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2 years ago

Lola rode her bike for 7 of an hour on Saturday and of an hour on

Mathematics

2 answers:

Paraphin [41]2 years ago

4 0

Answer:

3

Step by step explanation:

Alinara [238K]2 years ago

4 0

Answer:

Step-by-step explanation:

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If we change the -3 to a 3 in the equation what would happen to the graph

kirza4 [7]

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3 years ago

7(m+5) = 21 <br><br>What is it?

son4ous [18]

M= -2 because you divide both side by 7 to get m+5=3 then subtract 5 from both sides to get m=-2

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3 years ago

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Which one of these numbers is a composite number a.17 b.51 c.97 d.89

pantera1 [17]

Step-by-step explanation:

The correct answer here is 51

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2 years ago

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Write each expression as an algebraic (nontrigonometric) expression in u, u > 0.

max2010maxim [7]

Answer:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0$

Step-by-step explanation:

We want to write the trignometric expression:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)\text{ where } u>0$

As an algebraic equation.

First, we can focus on the inner expression. Let θ equal the expression:

$\displaystyle \theta=\sec^{-1}\left(\frac{u}{10}\right)$

Take the secant of both sides:

$\displaystyle \sec(\theta)=\frac{u}{10}$

Since secant is the ratio of the hypotenuse side to the adjacent side, this means that the opposite side is:

$\displaystyle o=\sqrt{u^2-10^2}=\sqrt{u^2-100}$

By substitutition:

$\displaystyle= \sin(2\theta)$

Using an double-angle identity:

$=2\sin(\theta)\cos(\theta)$

We know that the opposite side is √(u² -100), the adjacent side is 10, and the hypotenuse is u. Therefore:

$\displaystyle =2\left(\frac{\sqrt{u^2-100}}{u}\right)\left(\frac{10}{u}\right)$

Simplify. Therefore:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0$

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3 years ago

Soo I need help... can u help?

Luda [366]

Answer:

I think its (6042 = 1780 + a) but I'm not really sure though but I hope if that helped somehow

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3 years ago

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