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2 years ago

8

Jose weighed himself on monday and he weighed 170 pounds. Two weeks later he weighed 165 pounds. write a linera function in the

form y=mx+b to model jose's weight loss each week, where x is the number of weeks and y is joses weight.

1 answer:

sukhopar [10]2 years ago

7 0

Y= 2x-5 i believe lemme kno if it’s wrong

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The areas of the two watch faces have a ratio of 16:25 What is the ratio of the radius of the smaller watch face to the radius o

Leona [35]

Answer:

The ratio of the radius of the smaller watch face to the radius of the larger watch face is 4:5.

Step-by-step explanation:

Let the Area of smaller watch face be $A_1$

Also Let the Area of Larger watch face be $A_2$

Also Let the radius of smaller watch face be $r_1$

Also Let the radius of Larger watch face be $r_2$

Now given:

$\frac{A_1}{A_2} =\frac{16}{25}$

We need to find the ratio of the radius of the smaller watch face to the radius of the larger watch face.

Solution:

Since the watch face is in circular form.

Then we can say that;

Area of the circle is equal 'π' times square of the radius 'r'.

framing in equation form we get;

$A_1 = \pi {r_1}^2$

$A_2 = \pi {r_2}^2$

So we get;

$\frac{A_1}{A_2}= \frac{\pi {r_1}^2}{\pi {r_2}^2}$

Substituting the value we get;

$\frac{16}{25}= \frac{\pi {r_1}^2}{\pi {r_2}^2}$

Now 'π' from numerator and denominator gets cancelled.

$\frac{16}{25}= \frac{{r_1}^2}{{r_2}^2}$

Now Taking square roots on both side we get;

$\sqrt{\frac{16}{25}}= \sqrt{\frac{{r_1}^2}{{r_2}^2}}\\\\\frac{4}{5}= \frac{r_1}{r_2}$

Hence the ratio of the radius of the smaller watch face to the radius of the larger watch face is 4:5.

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Vanyuwa [196]

Answer:

Whats your question exactly?

Step-by-step explanation:

I don't understand what the question is

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3 years ago

What is the value of y in the equation 3y - 8 = 22?

ArbitrLikvidat [17]

10

Step-by-step explanation:

3y-8=22

3y=22+8

3y=30

y=30/3

y =10

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2 years ago

Read 2 more answers

Alex Ar [27]

k= 4/9 or 0.444 in decimal form.

hope this helps!

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2 years ago

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