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3 years ago

5

Mrs. Rivera has bars of chocolate in a box. The total mass of bars of chocolate in the box is 3 kilograms. What is the mass in g

rams

1 answer:

sergeinik [125]3 years ago

6 0

Answer:The chocolate in the box is 3000 grams.

Every Kilogram has 1000 grams

You multiply 1000×3=3000

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Claire is putting up a fence around her rectangular garden. The length is four less than double the width. The perimeter is 22 f

Doss [256]

Answer:

The dimensions are 6 by 5 feet.

Length = 6 feet.

Width = 5 feet.

Step-by-step explanation:

Let the length = L

Let the width = W

Perimeter of a rectangle = 2L + 2W

Translating the word problem into an algebraic equation, we have;

L = 2W - 4 ........equation 1

22 = 2L + 2W .......equation 2

Substituting the value of "L" into equation 2, we have;

22 = 2(2W - 4) + 2W

22 = 4W - 8 + 2W

22 + 8 = 6W

30 = 6W

W = 30/6

Width, W = 5 feet.

To find the length, L

Substituting the value of "W" into equation 1, we have;

L = 2W - 4

L = 2(5) - 4

L = 10-4

Length, L =6 feet

Therefore, the dimensions of the garden are 6 by 5 feet.

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2 years ago

A rope is at least 21 feet long. You want to cut it into 3 pieces. The second piece is to be twice as long as the first piece,an

Yuri [45]

Answer:

the first piece of rope would be 4 feet

Step-by-step explanation:

4 + 8 + 9 = 21

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3 years ago

A car travels at a constant speed 65 mph how far would it travel in 4 hours

Alina [70]

It would travel 65 times 4 = 260 miles

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3 years ago

After a snowstorm in the town of Golden Glen, the morning temperature was – 10°F. But by the afternoon, the temperature had rise

spayn [35]

Answer:

7°F

Step-by-step explanation:

We have the following information from the question;

Morning temperature = -10°F

Increase in temperate = 17°F

Afternoon temperature = the unknown

Since;

Increase in temperature = Afternoon temperature - morning temperature

Then;

Afternoon temperature = increase in temperature + morning temperature

Afternoon temperature = 17°F + (-10°F)

Afternoon temperature= 7°F

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3 years ago

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an

Lapatulllka [165]

Answer:

Step-by-step explanation:

Given that there is a function of x,

$f(x) = 2sin x + 2cos x,0\leq x\leq 2\pi$

Let us find first and second derivative for f(x)

$f'(x) = 2cosx -2sinx\\f"(x) = -2sinx-2cosx$

When f'(x) =0 we have tanx = 1 and hence

a) f'(x) >0 for I and III quadrant

Hence increasing in $(0, \pi/2) U(\pi,3\pi/2)\\$

and decreasing in $(\pi/2, \pi)U(3\pi/2,2\pi)$

$x=\frac{\pi}{4}, \frac{3\pi}{4}$

$f"(\pi/4)$

Hence f has a maxima at x = pi/4 and minima at x = 3pi/4

b) Maximum value = $2sin \pi/4+2cos \pi/4 =2\sqrt{2}$

Minimum value = $2sin 3\pi/4+2cos 3\pi/4 =-2\sqrt{2}$

c)

f"(x) =0 gives tanx =-1

$x= 3\pi/4, 7\pi/4$

are points of inflection.

concave up in (3pi/4,7pi/4)

and concave down in (0,3pi/4)U(7pi/4,2pi)

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3 years ago