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2 years ago

I mark as brainliest

Mathematics

1 answer:

Eddi Din [679]2 years ago

8 0

Answer:

the third one

Step-by-step explanation:

if im wrong you don't have to mark as braniest

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A machine at a bottling company fills water bottles. the number of bottles filled is proportional to the amount of time the mach

Whitepunk [10]

In the table shown below:

Let N be the number of bottles filled,

Let T be the time in hours.

Given that the number of bottles filled is proportional to the amount of time the machine runs, we have

$\begin{gathered} N\propto T \\ \text{Introducing a proportionality constant, we have } \\ N\text{ = kT} \\ \Rightarrow k\text{ = }\frac{N}{T} \\ \end{gathered}$

Let's evaluate the value of k for each day.

Thus, on monday,

$\begin{gathered} N\text{ = 9900} \\ T\text{ = 5.5} \\ \text{thus,} \\ k\text{ = }\frac{9900}{5.5} \\ \Rightarrow k\text{ = 1800} \end{gathered}$

Tuesday:

$\begin{gathered} N\text{ = }11160 \\ T\text{ = }6.2 \\ \text{thus,} \\ k\text{ = }\frac{11160}{6.2} \\ \Rightarrow k\text{ = 1800} \end{gathered}$

Wednesday:

$\begin{gathered} N\text{ = }12330 \\ T\text{ = }6.25 \\ \text{thus,} \\ k\text{ = }\frac{12330}{6.25} \\ \Rightarrow k\text{ = 1972.8} \end{gathered}$

Thursday:

$\begin{gathered} N\text{ = }10440 \\ T\text{ = }5.80 \\ \text{thus,} \\ k\text{ = }\frac{10440}{5.8} \\ \Rightarrow k=1800 \end{gathered}$

It is observed that all exept wednesday have the same value of k.

Thus, the amount of time required for the number of bottles filled on wednesday is evaluated as

$\begin{gathered} N\text{ = }12330 \\ k\text{ = 1800} \\ T\text{ = ?} \\ \text{but} \\ k\text{ = }\frac{N}{T} \\ 1800\text{ = }\frac{12330}{T} \\ \Rightarrow T\text{ = }\frac{\text{12330}}{1800} \\ T\text{ = 6.85} \end{gathered}$

Hence, the incorrect day is Wednesday. The amount of time for that many bottles should be 6.85 hours.

3 0

11 months ago

Solve the equation: x3−20=−15

stich3 [128]

Answer:

x = 1.71

Step-by-step explanation:

$x^3-20=-15\\x^3=-15+20\\x^3=5\\\sqrt[3]{x^3} =\sqrt[3]{5} \\x=1.71$

7 0

2 years ago

Read 2 more answers

Find the gradient of the line passing through (6,8) and (4,10)

Mariulka [41]

Question:

Find the gradient of the line passing through (6,8) and (4,10).

Answer:

-1 is the right answer.

Step-by-step explanation:

Slope of the line = The gradient of the line

Gradient of the line is known as change in the value of y-axis by change in the value of x-axis

Gradient = ∆y\∆x

$Gradient = \frac{y_ 2 - y_1}{x_2 - x_1} = \frac{(10 - 8)}{(4 - 6)} = \frac{(2)}{(- 2)} =\boxed{- 1}\\$

7 0

2 years ago

You have a gross income of $117,151 and are filing your tax return singly. You claim one exemption and can take a deduction of $

Grace [21]

Answer is $10,825 did that test too (C)

3 0

3 years ago

Read 2 more answers

1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.

xxTIMURxx [149]

First, rewrite the equation so that y is a function of x :

$x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}$

(If you were to plot the actual curve, you would have both $y=x^{2/3}$ and $y=-x^{2/3}$ , but one curve is a reflection of the other, so the arc length for 1 ≤ x ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

$\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx$

We have

$y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}$

Then in the integral,

$\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx$

Substitute

$u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx$

This transforms the integral to

$\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du$

and computing it is trivial:

$\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)$

We can simplify this further to

$\displaystyle \frac8{27} \left(10\sqrt{10} - \frac{13\sqrt{13}}8\right) = \boxed{\frac{80\sqrt{10}-13\sqrt{13}}{27}}$

7 0

2 years ago

I mark as brainliest ​

I mark as brainliest