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2 years ago

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Hakeem is training for a long distance race. He ran 6 1/4 miles on Monday, 5 3/4 miles on Wednsday , and 5 1/4 miles on Friday.

About how many miles has Hakeem ran all together

1 answer:

Vlada [557]2 years ago

3 0

17 1/4 I believe is correct

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Sharon tracks the number of cupcakes she sells each day for a week. The numbers of cupcakes she sold are 23, 25, 32, 35, 36, 25,

Ghella [55]

Answer:

27

Step-by-step explanation:

To get the median, we line the values up from lowest to greatest.

23, 25, 25, 27, 32, 35, 36

What's the number in the middle? Since there's an odd number of values, it's easier to see: the median is 27.

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2 years ago

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What is the value of y? <br><br><br> • 3 <br> • 4 <br> • 5<br> • 6

coldgirl [10]

I think the answer is 6 but I might not be right. Please don’t be mad if it’s wrong. :(

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2 years ago

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A baby bird jumps from a tree branch and flutters to the ground. The function "f" models the bird's height (in meters) above the

AURORKA [14]

The answer to this question will depend on the function f itself. Basically you will find the height in meters above the ground of the bird when it jumped when the time t=0s. This is substsitute every t in the function for a value of zero and that way you will get the bird's height at the time it jumped. If you were given a graph for this function, you can find the y-intercept of the graph and that will be the answer as well. The question could be written like this:

A baby bird jumps from a tree branch and flutters to the ground. The function " $f(t)-4.9t^{2}+25$ " models the bird's height (in meters) above the ground as a function of time (in seconds) after jumping. What is bird's height above the ground when it jumped.

Answer:

25m

Step-by-step explanation:

Once your function is given, you can substitute t=0 since 0s is the time measured at the moment the bird jumped. So our function will be:

$f(0)=-4.9(0)^2+25$

$f(0)=25m$

So the height of the bird above the ground when it jumped is 25m in this particular function.

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2 years ago

In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc

Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

$A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°$

Use sine rule,

$\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}$

Again consider,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Thus, the angle B is function of A is, $B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Now find $\frac{dB}{dA}$

Differentiate implicitly the function $\sin{B}=\sqrt{\frac{3}{2}}\sin{A}$ with respect to A to get,

$\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}$

b)

When $A=\frac{\pi}{4},B=\frac{\pi}{3}$ , the value of $\frac{dB}{dA}$ is,

$\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}$

c)

In general, the linear approximation at x= a is,

$f(x)=f'(x).(x-a)+f(a)$

Here the function $f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

At $A=\frac{\pi}{4}$

$f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}$

And,

$f'(A)=\frac{dB}{dA}=\sqrt{3}$ from part b

Therefore, the linear approximation at $A=\frac{\pi}{4}$ is,

$f(x)=f'(A).(x-A)+f(A)\\\\=f'(\frac{\pi}{4}).(x-\frac{\pi}{4})+f(\frac{\pi}{4})\\\\=\sqrt{3}.[x-\frac{\pi}{4}]+\frac{\pi}{3}$

d)

Use part (c), when $A=46°$ , B is approximately,

$B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°$

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2 years ago

The table represents a linear function

Feliz [49]

Answer:

3

Step-by-step explanation:

.......................

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