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2 years ago

Jazmine has 4 and 2/3 fluid Ounces of coffee

Mathematics

2 answers:

Paha777 [63]2 years ago

4 0

Jazmine can make 28 cups of coffee with her creamer.

Solution:
4 2/3 ÷ 1/6 = 28

Calculation Steps:
=14/3 ÷ 1/6
= 14/3 x 6/1
=84/3
=28

harina [27]2 years ago

3 0

Answer:

Deliciously creamy, our classic vanilla flavor is the perfect indulgent flavor for any cup of coffee. With no refrigeration necessary and the longest industry shelf life at nine months, Coffee mate liquid creamer singles are the most popular creamer format, great for all food service needs. Available in a 4 x 50 count dispenser box and a 180 count box.

Step-by-step explanation:

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a group of workers can plant 5 1/4 acres in four and a half days what is the unit rate in acres per day

dangina [55]

Divide to find the unit rate:

5.25 acres

---------------- = 1.1666... acres/day, or 1 1/6 acres/day.

4.5 days

7 0

2 years ago

A circular plate has a radius of 13 centimeters.

butalik [34]

To solve the problem, we need to get the formula for area of a circle -
Area = pi * r^2; r= radius
= 3.14 * (13cm)^2
= 3.14 * 169cm^2 or 3.14 * 13^2cm^2
= 530.66cm^2
Therefore, the area of the circular plate is 530.66cm^2

3 0

2 years ago

Read 2 more answers

Please someone help me with this word problem for my math class T.T

olga55 [171]

Answer:

<u>Perimeter</u>:

= 58 m (approximate)

= 58.2066 or 58.21 m (exact)

= 208 m² (approximate)

= 210.0006 or 210 m² (exact)

Step-by-step explanation:

Given the following dimensions of a rectangle:

length (L) = $\sqrt{252}$ meters

width (W) = $\sqrt{175}$ meters

The formula for solving the perimeter of a rectangle is:

P = 2(L + W) or 2L + 2W

The formula for solving the area of a rectangle is:

A = L × W

<h2>Approximate Forms:</h2>

In order to determine the approximate perimeter, we must determine the perfect square that is close to the given dimensions.

13² = 169

14² = 196

15² = 225

16² = 256

Among the perfect squares provided, 16² = 256 is close to 252 (inside the given radical for the length), and 13² = 169 (inside the given radical for the width). We can use these values to approximate the perimeter and the area of the rectangle.

P = 2(L + W)

P = 2(13 + 16)

P = 58 m (approximate)

A = L × W

A = 13 × 16

A = 208 m² (approximate)

<h2>Exact Forms:</h2>

L = $\sqrt{252}$ meters = 15.8745 meters

W = $\sqrt{175}$ meters = 13.2288 meters

P = 2(L + W)

P = 2(15.8745 + 13.2288)

P = 2(29.1033)

P = 58.2066 or 58.21 m

A = L × W

A = 15.8745 × 13.2288

A = 210.0006 or 210 m²

8 0

2 years ago

How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?

Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)$

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)

$f'(x) = \dfrac{1}{1+x}$

f'(0) $= \dfrac{1}{1+0}=1$

f ' ' (x) $= \dfrac{1}{(1+x)^2}$

f ' ' (x) $= \dfrac{1}{(1+0)^2}=-1$

f ' ' '(x) $= \dfrac{2}{(1+x)^3}$

f ' ' '(x) $= \dfrac{2}{(1+0)^3} = 2$

f ' ' ' '(x) $= \dfrac{6}{(1+x)^4}$

f ' ' ' '(x) $= \dfrac{6}{(1+0)^4}=-6$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+x)^5} = 24$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+0)^5} = 24$

Now, the next process is to substitute the above values back into equation (1)

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...$

$In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...$

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

To estimate the value of In(1.4), let's replace x with 0.4

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001

Hence, the estimate of In(1.4) to the term is $\dfrac{0.4^5}{5}$ is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

8 0

2 years ago

Can you help me on question eight?!

Irina-Kira [14]

Answer: A

Step-by-step explanation:

6 0

2 years ago