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3 years ago

A 13.5-gasoline tank is 4 / 5 full how many gallons will it take to fill the tank

Mathematics

1 answer:

Sedbober [7]3 years ago

8 0

2.7 gallons to fill the tank

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Can Someone help me plz????

bearhunter [10]

Answer:

4. C

5. 7

6. 3

Step-by-step explanation:

Four:

You have to multiply both of them. Just easy math!

Five:

This one is more complicated. You can see that it is asking you to plug in for a,

F(3):

$f(3)= 4(3)-5$

$f(3)=12-5$

$f(3)=7$

F(1.25):

$f(1.25)=4(1.25)-5$

$f(1.25)=5-5$ $f(1.25)=0$

So now just subtract the answers. 7-0 equals 7.

Six:

Just look at the graph and see what number corresponds with y(6), or the 6 on the y axis. It appears to be three.

Have a nice day! Hope this helps. Don't forget to mark brainliest!

4 0

3 years ago

Read 2 more answers

A baker is making bread dough. He uses 3 cups of flour for every 8 ounces of water. How many cups of flour will he use if he use

Alekssandra [29.7K]

Answer:

Step-by-step explanation:

This is the fraction we use to solve this problem $\frac{Ounces}{Flour}$

Fractions are $\frac{8}{3}$ and $\frac{96}{x}$

We now ask ourselves

$8 * x = 96\\x = 12$

Since whatever we do to the numerator we have to to the same to the denominator, we multiply 3 and 12

$3 * 12 = x\\x = 36$

Hope this helps!

PLZZZ give brainliest

4 0

3 years ago

<img src="https://tex.z-dn.net/?f=538%20%5Ctimes%204%20%5C%5C%20%20%5C%5C%20214%20%5Ctimes%208%20%5C%5C%20%20%5C%5C%203721%20%5C

Shalnov [3]

Answer:

Step-by-step explanation:

multiply $(538)(4)=2152\\(214)(8)=1712\\(3721)(7)=26047\\(7956)(8)=63648$

7 0

3 years ago

Help me please. This is my last question. :( •Msm •Mhanifa •Nayefex

UkoKoshka [18]

Answer:

shape of bases is rectangle.
name of solid figure is cuboid.
perimeter of base =2(1+2)=6m
the height of prism =4m
lateral surface area =perimeter of base × height of the prism.
the area of base=l*b=2*1=2 m²
For any prism surface area=lateral area + 2 area of base
the surface area=6*4+2*2=28m²

4 0

2 years ago

Read 2 more answers

The concentration C of certain drug in a patient's bloodstream t hours after injection is given by

frozen [14]

Answer:

a) The horizontal asymptote of $C(t)$ is $c = 0$ .

b) When t increases, both the numerator and denominator increases, but given that the grade of the polynomial of the denominator is greater than the grade of the polynomial of the numerator, then the concentration of the drug converges to zero when time diverges to the infinity. There is a monotonous decrease behavior.

c) The time at which the concentration is highest is approximately 1.291 hours after injection.

Step-by-step explanation:

a) The horizontal asymptote of $C(t)$ is the horizontal line, to which the function converges when $t$ diverges to the infinity. That is:

$c = \lim _{t\to +\infty} \frac{t}{3\cdot t^{2}+5}$ (1)

$c = \lim_{t\to +\infty}\left(\frac{t}{3\cdot t^{2}+5} \right)\cdot \left(\frac{t^{2}}{t^{2}} \right)$

$c = \lim_{t\to +\infty}\frac{\frac{t}{t^{2}} }{\frac{3\cdot t^{2}+5}{t^{2}} }$

$c = \lim_{t\to +\infty} \frac{\frac{1}{t} }{3+\frac{5}{t^{2}} }$

$c = \frac{\lim_{t\to +\infty}\frac{1}{t} }{\lim_{t\to +\infty}3+\lim_{t\to +\infty}\frac{5}{t^{2}} }$

$c = \frac{0}{3+0}$

$c = 0$

The horizontal asymptote of $C(t)$ is $c = 0$ .

c) From Calculus we understand that maximum concentration can be found by means of the First and Second Derivative Tests.

First Derivative Test

The first derivative of the function is:

$C'(t) = \frac{(3\cdot t^{2}+5)-t\cdot (6\cdot t)}{(3\cdot t^{2}+5)^{2}}$

$C'(t) = \frac{1}{3\cdot t^{2}+5}-\frac{6\cdot t^{2}}{(3\cdot t^{2}+5)^{2}}$

$C'(t) = \frac{1}{3\cdot t^{2}+5}\cdot \left(1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} \right)$

Now we equalize the expression to zero:

$\frac{1}{3\cdot t^{2}+5}\cdot \left(1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} \right) = 0$

$1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} = 0$

$\frac{3\cdot t^{2}+5-6\cdot t^{2}}{3\cdot t^{2}+5} = 0$

$5-3\cdot t^{2} = 0$

$t = \sqrt{\frac{5}{3} }\,h$

$t \approx 1.291\,h$

The critical point occurs approximately at 1.291 hours after injection.

Second Derivative Test

The second derivative of the function is:

$C''(t) = -\frac{6\cdot t}{(3\cdot t^{2}+5)^{2}}-\frac{(12\cdot t)\cdot (3\cdot t^{2}+5)^{2}-2\cdot (3\cdot t^{2}+5)\cdot (6\cdot t)\cdot (6\cdot t^{2})}{(3\cdot t^{2}+5)^{4}}$

$C''(t) = -\frac{6\cdot t}{(3\cdot t^{2}+5)^{2}}- \frac{12\cdot t}{(3\cdot t^{2}+5)^{2}}+\frac{72\cdot t^{3}}{(3\cdot t^{2}+5)^{3}}$

$C''(t) = -\frac{18\cdot t}{(3\cdot t^{2}+5)^{2}}+\frac{72\cdot t^{3}}{(3\cdot t^{2}+5)^{3}}$

If we know that $t \approx 1.291\,h$ , then the value of the second derivative is:

$C''(1.291\,h) = -0.077$

Which means that the critical point is an absolute maximum.

The time at which the concentration is highest is approximately 1.291 hours after injection.

5 0

2 years ago