Approximate each irrational number to two decimal places without a calculator?

3.5,-2,0,-7/3,-6 in order from greatest to least

icang [17]

Answer:

in order from greatest to least:

3.5 0 -2 -7/3 -6

Step-by-step explanation:

5 0

3 years ago

Mrs. Carlyle bought a bag of nuts for her children. Phillip, Joy, Brent, and Preston took some from the bag.Phillip took 1/3. Jo

BARSIC [14]

Answer:

Originally there was 324 nuts in the bag.

Phillip took 108, Joy took 54, Brent took 81 and Preston took 10.

Step-by-step explanation:

Let's call the total amount of nuts N, and the number of nuts each child took by the initial of their name (Phillip will be P1 and Preston will be P2).

So we can write the following equations:

P1 = N/3

After removing N/3, the remaining nuts is (N - N/3):

J = (N - N/3)/4 = (2N/3)/4 = N/6

After removing N/6 from (N - N/3), the remaining nuts is (N - N/3 - N/6):

B = (N - N/3 - N/6)/2 = (N/2)/2 = N/4

P2 = 10

In the final there were 71 nuts remaining, so we have that:

N - N/3 - N/6 - N/4 - 10 = 71

N - N/3 - N/6 - N/4 = 81

N/4 = 81

N = 324 nuts

The amount of nuts took by each child is:

P1 = N/3 = 108 nuts

J = N/6 = 54 nuts

B = N/4 = 81 nuts

P2 = 10 nuts

8 0

3 years ago

Tom has 28 milk chocolates and emily has 24 dark chocolates they have to divide these chocolates into small packets that each ha

Nutka1998 [239]

28+24=52
So we make a sum of what's going to equal 52 in small packets.
I found 4x13=52. 13 would be the number of packets......
BUT!! Since it MUST be equal dark and milk and we have 28 and 26,
you would have to eliminate 1 pack because all 4 would be milk chocolate.
So that is only 12 packets.
4 per pack so that would be 2 dark and 2 milk per packet.

8 0

3 years ago

a zoo charges an admission price of $85 for the first 12 members of a group and $6 for each additional member of the group. What

elena-14-01-66 [18.8K]

The largest group that can go to the zoo is 23 members

Step-by-step explanation:

The given is:

A zoo charges an admission price of $85 for the first 12 members of a group and $6 for each additional member of the group
We need to find the largest group that can go to the zoo and spend less than $155 on admission

Assume that the number of members in the group is x

∵ There are x members in the group

∵ The zoo charges an admission price of $85 for the first 12

members of a group and $6 for each additional member

of the group

- Take 12 members from x and multiply the rest members by

6 and then add 85 to the product

∴ The group will spend = 6(x - 12) + 85

∵ The group wants to spend less than 155 to go to the zoo

- use the symbol < between the expression above and 155

∴ 6(x - 12) + 85 < 155

- Simplify the left hand side

∴ 6x - 72 + 85 < 155

- Add like terms in the left hand side

∴ 6x + 13 < 155

- Subtract 13 from both sides

∴ 6x < 142

- Divide both sides by 6

∴ x < 23.67

- Take the largest integer less than 23.67

∵ 23 is the largest integer less than 23.67

∴ The group has 23 members

The largest group that can go to the zoo is 23 members

Learn more:

You can learn more about the inequalities in brainly.com/question/1465430

#LearnwithBrainly

6 0

2 years ago

Find all values of c in the open interval (a, b) such that f'(c)=(f(b)-f(a))/(b-a)

timama [110]

<h3>Answer: c = 7/4</h3>

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Work Shown:

Compute the function value at the endpoints

$f(x) = \sqrt{4-x}\\\\f(-5) = \sqrt{4-(-5)} = 3\\\\f(4) = \sqrt{4-4} = 0\\\\$

With a = -5 and b = 4, we have

$f'(c) = \frac{f(b)-f(a)}{b-a}\\\\f'(c) = \frac{f(4)-f(-5)}{4-(-5)}\\\\f'(c) = \frac{0-3}{9}\\\\f'(c) = -\frac{1}{3}\\\\$

So,

$f(x) = \sqrt{4-x}\\\\f'(x) = -\frac{1}{2\sqrt{4-x}}\\\\f'(c) = -\frac{1}{3}\\\\-\frac{1}{2\sqrt{4-c}} = -\frac{1}{3}\\\\$

Use algebra to solve for c

$-\frac{1}{2\sqrt{4-c}} = -\frac{1}{3}\\\\\frac{1}{2\sqrt{4-c}} = \frac{1}{3}\\\\3 = 2\sqrt{4-c}\\\\2\sqrt{4-c} = 3\\\\\sqrt{4-c} = \frac{3}{2}\\\\4-c = \frac{9}{4}\\\\c = 4-\frac{9}{4}\\\\c = \frac{16-9}{4}\\\\c = \frac{7}{4}\\\\$

6 0

2 years ago