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2 years ago

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A proportional relationship between the number of pounds of potatoes (x) and the price in dollars (y) is graphed, and the ordere

d pair (4, 3) is on the graphed line. Part A: What is the price of 1 pound of potatoes? Show your work. (8 points) Part B: What does the ordered pair (8, 6) on the graph represent? Explain in words. (2 points)
(they're is no graph)

1 answer:

vovangra [49]2 years ago

3 0

X = lbs of potatoes and y = price

(4,3)....this means 4 lbs of potatoes cost $ 3

3/4 = 0.75 cents per lb <== 1 lb of potatoes cost 0.75

(8,6)...this means that 8 lbs of potatoes cost $ 6

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What is the maximum percent of net spendable income that should be set aside for housing

stepladder [879]

Answer:

Answer;

-38 %

Step-by-step explanation:

Explanation;

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7 0

2 years ago

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On a road map, the locations A, B and C are collinear. Location C divides the road from location A to B, such that AC:CB = 1:2.

koban [17]

Answer:

C. (-1,-2)

Step-by-step explanation:

Since C internally divides AB in the ratio AC/CB = 1/2 = m/n where m = 1 and n = 2, we use the formula for internal division.

Let A = (x₁, y₁) = (5, 16), B = (x₂, y₂) and C = (x, y) = (3, 10)

So x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

Substituting the values of the coordinates, we have

x = (mx₂ + nx₁)/(m + n)

3 = (1 × x₂ + 2 × 5)/(2 + 1)

3 = (x₂ + 10)/3

multiplying through by 3, we have

9 = x₂ + 10

x₂ = 9 - 10

x₂ = -1

y = (my₂ + ny₁)/(m + n)

10 = (1 × y₂ + 2 × 16)/(2 + 1)

10 = (x₂ + 32)/3

multiplying through by 3, we have

30 = y₂ + 32

y₂ = 30 - 32

y₂ = -2

So, the coordinates of B are (-1, -2)

3 0

2 years ago

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

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2 years ago

Assume that x and y are both differentiable functions of t and find the required values of dy/dt and dx/dt. x2 + y2 = 25 (a) Fin

Artemon [7]

Answer:

(a) $\frac{dy}{dt}=-3\frac{3}{4}$

(b) $\frac{dx}{dt}=3\frac{3}{4}$

Step-by-step explanation:

$x^{2} +y^{2}=25$

Take $\frac{d}{dt}$ of of each term.

$\frac{d}{dt}(x^{2})+\frac{d}{dt}(y^{2})=\frac{d}{dt}(25)\\\\(\frac{d}{dx}(x^{2})*\frac{dx}{dt}) +(\frac{d}{dy}(y^{2})*\frac{dy}{dt})=\frac{d}{dt}(25)\\\\2x\frac{dx}{dt} +2y\frac{dy}{dt} = 0\\\\$

For Question a

$2y\frac{dy}{dt}=-2x\frac{dx}{dt}\\\\\frac{dy}{dt}=\frac{-2x\frac{dx}{dt}}{2y} \\\\\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

Given that x = 3, y = 4, and dx/dt = 5.

$\frac{dy}{dt}=-\frac{3}{4}*5=-\frac{15}{4}\\ \\\frac{dy}{dt}=-3\frac{3}{4}$

For Question b

$2x\frac{dx}{dt}=-2y\frac{dy}{dt}\\\\\frac{dx}{dt}=\frac{-2y\frac{dy}{dt}}{2x} \\\\\frac{dx}{dt}=-\frac{y}{x}\frac{dy}{dt}$

Given that x = 4, y = 3, and dx/dt = -5.

$\frac{dx}{dt}=-\frac{3}{4}*-5=\frac{15}{4}\\ \\\frac{dx}{dt}=3\frac{3}{4}$

5 0

2 years ago

Mustafa, Heloise, and Gia have written more than a combined total of 22 articles for the school newspaper.

Ghella [55]

Answer: The required inequality is $x+\dfrac{1}{4}x+\dfrac{3}{2}x>22.$ and its solution is $x>8.$

Step-by-step explanation: Given that Mustafa, Heloise, and Gia have written more than a combined total of 22 articles for the school newspaper.

Also, Heloise has written $\frac{1}{4}$ as many articles as Mustafa has and Gia has written $\frac{3}{2}$ as many articles as Mustafa has.

We are to write an inequality to determine the number of articles, m, Mustafa could have written for the school newspaper. Also, to solve the inequality.

Since m denotes the number of articles that Mustafa could have written. Then, according to the given information, we have

$x+\dfrac{1}{4}x+\dfrac{3}{2}x>22.$

And the solution of the above inequality is as follows :

$x+\dfrac{1}{4}x+\dfrac{3}{2}x>22\\\\\\\Rightarrow \dfrac{4x+x+6x}{4}>22\\\\\\\Rightarrow 11x>88\\\\\Rightarrow x>\dfrac{88}{11}\\\\\Rightarrow x>8.$

Thus, the required inequality is $x+\dfrac{1}{4}x+\dfrac{3}{2}x>22.$ and its solution is $x>8.$

4 0

2 years ago

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