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2 years ago

10

When someone donated 14 gallons of paint to DeAnza Elementary. The fifth grade decided to use it to paint murals. They split the

gallons equally among the four classes.
How much paint did each class have to paint there mural?

How much paint will three classes use?

1 answer:

Oksanka [162]2 years ago

8 0

Each class had 3.5 gallons of paint to paint with and 3 classes used 10.5 gallons of paint

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In geometry you can use deductive rules to

Licemer1 [7]

Answer:

To prove that a given statement or conjuncture is true or false using deductive reasoning.

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3 years ago

What is the value of y in the equation 2 + y = −3?

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The answer for your question is -5

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3 years ago

Louis has 7 1/2 ft of red ribbon. How many red bows can he make using 3/4 ft pieces of ribbon for each bow?

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So I got 10 bows he can make

7 1/2 = 15/2 = 7.5
3/4 = .75

1) 7.50-.75=6.75
2) 6.75-.75=6.00
3) 6.00-.75=5.25
4) 5.25-.75=4.50
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10 times total

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3 years ago

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→1 3x x − 1

Reika [66]

Answer:

<h2>3/2</h2>

Step-by-step explanation:

Given the limit of a function expressed as $\lim_{x \to 1} (\dfrac{3x}{x-1} - \dfrac{3}{lnx})$ , we are to evaluate it. To evaluate it, we will simply substitute x = 1 into the function since the variable x tends to 1.

$\lim_{x \to 1} (\dfrac{3x}{x-1} - \dfrac{3}{lnx})\\\\= (\dfrac{3(1)}{1-1} - \dfrac{3}{ln1})\\\\= \dfrac{3}{0} - \dfrac{3}{0}\\\\= \infty - \infty (ind)$

Since we got an indeterminate function, we will find the LCM of the function and solve again.

$= \lim_{x \to 1} (\dfrac{3x}{x-1} - \dfrac{3}{lnx})\\\\= \lim_{x \to 1} \dfrac{3xlnx-3(x-1)}{(x-1)lnx}\\\\\\= \dfrac{3(1)ln(1)-3(1-1)}{(1-1)ln1}\\\\= \frac{3(0)-3(0)}{0(0)} \\\\= \frac{0}{0} (ind)$

Applying L'hospital rule;

$\frac{x}{y} = \lim_{x \to 1} \dfrac{d/dx(3xlnx-3(x-1))}{d/dx((x-1)lnx)}\\\\= \lim_{x \to 1} \dfrac{3x(\frac{1}{x})+ 3lnx-3)}{(x-1)\frac{1}{x} +lnx}\\\\= \lim_{x \to 1} \dfrac{3 + 3lnx-3}{(x-1)\frac{1}{x} +lnx}\\\\= \frac{3ln1}{(1-1)\frac{1}{1} +ln1}\\\\= \frac{0}{0} (ind)$

Applying L'hospital rule again;

$= \lim_{x \to 1} \dfrac{\frac{d}{dx} (3lnx)}{\frac{d}{dx} ((x-1)\frac{1}{x} +lnx)}\\\\= \lim_{x \to 1} \dfrac{\frac{3}{x} }{(x-1)\frac{-1}{x^2} + \frac{1}{x} +\frac{1}{x} }\\\\= \dfrac{\frac{3}{1} }{(1-1)\frac{-1}{1^2} + \frac{1}{1} +\frac{1}{1} }\\\\= \frac{3}{0(-1)+2}\\ \\= \frac{3}{2-0}\\ \\= 3/2$

<em>Hence the limit of the function is 3/2.</em>

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3 years ago

A class's exam scores are normally distributed. If the average score is 65 and the standard deviation is 6, what percentage of s

tiny-mole [99]

Standardising X=71 to find z-score

$z-score= \frac{71-65}{6} =1$

reading from z-table

$P(Z\ \textless \ 1)=0.8413=84.13%$

Students score below 71 is 84.13%

5 0

3 years ago

Read 2 more answers