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6 months ago

13

Find the coordinates of the midpoint of a segment with the endpoints (7,-5) and (3, 3).

1 answer:

77julia77 [94]6 months ago

5 0

Answer:

(5,-1)

Step-by-step explanation:

I use the graphing calculator desmos go into functions and geometry and click midpoint and insert numbers so it looks like this ex. midpoint((x,y),(x,y)) hope this helps!

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luis has a cooler full of soda. For every 5 cans of coke there are 3 cans of sprite. if there are 25 cans of coke, how many cans

belka [17]

Answer: 15 cans of sprite

Step-by-step explanation:

5 cans of coke and 3 cans of sprite

then its

25 cans of coke and 15 cans of sprite

8 0

2 years ago

Caroline knows the height and the required volume of a cone-shaped vase she’s designing. Which formula can she use to determine

timama [110]

Answer:

Option B. $r=\sqrt{\frac{3V}{\pi h}}$

Step-by-step explanation:

we know that

The volume of a cone is equal to

$V=\frac{1}{3}\pi r^{2} h$

Solve for the radius r

That means-----> isolate the variable r

Multiply by 3 both sides

$3V=\pi r^{2} h$

Divide by $(\pi h)$ both sides

$r^{2}=\frac{3V}{\pi h}$

square root both sides

$r=\sqrt{\frac{3V}{\pi h}}$

4 0

3 years ago

I need help ASAP!!!!!!

Fynjy0 [20]

Answer:

from what I see the...function 2

5 0

3 years ago

2.The number of grams A of a certain radioactive substance present at time, in yearsfrom the present, t is given by the formulaA

professor190 [17]

Answer:

Given that,

The number of grams A of a certain radioactive substance present at time, in years

from the present, t is given by the formula

$A=45e^{-0.0045(t)}$

a) To find the initial amount of this substance

At t=0, we get

$A=45e^{-0.0045(0)}$ $A=45e^0$

We know that e^0=1 ( anything to the power zero is 1)

we get,

$A=45$

The initial amount of the substance is 45 grams

b)To find thehalf-life of this substance

To find t when the substance becames half the amount.

A=45/2

Substitute this we get,

$\frac{45}{2}=45e^{-0.0045(t)}$

$\frac{1}{2}=e^{-0.0045(t)}$

Taking natural logarithm on both sides we get,

$\ln (\frac{1}{2})=-0.0045(t)^{}$ $(-1)\ln (\frac{1}{2})=0.0045(t)$ $\ln (\frac{1}{2})^{-1}=0.0045(t)$ $\ln (2)=0.0045(t)$ $0.6931=0.0045(t)$ $t=\frac{0.6931}{0.0045}$ $t=154.02$

Half-life of this substance is 154.02

c) To find the amount of substance will be present around in 2500 years

Put t=2500

we get,

$A=45e^{-0.0045(2500)}$ $A=45e^{-11.25}$ $A=45\times0.000013=0.000585$ $A=0.000585$

The amount of substance will be present around in 2500 years is 0.000585 grams

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6 months ago

The points (-1,1), (0,0), and (1,1) translate the points __ of g(x).

wariber [46]

The answer would be -1,2

8 0

3 years ago