All categories

3 years ago

How do i solve this please help

Mathematics

1 answer:

german3 years ago

4 0

Step-by-step explanation:

I hope this helps! Have a good rest of your day:)

You might be interested in

Find the surface area

sattari [20]

Answer:

1734.1591

Step-by-step explanation:

8 0

3 years ago

Find the distance and midpoint for the points (5,5) and (1,2).

stealth61 [152]

Answer:

midpoint = (3,3.5)

distance = 5

Step-by-step explanation:

midpoint = (x1+X2/2, y2+y2/2)

=(5+1/2, 5+2/2)

=(6/2, 7/2)

=(3, 3.5)

distance= √(x2-x1)^2 -(y2-y1)^2

=√(1-5)^2 -(2-5)^2

= √ 16+9 =√25 =5

5 0

3 years ago

Please help me with question 3

alexira [117]

There's no picture I wish I could help

5 0

4 years ago

Find the final amount of money in an account if $4,700 is deposited at 2.5% interest compounded quarterly (every 3 months) and t

marysya [2.9K]

Answer:

$6030.23

Step-by-step explanation:

The future value formula is useful for this.

FV = P(1 +r/n)^(nt)

where P is the principal invested, r is the annual rate, n is the number of times per year interest is compounded, and t is the number of years. Filling in the given values, we can do the arithmetic to find the amount.

FV = $4700·(1 + 0.025/4)^(4·10) ≈ $6030.23

The amount in the account will be $6030.23.

8 0

3 years ago

The volume V of an ice cream cone is given by V = 2 3 πR3 + 1 3 πR2h where R is the common radius of the spherical cap and the c

Nuetrik [128]

Answer:

The change in volume is estimated to be 17.20 $\rm{in^3}$

Step-by-step explanation:

The linearization or linear approximation of a function $f(x)$ is given by:

$f(x_0+dx) \approx f(x_0) + df(x)|_{x_0}$ where $df$ is the total differential of the function evaluated in the given point.

For the given function, the linearization is:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh$

Taking $R_0=1.5$ inches and $h=3$ inches and evaluating the partial derivatives we obtain:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh$

substituting the values and taking $dx=0.1$ and $dh=0.3$ inches we have:

$V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3} + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}$

Therefore the change in volume is estimated to be 17.20 $\rm{in^3}$

4 0

3 years ago