-r/2=5 what does r equal

Use differentials to approximate the change in the volume of a cube when the side is decreased from 8 to 7.99 cm. (in cm3)

Firlakuza [10]

Answer:

-1.92 cm³

Step-by-step explanation:

The formula for the volume of a cube in terms of side length is ...

V = s³

Then the first-order approximation of volume for a small change ∆s in side length is ...

V(s +∆s) ≈ V(s) + V'(s)·∆s

and the change in volume is ...

∆V ≈ V(s +∆s) -V(s) = V'(s)·∆s

The derivative V'(s) is ...

V'(s) = 3s²

so the change in volume for s = 8 cm and ∆s = -0.01 cm is ...

∆V ≈ 3s²·∆s = 3(8 cm)²(-0.01 cm) = -1.92 cm³

4 0

3 years ago

75 percent of the sutdents in a school went on a field trip. If 285 students went on the trip, how any students are in the schoo

natita [175]

Answer:

380

Step-by-step explanation:

285 / x = 75 / 100

Cross multiply 28500 = 75x

Divide by 75 380

8 0

2 years ago

Geometry question. please help

Vinvika [58]

Answer:

The length of PQ is 18 feet.

The length of PR is 18 feet.

The length of QR is 24 feet.

Step-by-step explanation:

A way to set an equation up for this problem is:

$\frac{4}{3}x+x+x=60$

where x is the three lengths of the isosceles triangle, but the base QR is 4/3 the length of the other two congruent sides, length PQ and PR. The 60 represents the total length of the perimeter.

Then, solve for x from the equation, and you’ll get x=18. But your not done yet. Since the variable x in the equation stands for the sides of the isosceles triangle, so plug 18 into the equation and it should look like this:

$\frac{4}{3}(18)+18+18=60$

Don’t solve the whole equation, just solve the $\frac{4}{3}(18)$ part of the equation, which is equal to 24. So the final equation is this:

$24+18+18=60$

Conclusion: 24 is the length of QR, and 18 is the length of PQ and PR. And they all equal 60, which is the perimeter. This is very true because the length of PQ and PR are the same (length 18), since it’s an isosceles triangle, and the length of QR is 4/3 the length of PQ and PR (4/3 of 18= 24).

Sorry for the long explanation.

But hope this helps and answers your question :)

8 0

3 years ago

Slove algebraically 3x - 4y = -24 x +4y=8

Vanyuwa [196]

Solve algebrically 3x - 4y = -24 and x + 4y = 8 is x = -4 and y = 3

Solution:

We have been given two equations which are as follows:

3x - 4y = -24 ----- eqn 1

x + 4y = 8 -------- eqn 2

We have been asked to solve the equations which means we have to find the value of ‘x’ and ‘y’.

We rearrange eqn 2 as follows:

x + 4y = 8

x = 8 - 4y ------eqn 3

Now we substitute eqn 3 in eqn 1 as follows:

3(8 - 4y) -4y = -24

24 - 12y - 4y = -24

-16y = -48

y = 3

Substitute "y" value in eqn 3. Therefore the value of ‘x’ becomes:

x = 8 - 4(3)

x = 8 - 12 = -4

Hence on solving both the given equations we get the value of x and y as -4 and 3 respectively.

5 0

3 years ago

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

Therefore the half-life is between 600 and 700 years

5 0

3 years ago