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2 years ago

HELP PLEASE!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

UkoKoshka [18]2 years ago

8 0

Answer:

Slope= 0 , X= 3

Step-by-step explanation:

the slope goes up 1 over 0.

I hope this helps

Have a good day.

GalinKa [24]2 years ago

7 0

Answer:

Step-by-step explanation:

on x line it = 3 and there is no slope/no steep hill yk

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Simplify: -3152 – 10(4.32 + 5 = 5)]

zheka24 [161]

Answer:-523/52

Step-by-step explanation:

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3 years ago

Which of the following is equivalent to 1 cup

kipiarov [429]

1 cup=16 tablespoons

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3 years ago

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How can you analyze the numerator and/or the denominator in a rational expression to locate the x-intercepts?

andre [41]

Answer:

$x=0$ for $f(x)=\frac{P(x)}{Q(x)}$ , with $Q(x) \neq 0$ .

Step-by-step explanation:

To obtain x-intercepts, we need to give certain value, specifically $x=0$ .

However, in case of a rational function, we must make sure that $x=0$ doesn't makes the denominator null, beacuse in that case the function is undefined.

Therefore, for x-intercepts we must say

$x=0$ for $f(x)=\frac{P(x)}{Q(x)}$ , with $Q(x) \neq 0$ .

7 0

3 years ago

There were 240 students at a primary school. After some new students were transferred into the school, there were 282 students i

stepan [7]

Answer:

Numbers of students at a primary school before= 240

Number of increased students =42

Again,

Increased number of students in percent(%)= 42/240×100%= 17.5%

Therefore 17.5% students are increased this year.

3 0

3 years ago

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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

HELP PLEASE!!!!!!!!!!!!!!!!!​

HELP PLEASE!!!!!!!!!!!!!!!!!