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2 years ago

What is 76.684 rounded to the nearest hundredth

2 answers:

5 0

76.684

76.68

The first place after the decimal point is the tenths, the second place after the decimal point is the hundredths and the third place after the decimal point is the thousandths. In order to round a number to the nearest hundredth, you must look at the number in the thousandths place. If that number is greater than or equal to five then the number initially in the tenths place will increase by one. If the number is less than 5 than the number in the tenths place stays the same.

Marianna [84]2 years ago

4 0

It is 76.68 this was so easy lol

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Mrs. Kato has 6 bags of dried beans and a 2-pound bag of rice in one shopping bag. In another shopping bag she has a 5-pound bag

Salsk061 [2.6K]

Answer: 3 pounds

Step-by-step explanation:

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2 years ago

A rectangle is 4 cm longer than it is wide, and its area is 117 cm2 . find its dimensions.

Brilliant_brown [7]

The area of a rectangle is equal L x W

4 cm longer than it is wide L = 4 + W

L x W = 117 we replace L here

(4 + W ) x W = 117

4W + W ^2 = 117

4W + W ^2 -117 = 0
W ^2 +4 W -117 = 0

W² + 4W - 117 = 0


THEN u want to use the use the quadratic formula

OR Factoring gives us

(W + 13)(W - 9) = 0

W = -13 or 9

But it can't be negative, so

W = 9 and L= 9+4 = 13

3 0

2 years ago

Read 2 more answers

A child wanders slowly down a circular staircase from the top of a tower. With x,y,zx,y,z in feet and the origin at the base of

babymother [125]

Answer:

a) The tower is 90 feet tall

b) She reaches the bottom at t = 18 minutes.

c) Her speed at time t is 5 \sqrt[]{5} ft/minute

d) Her acceleration at time t is 10 ft/minute^2

Step-by-step explanation:

Consider the path described by the child as going down the tower to have the following parametrization $\gamma(t) = (10\cos t, 10 \sin t, 90-5t)$

a) Assuming that the child is at the top of the tower when she starts going down, we have that at the initial time (t=0) we will have the value of the height of the tower. That is z = 90-5*0 = 90 ft.

b) The child reaches the bottom as soon as z =0. We want to find the value of t that does that. Then we have 0 = 90-5t, which gives us t = 18 minutes.

c) Given the parametrization we are given, the velocity of the child at time t is given by $\frac{d\gamma}{dt}= (\frac{d}{dt}(10\cos t), \frac{d}{dt} (10 \sin t ), \frac{d}{dt}(90-5t)) = (-10 \sin t, 10 \cos t, -5)$ . The speed is defined as the norm of the velocity vector,

so, the speed at time t is given by $v = \sqrt[]{(-10 \sin t)^2+(10 \cos t)^2+(-5)^2} = \sqrt[]{100(\sin^2 t + \cos^2 t)+25} = \sqrt[]{125}= 5 \sqrt[]{5}$

d) ON the same fashion we want to know the norm of the second derivative of $\gamma$ .

We have that $\gamma ^{''}(t) =(-10\cost t, -10 \sin t , 0)$ so the acceleration is given by $\sqrt[]{100(\cos^2 t+ \sin^2 t )} = 10$

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2 years ago

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Answer: 0.73^9

Step-by-step explanation: Khan

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2 years ago