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3 years ago

8 meters increased by 25% is what

Mathematics

1 answer:

Rudiy273 years ago

3 0

Answer:

ten meters

Step-by-step explanation:

twenty-five percent of eight is equal to two.

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A 10 meter piece of wire is cut in to two pieces. One piece is 2 meters longer than the other. How long are the pieces?

Vilka [71]

Answer:

Step-by-step explanation:

total metre = 10m

If one piece is 4m then 2m more then that = 4 + 2 = 6m

So total = 4 + 6 = 10m

7 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

HELP ME ASAP!!!!!! 15 POINTS!!!MATH!!!

aniked [119]

Angle m∠8 =41° and x=10°

Step-by-step explanation:

Given that m∠3=139° then

m∠3=m∠1 = corresponding angles = 139° and m∠7=m∠5=m∠3=139°

m∠8= 180°-139° =41° -sum of angles on a straight line

m∠8=m∠6=m∠4=m∠2= 41°

Given that m∠3=8x+70° and m∠8=4x-10° then you can find the value of x as;

m∠3+m∠8=180°

8x+70°+4x-10°=180°

12x+60°=180°

12x=180°-60°

12x=120°

12x/12 =120°/12

x=10°

Hence,

m∠3=8x+70° = 8*10 +70° = 80°+70°=150°

m∠8= 4x-10° = 4*10 - 10° = 40°-10° =30°

m∠1 =m∠3 = m∠5 = m∠7 =150°

m∠2 = m∠ 4 = m∠6 = m∠8 =30°

Learn More

Angles : brainly.com/question/12157314

Keywords : parallel lines, transversal, angles

#LearnwithBrainly

7 0

3 years ago

What type of function is represented in the table?

Akimi4 [234]

This is D. linear. Hope this helps!(:

8 0

3 years ago

Read 2 more answers

If s(t)=4^3t, what is s(-1)?

larisa86 [58]

Answer:

Step-by-step explanation:

we can write -1 for;

s(-1)= 4^3(-1)

4^-3 = 1/64

8 0

3 years ago