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valentinak56 [21]

2 years ago

11

Elena rode her bike 2 miles in 10 minutes. She rode at a constant speed. Complete the table to show the time it took her to trav

el different distances at this speed. Write your answer as a decimal.

2 answers:

Komok [63]2 years ago

6 0

Answer:

0.2

Step-by-step explanation:

2/10 or 1/5

NeTakaya2 years ago

6 0

Answer:

0.2

Step-by-step explanation:

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Four times a number<br> decreased by ten is 34.<br> Write an equation & solve.

igomit [66]

4x - 10 = 34
4x +10 +10
4x = 44
4x/4 = 44/4
x = 11

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3 years ago

Line g is dilated by a scale factor of 1/2 from the origin to create line g'. Where are points E' and F' located after dilation,

Sholpan [36]

Answer:

The locations of E' and F' are E'(-2,0) and F'(0,1), and lines g and g' are parallel.

Step-by-step explanation:

I got it right on my test! :)

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Tiana's checking account charges a $7.75 monthly service fee and a $0.16 per-check fee. If Tiana writes 17 checks per month, sho

AlekseyPX

the answer to this question is B

3 0

3 years ago

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A stereo costs $320. in a sale it is reduced by 25% what is the sale price

Rudiy27

First, set up a proportion to find the reduction in price. 25/100=x/320. Cross multiply to get 8000=100x. Divide by 100 on both sides to get x=80. That's the reduction in price. Subtract from the total to find the new price 320-80=$240

6 0

3 years ago

If alpha and beta are the angles in the first quadrant tan alpha = 1/7 and sin beta =1/ root 10 then usind the formula sin (A +B

zysi [14]

Answer:

$\arcsin\left(\frac{129\sqrt{2}}{250}\right)\approx 0.8179$

Step-by-step explanation:

$\alpha \text{ and } \beta \text{ in Quadrant I}$

$\tan(\alpha)=\frac{1}{7} \text{ and } \sin(\beta)=\frac{1}{\sqrt{10}}=\frac{\sqrt{10} }{10}$

<u>Using Pythagorean Identities</u>:

$\boxed{\sin^2(\theta)+\cos^2(\theta)=1} \text{ and } \boxed{1+\tan^2(\theta)=\sec^2(\theta)}$

$\left(\frac{\sqrt{10} }{10} \right)^2+\cos^2(\beta)=1 \Longrightarrow \cos(\beta)=\sqrt{1-\frac{10}{100}} =\sqrt{\frac{90}{100}}=\frac{3\sqrt{10}}{10}$

$\text{Note: } \cos(\beta) \text{ is positive because the angle is in the first qudrant}$

$1+\left(\frac{1 }{7} \right)^2=\frac{1}{\cos^2(\alpha)} \Longrightarrow 1+\frac{1}{49}=\frac{1}{\cos^2(\alpha)} \Longrightarrow \frac{50}{49} =\frac{1}{\cos^2(\alpha)}$

$\Longrightarrow \frac{49}{50}=\cos^2(\alpha) \Longrightarrow \cos(\alpha)=\sqrt{\frac{49}{50} } =\frac{7\sqrt{2}}{10}$

$\text{Now let's find }\sin(\alpha)$

$\sin^2(\alpha)+\left(\frac{7\sqrt{2} }{10}\right)^2=1 \Longrightarrow \sin^2(\alpha) +\frac{49}{50}=1 \Longrightarrow \sin(\alpha)=\sqrt{1-\frac{49}{50}} = \frac{\sqrt{2}}{10}$

<u>The sum Identity is</u>:

$\sin(\alpha + \beta)=\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)$

I will just follow what the question asks.

$\text{Find the value of }\alpha+2\beta$

$\sin(\alpha + 2\beta)=\sin(\alpha)\cos(2\beta)+\sin(2\beta)\cos(\alpha)$

$\text{I will first calculate }\cos(2\beta)$

$\cos(2\beta)=\frac{1-\tan^2(\beta)}{1+\tan^2(\beta)} =\frac{1-(\frac{1}{7})^2 }{1+(\frac{1}{7})^2}=\frac{24}{25}$

$\text{Now }\sin(2\beta)$

$\sin(2\beta)=2\sin(\beta)\cos(\beta)=2 \cdot \frac{\sqrt{10} }{10}\cdot \frac{3\sqrt{10} }{10} = \frac{3}{5}$

Now we can perform the sum identity:

$\sin(\alpha + 2\beta)=\sin(\alpha)\cos(2\beta)+\sin(2\beta)\cos(\alpha)$

$\sin(\alpha + 2\beta)=\frac{\sqrt{2}}{10}\cdot \frac{24}{25} +\frac{3}{5} \cdot \frac{7\sqrt{2} }{10} = \frac{129\sqrt{2}}{250}$

But we are not done yet! You want

$\alpha + 2\beta$ and not $\sin(\alpha + 2\beta)$

You actually want the

$\arcsin\left(\frac{129\sqrt{2}}{250}\right)\approx 0.8179$

7 0

2 years ago

Read 2 more answers