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3 years ago

5) Narcy had 164 dodars to spend on 9 books. After

Mathematics

2 answers:

MArishka [77]3 years ago

7 0

Answer:

18$ each book

Step-by-step explanation:

timama [110]3 years ago

5 0

Answer:

each book cost 18.22

Step-by-step explanation:

164 dollars to spend on 9 books. After buying them he had 11 dollars. How much did each book cost?

164÷9=x

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Calculate the pay for the following day of a weekly time card given a wage of $12.50/hr.

Elenna [48]

The one day pay is $106.25 rounded to the nearest hundredth.

Step-by-step explanation:

From the table shown :

The timing shown in the morning is from 8:00 to 12:15
The number of hours worked in the morning = 4 hours 15 minutes.

It is given that, the pay is $12.5 per hour.

Therefore, the pay earned in the morning = No.of hours × pay per hour.

⇒ 4 hours × 12.5 = $50

⇒ (15 mins / 60 mins) × 12.5 = $3.125

⇒ 50+3.125

⇒ 53.125

The timing shown in the afternoon is from 8:00 to 12:15
The number of hours worked in the morning = 4 hours 15 minutes.

Therefore, the pay earned in the afternoon = No.of hours × pay per hour.

⇒ 4 hours × 12.5 = $50

⇒ (15 mins / 60 mins) × 12.5 = $3.125

⇒ 50+3.125

⇒ 53.125

The pay for 1 day = pay earned in the morning section + pay earned in the afternoon section.

⇒ 53.125 + 53.125

⇒ 106.25

∴ The one day pay is $106.25 rounded to the nearest hundredth.

7 0

3 years ago

Solve 6 + 5 √ 2 4 9 − 2 x = 7

NISA [10]

$6+5\sqrt{249}-2x=7 \\-2x=7-6-5\sqrt{249} \\-2x\approx-77.9 \\x\approx\frac{-77.9}{2}\approx38.95$

Hope this helps.

7 0

3 years ago

N a? triangle, the measure of the first angle is twicetwice the measure of the second angle. the measure of the third angle is 1

Mrac [35]

Skip work! get some booty! don't care about skool

3 0

3 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

Find the area of the rug. DO NOT ROUND. what is the area?

Neporo4naja [7]

Answer:

28.26 ft²

Explanation:

$\sf area \ of \ circle = \pi r^2$

Here given diameter: 6 ft

radius: d/2 = 6/2 = 3 ft

========

area of rug:

$\dashrightarrow \ \pi (3)^2$

$\dashrightarrow \ 9\pi$

$\dashrightarrow \ 9(3.14)$

$\dashrightarrow \ 28.26 \ ft^2$

7 0

3 years ago

Read 2 more answers