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3 years ago

How do you ungroup a 3 digit number? My nephew is working on his homework and I have no clue how to help him.

Mathematics

1 answer:

Talja [164]3 years ago

6 0

In monetary terms, you're making change. If you need more ones, you trade one 10 for 10 ones.

When I was in school, many years ago, we used small superscript-type numbers to show this:

7 3 0 ⇒ 7 2 ¹0 . . . . regrouping to gain 10 ones

7 2 ¹0 ⇒ 6 ¹2 ¹0 . . .regrouping again to gain 10 more tens

You'd have to look at the example problems that precede this page in order to see what the meaning of "magnifying glass" is.

You might be interested in

(-x2y3)*(12xy3) (-2xy2)3

Leokris [45]

Assuming the numbers are exponents expect the coefficient of 2, top expression, -12x^3y^6, multiply bottom
and= 72x^4y^8

4 0

2 years ago

Given the Arithmetic series A1+A2+A3+A4 7 + 11 + 15 + 19 + . . . + 91 What is the value of sum?

aivan3 [116]

Answer:

The value of the sum is 1078

Step-by-step explanation:

* Lets revise how to find the sum of the arithmetic series

- In the arithmetic series there is a constant difference between each

two consecutive terms

- Ex:

# 4 , 9 , 14 , 19 , 24 , .......... (+ 5)

# 25 , 15 , 5 , -5 , -15 , .......... (-10)

- So if the first term is a and the constant difference between each two

consecutive terms is d, then

U1 = a , U2 = a + d , U3 = a + 2d , U4 = a + 3d , ..........

- Then the nth term is Un = a + (n - 1)d

- The sum of n terms is Sn = n/2[a + L] , where L is the last term in

the series

* Lets solve the problem

∵ The arithmetic series is 7 + 11 + 15 + 19 + ......................... + 91

∴ The first term (a) is 7

∴ The last term is (L) 91

- Lets find the constant difference

∵ 11 - 7 = 4 and 15 - 11 = 4

∴ The constant difference (d) is 4

- Lets find the sum of the series from 7 to 91

∵ Sn = n/2[a + L]

∵ a = 7 and L = 91

∴ Sn = n/2[7 + 91]

- We need to know how many terms in the series

∵ L is the last term and equals 91 lets find its position in the series

∵ Un = a + (n - 1)d

∵ a = 7 , d = 4 and Un = 91

∴ 91 = 7 + (n - 1)(4) ⇒ subtract 7 from both sides

∴ 84 = (n - 1)(4) ⇒ divide both sides by 4

∴ 21 = n - 1 ⇒ add 1 to both sides

∴ n = 22

∴ The number of the terms in the series is 22

- Lets find the sum of the 22 terms (S22)

∴ S22 = 22/2[7 + 91]

∴S22 = 11[98] = 1078

* The value of the sum is 1078

4 0

3 years ago

Solve the inequality 4t^2 ≤ 9t-2 please show steps and interval notation. thank you!

Misha Larkins [42]

Answer:

[0.25, 2]

Step-by-step explanation:

We have

4t² ≤ 9t-2

subtract 9t-2 from both sides to make this a quadratic

4t²-9t+2 ≤ 0

To solve this, we can solve for 4t²-9t+2=0 and do some guess and check to find which values result in the function being less than 0.

4t²-9t+2=0

We can see that -8 and -1 add up to -9, the coefficient of t, and 4 (the coefficient of t²) and 2 multiply to 8, which is also equal to -8 * -1. Therefore, we can write this as

4t²-8t-t+2=0

4t(t-2)-1(t-2)=0

(4t-1)(t-2)=0

Our zeros are thus t=2 and t = 1/4. Using these zeros, we can set up three zones: t < 1/4, 1/4<t<2, and t>2. We can take one random value from each of these zones and see if it fits the criteria of

4t²-9t+2 ≤ 0

For t<1/4, we can plug in 0. 4(0)²-9(0) + 2 = 2 >0 , so this is not correct

For 1/4<t<2, we can plug 1 in. 4(1)²-9(1) +2 = -3 <0, so this is correct

For t > 2, we can plug 5 in. 4(5)²-9(5) + 2 = 57 > 0, so this is not correct.

Therefore, for 4t^2 ≤ 9t-2 , which can also be written as 4t²-9t+2 ≤ 0, when t is between 1/4 and 2, the inequality is correct. Furthermore, as the sides are equal when t= 1/4 and t=2, this can be written as [0.25, 2]

8 0

3 years ago

How do you solve this? My algebra 2 teacher taught it to us in class but it still confuses me! Thank you!

nataly862011 [7]

$\bf \cfrac{\frac{1}{8}-\frac{1}{2x}}{\frac{1}{12x}-\frac{1}{3x^2}} \begin{array}{llll} \leftarrow \textit{our LCD here will just be }8x\\[0.9em] \leftarrow \textit{our LCD down here will be }12x^2 \end{array}$

$\bf \cfrac{~~\frac{(x)1-(4)1}{8x}~~}{\frac{(x)1-(4)1}{12x^2}}\implies \cfrac{~~\frac{x-4}{8x}~~}{\frac{x-4}{12x^2}}\implies \cfrac{~~\begin{matrix} x-4 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{\underset{2}{~~\begin{matrix} 8x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}\cdot \cfrac{\stackrel{3x}{~~\begin{matrix} 12x^2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}{~~\begin{matrix} x-4 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }\implies \cfrac{3x}{2}$

8 0

3 years ago

How do u guys solved this?

TEA [102]

Answer:

3 boxes

Step-by-step explanation:

First, we need to find the area of the whole square, so we can find out how much to subtract from. We can easily do this by multiplying 16 by 16, since it is a square.

16*16=256 square feet.

Now, we just need to find the area of the circular rug, and subtract it from 256 to find the area not covered by the rug that needs to be tiled. Since the sides of the rug touch each side of the floor, than the diameter will be the same as the side length of the square. Thus, the diameter is 16.

Now we can use the equation A=pir^2 or Area=pi*radius squared to find the area of the circle. The radius is halve of the diameter, so divide 16 by 2 to get the radius, or 8. Now all we have to do is square 8 and multiply it by 3.14.

8^2=64

Now multiply.

64*3.14=200.96.

The area of the circle is 200.96, so subtract 200.96 from 256.

256-200.96=55.04.

So they will need 3 boxes to cover the area left.

Hope this helps!

3 0

3 years ago