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3 years ago

PLZ ANSWER VERY IMPORTANT GRADE

Mathematics

1 answer:

olchik [2.2K]3 years ago

8 0

Pretty sure it is Tawni. if you divide all of the events by the times to get the unit rate, Tani had the fastest time.

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What is X-24 = 58; x= 82

Alecsey [184]

But that doesn't really make sense, because if x has already been solved, then the equation makes sense, because if we substitute x into it, then we have 82-24, which is 58. I could be wrong... sooo... but I hope it helps?

5 0

3 years ago

5. Find the first, fourth, and tenth terms of the arithmetic

Lubov Fominskaja [6]

The first, fourth and tenth terms of the arithmetic sequence is $-6, -\frac{27}{5}$ and $-\frac{21}{5}$

Explanation:

The given rule for the arithmetic sequence is $A(n)=-6+(n-1)(\frac{1}{5} )$

We need to determine the first, fourth and tenth terms of the sequence.

To find the first, fourth and tenth terms, let us substitute $n=1,4,10$ in the general rule for the arithmetic sequence.

To find the first term, substitute $n=1$ in $A(n)=-6+(n-1)(\frac{1}{5} )$ , we get,

$A(1)=-6+(1-1)(\frac{1}{5} )$

$A(1)=-6+(0)(\frac{1}{5} )$

$A(1)=-6$

Thus, the first term of the arithmetic sequence is -6.

To find the fourth term, substitute $n=4$ in $A(n)=-6+(n-1)(\frac{1}{5} )$ , we get,

$A(2)=-6+(4-1)(\frac{1}{5} )$

$A(2)=-6+(3)(\frac{1}{5} )$

$A(2)=\frac{-30+3}{5}$

$A(2)=\frac{-27}{5}$

Thus, the fourth term of the arithmetic sequence is $-\frac{27}{5}$

To find the tenth term, substitute $n=10$ in $A(n)=-6+(n-1)(\frac{1}{5} )$ , we get,

$A(10)=-6+(10-1)(\frac{1}{5} )$

$A(10)=-6+(9)(\frac{1}{5} )$

$A(10)=-6+\frac{9}{5}$

$A(10)=-\frac{21}{5}$

Thus, the tenth term of the arithmetic sequence is $-\frac{21}{5}$

Hence, the first, fourth and tenth terms of the arithmetic sequence is $-6, -\frac{27}{5}$ and $-\frac{21}{5}$

3 0

4 years ago

Original amount: 27<br> End amount: 38.61<br><br> Increase or decrease

vichka [17]

Answer:

increase

Step-by-step explanation:

11.61 was added to 27

5 0

3 years ago

Use differentials to estimate the amount of material in a closed cylindrical can that is 60 cm high and 24 cm in diameter if the

alexira [117]

Answer:

$dV=542.9 cm^{3}$

Step-by-step explanation:

Let's start with the equation of the volume of a cylinder:

$V=\pi r^{2}*h$

Where:

r is the radius of the cylinder (r=D/2=24/2=12 cm)
h is the high of the cylinder (h = 60 cm)

We can us partial derivatives to find the differential of this volume. So we will have:

$dV=\frac{\partial V}{\partial r}dr+\frac{\partial V}{\partial h}dh$ (1)

Now:

$\frac{\partial V}{\partial r}=2\pi r*h=2\pi 12*60=4523.9 cm^{2}$

$\frac{\partial V}{\partial h}=\pi 12^{2}=452.4 cm^{2}$

dr is a differential of the radius, so in our case it is 0.1 cm and dh, differential of the high, is 0.1*2 cm. We multiply by 2 because we need to consider the top and the bottom of the cylinder.

Now we just need to put all of this definitions in the equation (1).

$dV=4523.9*0.1+452.4*0.2=542.9 cm^{3}$