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1 year ago

A gardener uses 1/3 of a liter of water to water 2/7 of a garden.

Mathematics

1 answer:

Zielflug [23.3K]1 year ago

4 0

1 1 / 6 litres of water is needed to watering the entire garden at the given rate.

<h3>How to find the amount of water using rate?</h3>

We need to find how many litres of water is needed to watering the entire garden at this rate.

Let the total garden be x parts then,

Therefore,

2 / 7 x = 1 / 3

cross multiply

2x × 3 = 7

6x = 7

divide both sides by 6

6x / 6 = 7 / 6

x = 1 1 / 6

Therefore, Thus, 1 1 / 6 litres of water is needed to watering the entire garden at the given rate.

learn more on rate here: brainly.com/question/3920797

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X^2 -50x+456=0 solve this algebraiclly find x

bagirrra123 [75]

Answer:

$x=-406$

Step-by-step explanation:

$x^2-50x+456=0$

$x^2-50x=-456$

This is because we subtract 456 from each side.

$x-50=-456$

We then divide the whole equation by x .

$x=-406$

Adding 50 to both sides gives us x by itself, hope this helps :D

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2 years ago

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The functions q and r are defined as follows. HELP PLEASE

dusya [7]

Answer:

-22

Step-by-step explanation:

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3 years ago

find the area of the trapezium whose parallel sides are 25 cm and 13 cm The Other sides of a Trapezium are 15 cm and 15 CM

Snezhnost [94]

$\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}$

Given - A trapezium ABCD with non parallel sides of measure 15 cm each ! along , the parallel sides are of measure 13 cm and 25 cm

To find - Area of trapezium

Refer the figure attached ~

In the given figure ,

AB = 25 cm

BC = AD = 15 cm

CD = 13 cm

Construction -

$draw \: CE \: \parallel \: AD \: \\ and \: CD \: \perp \: AE$

Now , we can clearly see that AECD is a parallelogram !

$\therefore$ AE = CD = 13 cm

Now ,

$AB = AE + BE \\\implies \: BE =AB - AE \\ \implies \: BE = 25 - 13 \\ \implies \: BE = 12 \: cm$

Now , In ∆ BCE ,

$semi \: perimeter \: (s) = \frac{15 + 15 + 12}{2} \\ \\ \implies \: s = \frac{42}{2} = 21 \: cm$

Now , by Heron's formula

$area \: of \: \triangle \: BCE = \sqrt{s(s - a)(s - b)(s - c)} \\ \implies \sqrt{21(21 - 15)(21 - 15)(21 - 12)} \\ \implies \: 21 \times 6 \times 6 \times 9 \\ \implies \: 12 \sqrt{21} \: cm {}^{2}$

Also ,

$area \: of \: \triangle \: = \frac{1}{2} \times base \times height \\ \\\implies 18 \sqrt{21} = \: \frac{1}{\cancel2} \times \cancel12 \times height \\ \\ \implies \: 18 \sqrt{21} = 6 \times height \\ \\ \implies \: height = \frac{\cancel{18} \sqrt{21} }{ \cancel 6} \\ \\ \implies \: height = 3 \sqrt{21} \: cm {}^{2}$

Since we've obtained the height now , we can easily find out the area of trapezium !

$Area \: of \: trapezium = \frac{1}{2} \times(sum \: of \:parallel \: sides) \times height \\ \\ \implies \: \frac{1}{2} \times (25 + 13) \times 3 \sqrt{21} \\ \\ \implies \: \frac{1}{\cancel2} \times \cancel{38 }\times 3 \sqrt{21} \\ \\ \implies \: 19 \times 3 \sqrt{21} \: cm {}^{2} \\ \\ \implies \: 57 \sqrt{21} \: cm {}^{2}$