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bekas [8.4K]
3 years ago
6

What is the total amount johann had repaid his parents at the end of the first week

Mathematics
1 answer:
makvit [3.9K]3 years ago
5 0

There is no image , word problem or equation. There for this question is incomplete.

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What is the area to this shape and tell me why pls I need help asap tonight
LUCKY_DIMON [66]
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3 years ago
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What is the value of x in the equation ? 2.5(6x-4)=10+4(1.5+.05x)
natima [27]

Answer:

x=2

Step-by-step explanation:

Distribute

15x-10=10+6+2x

simplify

15x-10=16+2x

subtract 2x to get x on one side

13x-10=16

add 10

13x=26

divide

x=2

8 0
3 years ago
Zamba has found a little black dress on sale for 50% off the original price of $239.99. She also has a coupon offering free ship
Yuri [45]

Answer: $139.23

Step-by-step explanation: Take 50% of original dress price and add price of shoes. Subtract 10% of that.

3 0
3 years ago
circumference of a circular field is 197.82 yards. What is the diameter of the field? Use 3.14 for it and do not round your answ
storchak [24]

Answer:

63 yards.

Step-by-step explanation:

The circumference of a circle is its diameter multiplied by the constant pi.C=\pi d. We are told to use \pi=3.14 and C=197.82, so we rearrange: \frac{197.82}{3.14}=d

We find that d=63.

7 0
3 years ago
The shaded region R in diagram below is enclosed by y-axis, y = x^2 - 1 and y = 3.
Lostsunrise [7]

Cross sections of the volume are washers or annuli with outer radii <em>x(y)</em> + 1, where

<em>y</em> = <em>x(y) </em>² - 1   ==>   <em>x(y)</em> = √(<em>y</em> + 1)

and inner radii 1. The distance between the outermost edge of each shell to the axis of revolution is then 1 + √(<em>y</em> + 1), and the distance between the innermost edge of <em>R</em> on the <em>y</em>-axis to the axis of revolution is 1.

For each value of <em>y</em> in the interval [-1, 3], the corresponding cross section has an area of

<em>π</em> (1 + √(<em>y</em> + 1))² - <em>π</em> (1)² = <em>π</em> (2√(<em>y</em> + 1) + <em>y</em> + 1)

Then the volume of the solid is the integral of this area over [-1, 3]:

\displaystyle\int_{-1}^3\pi y\,\mathrm dy = \frac{\pi y^2}2\bigg|_{-1}^3 = \boxed{4\pi}

\displaystyle\int_{-1}^3 \pi\left(2\sqrt{y+1}+y+1\right)\,\mathrm dy = \pi\left(\frac43(y+1)^{3/2}+\frac{y^2}2+y\right)\bigg|_{-1}^3 = \boxed{\frac{56\pi}3}

8 0
3 years ago
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