Answer:
I think true I'm so sorry if IT'S wrong
You would use long division to determine that 8/12 = 0.667 approximately. The steps would go something like this
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Step 1) Write 8 under a long division bar and place 12 to the left and outside the bar
Step 2) Change the 8 into 8.0000 (you can have as many trailing zeros as you want)
Step 3) While 12 doesn't go into 8, 12 does go into 80. So you can think of 8.0 as 80. Ask yourself "how many times does 12 go into 80?". The answer is 6 times because 6*12 = 72 (note how 7*12 = 84 is too big). So we write a "6" above the 8.0000 such that it is above the first "0". So far we have 0.6 as the quotient.
Step 4) Multiply that 6 by 12 to get 72. We write the 72 under the 80 and then subtract to get 80 - 72 = 8. Pull down the second 0 from "8.0000" to change that 8 into 80.
Step 5) We effectively repeat steps 3 and 4 again. So thats why we get another 6 as the next part of the quotient, and why the 6's repeat forever. The only reason why there's a 7 at the end is because of rounding.
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The long division is shown in the image attachment below. The gridlines are there to show how things line up though when you write out the long division steps (to show to your teacher) you wont use any grid lines.
side note: the fraction 8/12 reduces to 2/3. You divide both top and bottom by the GCF 4. So 8/4 = 2 and 12/4 = 3.
<u>The correct answer</u> is B.
Explanation:
89.65 is nearly 90.
4.75 is nearly 5.
To estimate the product, we will multiply 90*5, which equals 450.
Answer:
It didn't saying anything about rounding but the the dimensions in there exact form are:
![L=\frac{-5+\sqrt{249}}{2} \text{ cm}](https://tex.z-dn.net/?f=L%3D%5Cfrac%7B-5%2B%5Csqrt%7B249%7D%7D%7B2%7D%20%5Ctext%7B%20cm%7D)
and
.
Now if they said to round to the nearest hundredths the dimensions in this rounded form would be:
L=5.39 cm and W=5.19 cm
(Check your question again and see if you meant what you have)
Step-by-step explanation:
L is 5 less than twice W
L = 2W-5
Area of rectangle=L*W
Area is 28 means L*W=28.
I'm going to insert L=2W-5 into L*W=28 giving me
(2W-5)*W=28
Distribute
2W^2-5W=28
Subtract 28 on both sides
2W^2-5W-28=0
a=2
b=-5
c=-28
We are going to use the quadratic formula.
Plug in...
![W=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\W=\frac{5 \pm \sqrt{(-5)^2-4(2)(-28)}}{2(2)}\\\\W=\frac{5 \pm \sqrt{25+224}}{4}\\\\W=\frac{5 \pm \sqrt{249}}{4}](https://tex.z-dn.net/?f=W%3D%5Cfrac%7B-b%20%5Cpm%20%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D%5C%5C%5C%5CW%3D%5Cfrac%7B5%20%5Cpm%20%5Csqrt%7B%28-5%29%5E2-4%282%29%28-28%29%7D%7D%7B2%282%29%7D%5C%5C%5C%5CW%3D%5Cfrac%7B5%20%5Cpm%20%5Csqrt%7B25%2B224%7D%7D%7B4%7D%5C%5C%5C%5CW%3D%5Cfrac%7B5%20%5Cpm%20%5Csqrt%7B249%7D%7D%7B4%7D)
W needs to be positive so ![W=\frac{5 + \sqrt{249}}{4}](https://tex.z-dn.net/?f=W%3D%5Cfrac%7B5%20%2B%20%5Csqrt%7B249%7D%7D%7B4%7D)
And ![L=2W-5=2(\frac{5 + \sqrt{249}}{4})-5](https://tex.z-dn.net/?f=L%3D2W-5%3D2%28%5Cfrac%7B5%20%2B%20%5Csqrt%7B249%7D%7D%7B4%7D%29-5)
![L=\frac{5+\sqrt{249}}{2}-5](https://tex.z-dn.net/?f=L%3D%5Cfrac%7B5%2B%5Csqrt%7B249%7D%7D%7B2%7D-5)
![L=\frac{5+\sqrt{249}}{2}-\frac{10}{2}](https://tex.z-dn.net/?f=L%3D%5Cfrac%7B5%2B%5Csqrt%7B249%7D%7D%7B2%7D-%5Cfrac%7B10%7D%7B2%7D)
![L=\frac{-5+\sqrt{249}}{2}](https://tex.z-dn.net/?f=L%3D%5Cfrac%7B-5%2B%5Csqrt%7B249%7D%7D%7B2%7D)
Does L*W=28?
Let's check.
![L \cdot W=\frac{-5+\sqrt{249}}{2} \cdot \frac{5 + \sqrt{249}}{4}](https://tex.z-dn.net/?f=L%20%5Ccdot%20W%3D%5Cfrac%7B-5%2B%5Csqrt%7B249%7D%7D%7B2%7D%20%5Ccdot%20%5Cfrac%7B5%20%2B%20%5Csqrt%7B249%7D%7D%7B4%7D)
![L \cdot W=\frac{249-25}{8}](https://tex.z-dn.net/?f=L%20%5Ccdot%20W%3D%5Cfrac%7B249-25%7D%7B8%7D)
In the last step, I was able to do a quick multiplication because I was multiplying conjugates. (a+b)(a-b)=a^2-b^2.
![L \cdot W=\frac{ 224}{8}](https://tex.z-dn.net/?f=L%20%5Ccdot%20W%3D%5Cfrac%7B%20224%7D%7B8%7D)
![L \cdot W=28](https://tex.z-dn.net/?f=L%20%5Ccdot%20W%3D28)