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2 years ago

Please help ASAP please help

Mathematics

1 answer:

finlep [7]2 years ago

3 0

Answer:

(4.5, 6)

Step-by-step explanation:

The midpoint is the average of the end points:

((8, 2) +(1, 10))/2 = (8+1, 2+10)/2 = (9, 12)/2 = (4.5, 6)

The midpoint is (4.5, 6).

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Quently As

Anna71 [15]

Answer:

A) 18.3

Step-by-step explanation:

Since adding b and a gives 19.

c must be less than 19.

So, A) 18.3 is the correct answer

5 0

2 years ago

Read 2 more answers

PLEASE HELP IMAGE IS BELOW!!

guajiro [1.7K]

Answer:

Number of small boxes shipped = 7

Number of large boxes shipped = 15

Step-by-step explanation:

Let the number of small boxes = s

And the number of large boxes = l

Weight of small box = 25 pound

Weight of the large box = 50 pounds

Total weight of the shipment = 925 pounds

Therefore, equation for the weight of shipment will be,

25s + 50l = 925

s + 2l = 37 ----- (1)

Total number of boxes shipped = 22 boxes

Therefore, equation will be,

s + l = 22 ------(2)

Subtract equation (2) from equation (1)

(s + 2l) - (s + l) = 37 -22

l = 15

From equation (2)

s + 15 = 22

s = 7

Therefore, number of small boxes shipped = 7

Number of large boxes shipped = 15

6 0

2 years ago

Can someone help me on both questions

AURORKA [14]

20. is 71 hope this helos

5 0

3 years ago

Read 2 more answers

. If y=3x + 2, find the value of y <br> when x = -3

emmainna [20.7K]

Y=3x-3+2
y=-9+2
y= -7

Hope that helped

5 0

3 years ago

Read 2 more answers

Suppose a certain population satisfies the logistic equation given by dP

Ksenya-84 [330]

Answer:

The population when t = 3 is 10.

Step-by-step explanation:

Suppose a certain population satisfies the logistic equation given by

$\frac{dP}{dt}=10P-P^2$

with P(0)=1. We need to find the population when t=3.

Using variable separable method we get

$\frac{dP}{10P-P^2}=dt$

Integrate both sides.

$\int \frac{dP}{10P-P^2}=\int dt$ .... (1)

Using partial fraction

$\frac{1}{P(10-P)}=\frac{A}{P}+\frac{B}{(10-P)}$

$A=\frac{1}{10},B=\frac{1}{10}$

Using these values the equation (1) can be written as

$\int (\frac{1}{10P}+\frac{1}{10(10-P)})dP=\int dt$

$\int \frac{dP}{10P}+\int \frac{dP}{10(10-P)}=\int dt$

On simplification we get

$\frac{1}{10}\ln P-\frac{1}{10}\ln (10-P)=t+C$

$\frac{1}{10}(\ln \frac{P}{10-P})=t+C$

We have P(0)=1

Substitute t=0 and P=1 in above equation.

$\frac{1}{10}(\ln \frac{1}{10-1})=0+C$

$\frac{1}{10}(\ln \frac{1}{9})=C$

The required equation is

$\frac{1}{10}(\ln \frac{P}{10-P})=t+\frac{1}{10}(\ln \frac{1}{9})$

Multiply both sides by 10.

$\ln \frac{P}{10-P}=10t+\ln \frac{1}{9}$

$e^{\ln \frac{P}{10-P}}=e^{10t+\ln \frac{1}{9}}$

$\frac{P}{10-P}=\frac{1}{9}e^{10t}$

Reciprocal it

$\dfrac{10-P}{P}=9e^{-10t}$

$P(t)=\dfrac{10}{1+9e^{-10t}}$

The population when t = 3 is

$P(3)=\dfrac{10}{1+9e^{-10\cdot 3}}$

Using calculator,

$P=9.999\approx 10$

Therefore, the population when t = 3 is 10.

8 0

2 years ago