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2 years ago

7

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

d pair (5, 4) 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 (10, 8) on the graph represent? Explain in words. (2 points)

1 answer:

olga55 [171]2 years ago

4 0

Answer:

Part A: The price of 1 pound of potatoes is 1 dollar and 25 cents (1.25) because 5 / 4 is 1.25 and that's 1 pound

Part B: The ordered pair (10, 8) represents that 10 punds of potatoes cost 8 dollars

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A farmer has 300 ft of fence to enclose 2 adjacent rectangular pens bordering his barn. Find the maximum area he can close.

Margarita [4]

Answer:

Width of 37.5 feet and length of 50 feet will maximize the area.

Step-by-step explanation:

Let w represent width and l represent length of the pen.

We have been given that a farmer has 300 ft of fence to enclose 2 adjacent rectangular pens bordering his barn. We are asked to find the dimensions that will maximize the area.

We can see from the attachment that the perimeter of the pens would be $4w+3l$ .

We can set our given information in an equation $4w+3l=300$ .

The area of the two pens would be $A=2l\cdot w$ .

From perimeter equation, we will get:

$w=\frac{300-3l}{4}$

Substituting this value in area equation, we will get:

$A=2l\cdot (\frac{300-3l}{4})$

Since we need to maximize area, so we need to find derivative of area function as:

$A=\frac{600l-6l^2}{4}$

Bring out the constant:

$A=\frac{1}{4}*\frac{d}{dl}(600l-6l^2)$

$A=\frac{1}{4}*(600-12l)$

$A=150-3l$

Now, we will set our derivative equal to 0 as:

$150-3l=0$

$150=3l$

$\frac{150}{3}=\frac{3l}{3}$

$50=l$

Now, we will substitute $l=50$ in equation $w=\frac{300-3l}{4}$ to solve for width as:

$w=\frac{300-3(50)}{4}$

$w=\frac{300-150}{4}$

$w=\frac{150}{4}$

$w=37.5$

Therefore, width of 37.5 feet and length of 50 feet will maximize the area.

8 0

3 years ago

Which of the following is a fifth degree binomial?

Anna11 [10]

Answer:

C)

Step-by-step explanation:

$2xy^{2}-5y^{5}$

Binomial should have two terms. Five degree binomial means highest degree of the binomial should be 5

5 0

3 years ago

Which of the following of x and y would make up the following expression represent a real number? (4+5i)(x+yi)

salantis [7]

Answer:

x = 4 and y = - 5

Step-by-step explanation:

note that the product of a complex number and it's conjugate is real

That is

(a + bi)(a - bi) where a, b are real

= a² - abi + abi - b²i²

= a² + b² ← a real number

For (4 + 5i)(x + yi) to be real

we require (x + yi) to be the conjugate of 4 + 5i , that is 4 - 5i

(4 + 5i)(4 - 5i) ⇒ x = 4 and y = - 5

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

8 0

3 years ago

Divided it by 7 and added 25. the result was 34 what was my number?

AfilCa [17]

Answer:63

Step-by-step explanation:

63/7+25=34

6 0

2 years ago

Read 2 more answers