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4 years ago

How do you change a decimal tenth to a decimal hundredth

Mathematics

1 answer:

antiseptic1488 [7]4 years ago

7 0

Answer: Divide the number of tenths by 0.1 to convert to hundredths.

Step-by-step explanation:

For example, if you have 5 tenths, you divide 5 by 0.1 to get 50 hundredths.

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If JM = 7, MK = 21, and LN = 24, what is NJ?

SSSSS [86.1K]

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3 years ago

Abby can buy an 8 pound bag of dog food for $7.40 or a 4 pound bag of the same dog food for $5.38. Which is the better buy?

nydimaria [60]

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4 years ago

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What is 2 3/4 as a decimal?

Vesna [10]

Answer:

2.75

Step-by-step explanation:

first convert it to a improper fraction

2 3/4 = 11/4

then dividing the numerator by the denominator.

11/4 = 2.75

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3 years ago

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For the function given below, find a formula for the Riemann sum obtained by dividing the interval [0,5] into n equal subinterva

sergij07 [2.7K]

Given

we are given a function

$f(x)=x^2+5$

over the interval [0,5].

Required

we need to find formula for Riemann sum and calculate area under the curve over [0,5].

Explanation

If we divide interval [a,b] into n equal intervals, then each subinterval has width

$\Delta x=\frac{b-a}{n}$

and the endpoints are given by

$a+k.\Delta x,\text{ for }0\leq k\leq n$

For k=0 and k=n, we get

$\begin{gathered} x_0=a+0(\frac{b-a}{n})=a \\ x_n=a+n(\frac{b-a}{n})=b \end{gathered}$

Each rectangle has width and height as

$\Delta x\text{ and }f(x_k)\text{ respectively.}$

we sum the areas of all rectangles then take the limit n tends to infinity to get area under the curve:

$Area=\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\Delta x.f(x_k)$

Here

$f(x)=x^2+5\text{ over the interval \lbrack0,5\rbrack}$ $\Delta x=\frac{5-0}{n}=\frac{5}{n}$ $x_k=0+k.\Delta x=\frac{5k}{n}$ $f(x_k)=f(\frac{5k}{n})=(\frac{5k}{n})^2+5=\frac{25k^2}{n^2}+5$

Now Area=

$\begin{gathered} \lim_{n\to\infty}\sum_{k\mathop{=}1}^n\Delta x.f(x_k)=\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\frac{5}{n}(\frac{25k^2}{n^2}+5) \\ =\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\frac{125k^2}{n^3}+\frac{25}{n} \\ =\lim_{n\to\infty}(\frac{125}{n^3}\sum_{k\mathop{=}1}^nk^2+\frac{25}{n}\sum_{k\mathop{=}1}^n1) \\ =\lim_{n\to\infty}(\frac{125}{n^3}.\frac{1}{6}n(n+1)(2n+1)+\frac{25}{n}n) \\ =\lim_{n\to\infty}(\frac{125(n+1)(2n+1)}{6n^2}+25) \\ =\lim_{n\to\infty}(\frac{125}{6}(1+\frac{1}{n})(2+\frac{1}{n})+25) \\ =\frac{125}{6}\times2+25=66.6 \end{gathered}$

So the required area is 66.6 sq units.

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1 year ago

Help me, please!!!!!!

Lelu [443]

12^2 + 9^2 = 225

the square root of 225 = 15

x = 15

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4 years ago

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How do you change a decimal tenth to a decimal hundredth​

How do you change a decimal tenth to a decimal hundredth