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3 years ago

AWARDING BRAINLIEST THIS IS FROM RSM PLEASE HELP

Mathematics

1 answer:

bogdanovich [222]3 years ago

4 0

Hi I'm from RSM and I got this question right. The answer is KL=MN=32 units.

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A small office purchases bottled drinking water. The water comes in containers that hold gallons. Two-thirds of the office conta

Zarrin [17]

Answer:

213.333 ounces

Step-by-step explanation:

A small office purchases bottled drinking water. The water comes in containers that hold 5 gallons. Two-thirds of the office container was drained last week to put out a fire in the wastebasket.

a. How much water (in ounces) is left in the container?

b. How many 12-ounce glasses can be filled with the water that remains in the container

Container = 5 gallons

Two-thirds was drained last week to put out a fire in the wastebasket

That is 2/3 of 5

= 2/3 × 5

= 10 / 3 gallons

Water left in the container = 5 - 10/3

= 15-10/3

= 5/3 gallons

Convert 5/3 gallons to ounces

5/3 gallons = 213.333 ounces

B. Number of 12 ounces glass that can be filled from the water left in the container = 213.333 ounces / 12

= 17.77775

Approximately 18

5 0

3 years ago

Pure acid is to be added to a 10% acid solution to obtain 117 L of a 20% acid solution. What amounts of each should be used?

creativ13 [48]

Answer:

A typical mixture equation

x + .10(54-x) = .20(54)

x + 5.4 - .10x = 10.8

x - .1x = 10.8 - 5.4

.9x = 5.4

x = 5.4%2F.9

x = 6 L of pure acid to be combined with: 54-6 = 48 L of 10% solution

Step-by-step explanation:

does this help?

8 0

3 years ago

Andrew is flying a kite at a height of 120ft. While the wind is carrying the kite horizontally away from her at the rate of 24ft

Paha777 [63]

Answer:

The string of the kite is being let out at the rate of 19.2 ft/sec.

Step-by-step explanation:

See the diagram attached.

The vertical height (BK) of the kite is 120 ft.

The kite string is of length (AK) of 200 ft.

Therefore, the horizontal distance (AB) from Andrew to kite is = $\sqrt{200^{2} - 120^{2}} = 160$ ft.

Now, applying the Pythagoras Theorem,

AB² + BK² = AK²

⇒ l² + h² = s² {Where, l is the horizontal length, h is the height and s is the string length}

Differentiating with respect to time t both sides

$2l\frac{dl}{dt} + 0 = 2s\frac{ds}{dt}$ {Since, height of kite is constant}

⇒ $2 \times 160 \times 24 = 2 \times 200 \times \frac{ds}{dt}$

⇒ $\frac{ds}{dt} = 19.2$ ft per sec.

Therefore, when the wind is carrying the kite horizontally at the rate of 24 ft/sec, then the string of the kite is being let out at the rate of 19.2 ft/sec. (Answer)