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4 years ago

6.1 m A 85.4 sq m ⓅⓁⓈ ⒽⒺⓁⓅ

Mathematics

1 answer:

Mrac [35]4 years ago

8 0

The teacher probably wants you to multiply cause area refers to that

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Which of the following statements provides the correct freezing and boiling points of water on the Celsius and Fahrenheit temper

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Answer: i need a pic

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3 years ago

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OverLord2011 [107]

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3 years ago

How would the number of times edgardo uses a container to fill the 10 gallon cooler change if he uses a one cup container? Expla

olya-2409 [2.1K]

Answer:

Edgardo would use the container 160 times to fill.

Step-by-step explanation:

Given:

Edgardo uses a container to fill the 10 gallon cooler.

If he uses a one cup container.

Now, to find the times it would take to fill the 10 gallon container.

As, the capacity of container = 1 cup.

And, the capacity of cooler = 10 gallon.

So, we use conversion factor to find the times the container to fill the cooler:

1 gallon = 16 cups.

10 gallon = $10\times 16\ cups.$

10 gallon = 160 cups.

Thus, to fill the 10 gallon cooler 160 cups of container needed.

And, the container is of 1 cup.

So, to get the times he would use the container we divide 160 by 1:

$160\div 1=160.$

Hence, 160 times Edgardo would use the 1 cup container to fill the 10 gallon cooler.

Therefore, Edgardo would use the container 160 times to fill.

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3 years ago

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Serga [27]

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3 years ago

While traveling to and from a certain destination, you realized increasing your speed by 40 mph saved 4 hours on your return. If

Luden [163]

Answer:

The speed driven while returning is 88 mph.

Step-by-step explanation:

We are given that while traveling to and from a certain destination, you realized increasing your speed by 40 mph saved 4 hours on your return.

Also, the total distance of the roundtrip was 420 miles.

Let the speed driven while returning be 'x mph' which means that the speed driven while going was '(x - 40) mph' because it has been given that while returning we have increased the speed by 40 mph.

As we know that the Speed-Distance-Time formula is given by;

$\text {Speed} = \frac{\text{Distance}}{\text{Time}}$ or $\text {Time} = \frac{\text{Distance}}{\text{Speed}}$

So, according to the question;

$\frac{420}{x-40} -\frac{420}{x} = 4 \text{ hours}$ where Distance = 420 miles

$\frac{420x-420(x-40)}{x(x-40)} = 4$

$\frac{420x-420x+16800}{x^{2} -40x} = 4$

$\frac{16800}{x^{2} -40x} = 4$

$4x^{2} -160x= 16800$

$4x^{2} -160x- 16800=0$

$x^{2} -40x- 4200=0$

Now finding the roots of the above equation;

Here a = 1, b = -40 and c = -4200

$D = b^{2} -4ac$

= $(-40)^{2} -4(1)(-4200)$ = 18400

Now, the roots of a quadratic equation is given by;

$x = \frac{-b\pm \sqrt{D} }{2a}$

$x = \frac{-(-40)\pm \sqrt{18400} }{2\times 1}$

So, the two roots of x are : $x = \frac{40-\sqrt{18400} }{2}$ and $x = \frac{40+\sqrt{18400} }{2}$

Solving these two we get; $x = -47.8$ and $x = 87.8$

Here we ignore the negative value of x, so the speed driven while returning is 87.8 ≈ 88 mph.

8 0

3 years ago