickets sold.

Step-by-step explanation:

Let the number of tickets sold to the individuals = s

Let the number of tickets sold to the couples = c

According to the question,

s + c = 46 ( Equation 1)

Since each individual's ticket is $6, the total amount of money made by selling tickets to individuals is 6s.

Similarly, since each ticket sold to couples is $8, the total amount of money made by selling tickets to couples is 8c.

So,

6 s + 8 c = 200 ( Equation 2)

On solving both the equations, we get

c = 38 and s = 8

Therefore, 8 tickets were sold to individuals and 38 tickets were sold to the couples.

5 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:

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