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Likurg_2 [28]
3 years ago
6

Rationalize the denominator in the expression.

" id="TexFormula1" title=" \frac{4}{\sqrt{2} } " alt=" \frac{4}{\sqrt{2} } " align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
Dovator [93]3 years ago
7 0

Answer:

2\sqrt{2}

Step-by-step explanation:

See the photo for the steps. :)

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A recipe calls for 4 parts chocolate chips for every 6 parts of nuts. Callum makes the recipe for a party and uses 18 tablespoon
ahrayia [7]
12 is your answer because 18 divided by 6 =3 then 4x3=12
7 0
3 years ago
How do you solve this? A and B
svp [43]
An adult polar bear is 8 times longer than a newborn.
Newborn= 12 1/2 ft
Multiply this by 8.
But first you have to turn the fraction
12 1/2 into an improper fraction. To do this you have to multiply the denominator, 2, by the whole number, 12, and add the numerator to the product, which totals to 25. The fraction is 25/2. (25/2 * 8/1)= 100 in simplified. 100 inches= 8 1/3 ft long
A= 8 1/3 ft long.
For B, you are dividing since you are trying to figure out the length of a newborn, not an adult polar bear. 9/8 turned into a mixed fraction is
1 1/8 ft. But, it asks in inches. 
B= 13 1/2 inches
I know this answer is very long but I hope it helps you understand better! 
:)

8 0
4 years ago
Read 2 more answers
Make b the subject of the formula<br><br>A=½(a+b)h​
kenny6666 [7]
B=A/(1/2h) -a
work shown below

3 0
3 years ago
Read 2 more answers
What is the greatest integer a, such that a^2 + 3b is less than (2b)^2, assuming that b is 5? Please answer and have a nice day!
solniwko [45]
<h3>Answer: Largest value is a = 9</h3>

===================================================

Work Shown:

b = 5

(2b)^2 = (2*5)^2 = 100

So we want the expression a^2+3b to be less than (2b)^2 = 100

We need to solve a^2 + 3b < 100 which turns into

a^2 + 3b < 100

a^2 + 3(5) < 100

a^2 + 15 < 100

after substituting in b = 5.

------------------

Let's isolate 'a'

a^2 + 15 < 100

a^2 < 100-15

a^2 < 85

a < sqrt(85)

a < 9.2195

'a' is an integer, so we round down to the nearest whole number to get a \le 9

So the greatest integer possible for 'a' is a = 9.

------------------

Check:

plug in a = 9 and b = 5

a^2 + 3b < 100

9^2 + 3(5) < 100

81 + 15 < 100

96 < 100 .... true statement

now try a = 10 and b = 5

a^2 + 3b < 100

10^2 + 3(5) < 100

100 + 15 < 100 ... you can probably already see the issue

115 < 100 ... this is false, so a = 10 doesn't work

5 0
3 years ago
Mr. Notardonato is going to run a half marathon, which is 13.1 miles. How many inches are in 13.1 miles? (
lakkis [162]

Answer:

830,016 inches

Step-by-step explanation:

this is how many inches

3 0
3 years ago
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