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andre [41]
3 years ago
9

If Josh can read 18 pages in 40 minutes, how long will it take him to read 90 pages?

Mathematics
2 answers:
Dvinal [7]3 years ago
3 0

Answer:

18=40

90? 40×90=3600÷18=200 pages.

Andrei [34K]3 years ago
3 0

Answer:

It will take him 200 minutes to read 90 pages.

Step-by-step explanation:

Josh can read 18 pages in 40 minutes.

Ratio of pages to minutes = 18 : 40

Number of pages to read = 90

Minutes = x

Ratio of pages to minutes = 90 : x

The relationship is proportional

18 : 40 :: 90 : x

Product of extremes = Product of means

18x = 40*90

18x = 3600

Dividing both sides by 18

\frac{18x}{18}=\frac{3600}{18}

x = 200

Therefore,

It will take him 200 minutes to read 90 pages.

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3 years ago
What’s the answer to this?
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y=3/2x+0

Step-by-step explanation:

The formula for slope intercept form formula is y=mx+b where m is the slope and b is the y intercept and since the slope is rise over run ( or rise/run just put it into fraction form) we count from the y intercept up until we can see the line reach a point where it touches a actual cross point ( in this case from the y intercept we see it goes up three) Then we count over how many to that cross point ( the full point, not just a random place on the chart) (in this case 2) and that creates 3/2. Now for the y intercept. Where does the line intercept the vertical line? That's your y intercept. In this case it's 0. Now you can see where we count up from three ( for the slope) and over two. Right onto that point. Hope this makes sense! If not look up Khan academy for some extra tutoring that is free.

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part 2. Find the value of the trig function indicated, use for that Pythagorean theorem to find the third side if you need it.​
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Answer:  \bold{5)\ \cos \theta=\dfrac{\sqrt{11}}{6}\qquad 6)\ \tan \theta =\dfrac{8}{17}\qquad 7)\ \cos \theta = \dfrac{4}{3}\qquad 8)\ \cos \theta = \dfrac{\sqrt{10}}{10}}

<u>Step-by-Step Explanation:</u>

Pythagorean Theorem is: a² + b² = c²  , <em>where "c" is the hypotenuse</em>

5)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{3\sqrt{11}}{18}\quad \rightarrow \large\boxed{\dfrac{\sqrt{11}}{6}}

Note: (15)² + (3√11)² = hypotenuse²   →   hypotenuse = 18

6)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{8}{17}\quad =\large\boxed{\dfrac{8}{17}}

Note: 8² + 15² = hypotenuse²   →   hypotenuse = 17

7)\ \tan \theta=\dfrac{\text{side opposite to}\ \theta}{\text{side adjacent to}\ \theta}=\dfrac{20}{15}\quad \rightarrow \large\boxed{\dfrac{4}{3}}

Note: hypotenuse not needed for tan

8)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{2}{2\sqrt{10}}\quad =\large\boxed{\dfrac{\sqrt{10}}{10}}

Note: 2² + 6² = hypotenuse²   →   hypotenuse = 2√10

8 0
2 years ago
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