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3 years ago

I can’t get this answer right

Mathematics

2 answers:

statuscvo [17]3 years ago

8 0

Answer:

Step-by-step explanation:

I believe this is correct, if not feel free to let me know and I will fix it. I'm sorry in advance if it is incorrect.

scZoUnD [109]3 years ago

7 0

I think it’s D

-xoxo<3

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Please help me question attached below:)))

atroni [7]

Answer:

D(5, 4), E(14, 7), M(9.5, 5.5)

Step-by-step explanation:

As AD = 1/4AB and DE ║ AC, the ratio CE/CB = 1/4, or CE = 1/4CB

Find the coordinates of D:

x = 1 + 1/4(17 - 1) = 1 + 4 = 5
y = 5 + 1/4(1 - 5) = 5 - 1 = 4

Find the coordinates of E:

x = 13 + 1/4(17 - 13) = 13 + 1 = 14
y = 9 + 1/4(1 - 9) = 9 - 2 = 7

Find the coordinates of the midpoint M of DE:

x = (5 + 14)/2 = 19/2 = 9.5
y = (4 + 7)/2 = 11/2 = 5.5

8 0

2 years ago

In a game,the probability that a spinner will land on a 3 is 1/5.How many times would you expect to land on a 3 if you spin the

schepotkina [342]

Answer:

3 times

Step-by-step explanation:

15 * 1/5 = 15/5

3 0

3 years ago

13x+15=11x-25 5t-5=5t+7 Please show steps for both Thank you

Tanzania [10]

I hope this helps you

13x-11x=-25-15

2x=-40

x=-20

5t-5t=7+5

0=12 false

3 0

3 years ago

Please help me is this statement true or false 3(36 - 11) = 45?

viva [34]

Hi there! The answer is FALSE

To find the answer we need to work out the parenthesis.
$3(36 - 11) =$
$3 \times 36 - 3 \times 11 =$
$108 - 33 =$
$75$

75 does not equal 45 and therefore the statement is false.

3 0

3 years ago

A tank initially has 300 gallons of a solution that contains 50 lb. of dissolved salt. A brine solution with a concentration of

belka [17]

Let s(t) be the amount of salt in the tank at time t. Then s(0) = 50 lb.

Salt flows into the tank at a rate of

(2 gal/min) (6 lb/gal) = 12 lb/min

and flows out at a rate of

(2 gal/min) (s(t)/300 lb/gal) = s(t)/150 lb/min

so that the net rate at which salt is exchanged through the tank is

ds(t)/dt = 12 - s(t)/150 … (lb/min)

Solve for s(t). The DE is separable, so we have

ds/dt = 12 - s/150

150 ds/dt = 1800 - s

150/(1800 - s) ds = dt

Integrate both sides to get

-150 ln|1800 - s| = t + C

Solve for s :

ln|1800 - s| = -t/150 + C

1800 - s = exp(-t/150 + C )

1800 - s = C exp(-t/150)

s = 1800 - C exp(-t/150)

Now given that s(0) = 50, we solve for C :

50 = 1800 - C exp(-0/150)

50 = 1800 - C

C = 1750

Then the amount of salt in the tank at any time t ≥ 0 is

s(t) = 1800 - 1750 exp(-t/150)

To find the time it takes for the tank to hold 100 lb of salt, solve for t in

100 = 1800 - 1750 exp(-t/150)

1700 = 1750 exp(-t/150)

34/35 = exp(-t/150)

ln(34/35) = -t/150

t = -150 ln(34/35) ≈ 4.348

So it would take approximately 4.348 minutes.

By the way, we didn't have to solve for s(t), we could have instead stopped with

-150 ln|1800 - s| = t + C

Solve for C - this C is not the same as the one we found using the other method. s(0) = 50, so

-150 ln|1800 - 50| = 0 + C

C = -150 ln|1750|

==> t = 150 ln(1750) - 150 ln|1800 - s|

Then s(t) = 100 lb when

t = 150 ln(1750) - 150 ln(1700)

t = 150 ln(1750/1700)

t = 150 ln(35/34)

t ≈ 4.348

5 0

3 years ago