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Ksju [112]
3 years ago
11

Please help with the question

Mathematics
1 answer:
Artyom0805 [142]3 years ago
4 0

I find it convenient to look at the differences and the rate at which those differences are made up.

8. Jim is closing the $150 gap at the rate of $7.50 per week. He will catch up in

... 150/(7.5/week) = 20 weeks

9. At noon, the price of Stock A has increased by 0.05×3 = 0.15, so is now $15.90, which is $0.63 more than Stock B at that time. The prices are closing the gap at the rate of $0.05 +0.13 = $0.18 per hour, so will be the same after

... $0.63/($0.18/hour) = 3.5 hours . . . . after noon, at 3:30 pm

_____

You can also write and solve equations for the prices of the stocks. Or you can use a graphing calculator to tell you the solution. When equations are involved, I like to solve them the simplest possible way: let technology do it.

You are given the value at a time, and the rate of change of that value, so the equations are easily written in point-slope form. You will note that the common price at 3:30 pm (15.5 hours after midnight) is one that is not a whole number of cents. (That's usually OK for when trading stocks.)

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How many loads of laundry can you wash if you work 3 shifts? express your answer numerically as an integer?
Likurg_2 [28]

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Hello, may I please have some help on these 4 questions? Thanks very much! {Also, if you could answer them quick, that would be
wlad13 [49]
1.

Parallelogram with base a = 20 in. and height h = 16 in.
Triangle with base a = 20 in. and height b = 8 in.

so:

A = A_P+A_T=a\cdot h+\frac{1}{2}\cdot a\cdot b=20\cdot16+\frac{1}{2}\cdot20\cdot8=320+80=\\\\=\boxed{400\text{ in}^2}

2.

Trapezoid with base b₁ = 14 cm, base b₂ = 4 cm and height h = 10 cm
Triangle with base b₁ = 14 cm and height x = 18 cm - 10 cm = 8 cm

so:

A=A_T+A_\Delta=\frac{1}{2}\cdot(b_1+b_2)\cdot h+\frac{1}{2}\cdot b_1\cdot x=\\\\=
\frac{1}{2}\cdot(14+4)\cdot 10+\frac{1}{2}\cdot 14\cdot8=\frac{1}{2}\cdot18\cdot 10+\frac{1}{2}\cdot 14\cdot8=90+56=\\\\=\boxed{146\text{ cm}^2}

3.

We have three rectangles:

A=A_1+A_2+A_3=1\cdot 4+6\cdot5+18\cdot6=4+30+108=\boxed{142\text{ ft}^2}

4.

Area of a circle:

A_\circ=\pi\cdot r^2=3.14\cdot5^2=3.14\cdot25=78.5 \text{ cm}^2

Area of a rectangle:

A_R=8\cdot 6=48\text{ cm}^2

Area of the shaded region:

A=A_\circ-A_R=78.5-48=\boxed{30.5\text{ cm}^2}


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3 years ago
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