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2 years ago

A store sells 24 greeting cards for $12. Which is NOT a unit rate to describe this sale?

Mathematics

2 answers:

ahrayia [7]2 years ago

3 0

B,C,D are not a unit rate

Dmitry [639]2 years ago

3 0

A unit rate is the cost per 1 unit of product
12 dollars for 24 cards
0.50 dollars for 1 card
that should be the only answer
all else are false
ACD are false
but if you can only pick 1, pick C since it is per 5 dollars

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Determine how many solutions each equation has without using inverse operations. Explain your reasoning for each.

GuDViN [60]

Answer:

an infinite number

Step-by-step explanation:

The left side of the equation is a restatement of the right side of the equation. It is true for all values of x. There are an infinite number of solutions.

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3 years ago

Which function is increasing?

DerKrebs [107]

Answer:

A, it's the only one with a number greater then 1

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2 years ago

Solve each inequality. p + 5 < 10 A. p < –15 B. p < 5 C. p < 15 D. p < –5

EastWind [94]

Answer:

B) p<5

Step-by-step explanation:

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3 years ago

Mrs. Simmons gave a history test worth 92 points. There were only two types of questions on it: 2-point true/false questions and

SVETLANKA909090 [29]

Answer:

There are 8 true/false questions and 26 fill-in-the-blank questions.

Step-by-step explanation:

Let "t" be the number of 2-point true/false questions and "f" the number of 5-point fill-in-the-blank questions.

Mrs. Simmons gave a history test worth 92 points. Symbollically,

2 t + 5 f = 92 [1]

There were a total of 34 questions in the test. Symbollicaly,

t + f = 34

t = 34 - f [2]

If we replace [2] in 1, we get

2 (34 - f) + 5 f = 92

68 - 2 f + 5 f = 92

3 f = 24

f = 8

We replace f = 8 in [2],

t = 34 - 8 = 26

There are 8 true/false questions and 26 fill-in-the-blank questions.

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2 years ago

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago