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Marina86 [1]
3 years ago
10

The resting heart rate of an adult is usually about 60 beats per minute. Use your completed table to find approximately how many

minutes it will take for the average resting heart to beat 1,000,000 times.
Mathematics
1 answer:
Dvinal [7]3 years ago
3 0
60x=1,000,000
----- ---------------
60 60
X=16,666 2/3 or
X=16,666.6667 it would take approximately 16,667 minutes for the average resting heart to beat 1,000,000 times.
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Gemiola [76]

Answer: C. 6x+18

Step-by-step explanation:

Use the distributive property

First multiply 6 times x which is 6x

Then multiply 6 times 3 which is 18

Keep the sign the same

Therefore, this should give you 6x+18

7 0
2 years ago
12.5% of $100 is<br> what number?
Westkost [7]

Answer:100$-100%               100*12.5=1250

                                                    1250/100=12.5

                 ?-12.5%

Step-by-step explanation:

7 0
2 years ago
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For the function y=-1+6 cos(2pi/7(x-5)) what is the minimum value?
Ad libitum [116K]
Usually one will differentiate the function to find the minimum/maximum point, but in this case differentiating yields:
\frac{2\pi}{7(x-5)^{2}}\sin{\frac{2\pi}{7(x-5)}}}
which contains multiple solution if one tries to solve for x when the differentiated form is 0.

I would, though, venture a guess that the minimum value would be (approaching) 5, since the function would be undefined in the vicinity.

If, however, the function is
-1+\cos{\frac{2\pi}{7}(x-5)}}
Then differentiating and equating to 0 yields:
\sin{\frac{2\pi}{7}(x-5)}}=0
which gives:
x=5 or 8.5

We reject x=5 as it is when it ix the maximum and thus,
x=8.5\pm7n, for n=0,\pm 1,\pm 2, ...
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Simplify .4 ^ square root of 6
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Evaluate the double integral.
Fynjy0 [20]

Answer:

\iint_D 8y^2 \ dA = \dfrac{88}{3}

Step-by-step explanation:

The equation of the line through the point (x_o,y_o) & (x_1,y_1) can be represented by:

y-y_o = m(x - x_o)

Making m the subject;

m = \dfrac{y_1 - y_0}{x_1-x_0}

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

m= \dfrac{2-1}{1-0}

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

m = \dfrac{1-2}{4-1}

m = \dfrac{-1}{3}

∴

y-2 = -\dfrac{1}{3}(x-1)

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

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Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy

\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg)   dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ([y^2(-4y+8)] \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( -4y^3+8y^2 \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1

\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1

\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [  \dfrac{-45+56}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [  \dfrac{11}{3}\bigg]

\iint_D 8y^2 \ dA = \dfrac{88}{3}

4 0
2 years ago
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