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3 years ago

4. A man who is 40 years old wants to begin exercising daily. Calculate his target heart

Mathematics

1 answer:

Andrej [43]3 years ago

3 0

200- your age = your target heart rate.

Hence, for a 40 year old man, it’s 200-40= 160.

Hope this helps!

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Which is equivalent to (–9n + 12)?

Stella [2.4K]

-3(3n+4)

Step-by-step explanation:

answer from mathematics education book

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3 years ago

A radioactive material has a half-life of 15 days if you start with 160 mg after how long will they be 5mg left?

SOVA2 [1]

Answer:

75 days

Step-by-step explanation:

The multiplier each day is ...

(1/2)^(t/15)

Then we want to find t such that ...

5 = 160·(1/2)^(t/15) . . . . . . 5 mg are left after t days

5/160 = (1/2)^(t/15) . . . . . . divide by 160

1/32 = (1/2)^5 = (1/2)^(t/15) . . . . rewrite the left side

Equating exponents (equivalent to taking logs), we get ...

5 = t/15

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After 75 days, there will be 5 mg left.

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4 years ago

Arrange the polynomial so that the powers of x are in ascending order.

marshall27 [118]

4 + 3ax⁵ + 2ax² - 5a⁷x
4 - 5a⁷x + 2ax² + 3ax⁵

7 0

3 years ago

Find the vaule of x and y

ra1l [238]

5y=95
y=95/5(vertically opppsite angles)

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6 0

3 years ago

A chain lying on the ground is 10 m long and its mass is 70 kg. How much work (in J) is required to raise one end of the chain t

DerKrebs [107]

Answer:

$W= 34.3 \frac{kg}{s^2} (4^2-0^2)m^2 =548.8 \frac{kg m^2}{s^2} =548.8 J$

Step-by-step explanation:

Data Given: m = 70 kg , g = 9.8 ms^-2, h =10m.

For this case we can use the following formula:

$W = \int_{x_i}^{x_f} F(x) dx$

For this case we need to find an expression for the force in terms of the distance. And since on this case the total distance is 10 m long we can write the expression like this:

$F(x) = \frac{ma}{10m}= \frac{mg}{10m} x$

The only acceleration on this case is the gravity and if we replace the values given we got:

$\frac{70 kg *9.8 m/s^2}{10m} x=68.6 x\frac{kg}{s^2}$

Now we can find the required work with the following integral:

$W= 68.6 \frac{kg}{s^2} \int_{0}^4 x dx$

$W= 34.3 \frac{kg}{s^2} x^2 \Big|_0^4$

$W= 34.3 \frac{kg}{s^2} (4^2-0^2)m^2 =548.8 \frac{kg m^2}{s^2} =548.8 J$

7 0

4 years ago