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3 years ago

A pound of coffee costs $14.99what is the cost per ounce

Mathematics

1 answer:

alexdok [17]3 years ago

6 0

There are 16 oz in a Pound. Take 14.99 and divide it by 16. This will give you the cost per oz. 14.99/16= .936875

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Please help me Will name Brainliest

Anarel [89]

28÷1.5=18.67

meaning that more, than 17 discs can be placed in the box.

8 0

3 years ago

Read 2 more answers

if you repeat the perpendicular line segment construction twice using paper folding, you can construct:

ehidna [41]

Answer:

The correct answer is option C.

The mid point of the line segment.

Step-by-step explanation:

the perpendicular line segment construction twice using paper folding

we have to find the mid point of the given line segment.

We get the midpoint easily when fold the paper correctly

Therefore the correct answer is option C.

The mid point of the line segment.

6 0

3 years ago

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which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=7%20-%209w%20%3D%20%20-%2029%20" id="TexFormula1" title="7 - 9w = - 29 " alt="7 - 9w = - 29

Oksana_A [137]

7 - 9w = - 29

Subtract 7 from both sides

- 9w = - 36

Divide by - 9 on both sides

w = 4

5 0

3 years ago

Read 2 more answers

Suppose f(x)=2×+5 find the value of -2f(x+1)

loris [4]

F(x) = 2x + 5

-----------------------------------
Find f(x + 1) :
-----------------------------------
f(x +1) = 2(x + 1) + 5
f(x +1) = 2x + 2 + 5
f(x + 1) = 2x + 5

-----------------------------------
Find -2f(x+1):
-----------------------------------
-2(fx+1) = -2(2x + 5)
-2(fx+1) = -4x - 10

----------------------------------------------------------------------
Answer: -2(fx+1) = -4x - 10
----------------------------------------------------------------------

3 0

3 years ago