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taurus [48]
3 years ago
7

A plumber has a 2-meter length of pipe. He needs to cut it into sections that are 10 centimeters long. How many sections will he

be able to cut? *
Mathematics
1 answer:
musickatia [10]3 years ago
5 0

Answer:

  20

Step-by-step explanation:

For cutting problems, we make the assumption that there is no loss of length due to the material moved or removed by cutting.

  2 m = 200 cm

so there are ...

  (200 cm)/(10 cm/section) = 20 sections

that can be cut from the pipe.

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What are the circumference and area of a circle with radius 7in. Round to the nearest tenth.
Natalka [10]

Circumference is 40 in and area is 150 in²

Step-by-step explanation:

  • Step 1: Find the circumference of the circle when radius = 7 in

Circumference = 2πr = 2 × 22/7 × 7 = 44 in ≈ 40 in

  • Step 2: Find the area of the circle.

Area = πr² = 22/7 × 7² = 154 in² ≈ 150 in²

4 0
3 years ago
Write the expanded form for 34.75
S_A_V [24]

Answer:

30+4+0.7+0.05

Step-by-step explanation:

Because the expanded form is added the number.

3=30

4=4.

7=0.7

and 5=0.05 so all together is 30+4+0.7+0.05

7 0
2 years ago
(x+3)^2 = x^2-3(1-x)
Oxana [17]
(x+3)(x+3) = x^2 -3 + 3x
x^2 + 3x + 3x + 9 = x^2 + 3x - 3
x^2 + 6x + 9 = x^2 + 3x - 3
6x + 9 = 3x - 3
3x + 9 = -3
3x = -12
x = -4

Check:
(-4+3)^2 = (-4)^2 - 3(1+4)
(-1)^2 = 16 - 3(5)
1 = 16-15
1 = 1 :)
7 0
3 years ago
One serving of whole milk contains 8 grams of fat. Reduced-fat milk contains 41.25% less fat per serving than whole milk. How ma
KiRa [710]
<span><span>We are told the reduced fat milk has 41.25% less fat than whole milk.

So the reduced fat milk has 58.75% of the fat of whole milk (that is, 100% -41.25%).
A serving of whole milk has 8 grams of fat.
So a serving of reduced fat milk has .5875 * 8 g = 4.7 g of fat.</span><span>
</span></span>
6 0
3 years ago
Jill invested $20,000 in an account that earned 5.5% annual interest, compounded annually. What is the value of this account aft
attashe74 [19]

\bf ~~~~~~ \textit{Compound Interest Earned Amount}&#10;\\\\&#10;A=P\left(1+\frac{r}{n}\right)^{nt}&#10;\quad \begin{cases}&#10;A=\textit{accumulated amount}\\&#10;P=\textit{original amount deposited}\dotfill &\$20000\\&#10;r=rate\to 5.5\%\to \frac{5.5}{100}\dotfill &0.055\\&#10;n=&#10;\begin{array}{llll}&#10;\textit{times it compounds per year}\\&#10;\textit{annually, thus once}&#10;\end{array}\dotfill &1\\&#10;t=years\dotfill &10&#10;\end{cases}


\bf A=20000\left(1+\frac{0.055}{1}\right)^{1\cdot 10}\implies A=20000(1.055)^{10}&#10;\\\\\\&#10;A\approx 34162.88916707\implies A\approx \stackrel{\textit{rounded up}}{34162.89}

4 0
3 years ago
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