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3 years ago

Help on theese 2 for brainly tell me why we need to regroup

Mathematics

1 answer:

svlad2 [7]3 years ago

4 0

Answer:

You will need to regroup because 2/3 is less than 3/4.

Step-by-step explanation:

DON'T IGNORE THIS PART!!!

Let's find the common denominator for 10 2/3 and 6 3/4.

$10 \frac{2}{3} - 6 \frac{3}{4}$

What is the least common multiple for 3 and 4? Well, let's make a list that would help!

Multiples of 3: 3, 6, 9, 12, 15, 18, 21...

Multiples of 4: 4, 8, 12, 16, 20, 24, 28...

What is the first multiple that 3 and 4 matches with? It is 12!

Now, let's convert the fractions.

$10 \frac{2}{3} - 6 \frac{3}{4}$ = $10 \frac{8}{12} - 6 \frac{9}{12}$

Woah! How do you subtract 9/12 from 8/12? Don't worry--there is a solution to this problem! And--it is... regrouping!

You have to add a whole to 8/12. This means that you have to add 8/12 + 12/12, which is 20/12. And--you will have to subtract one from ten.

$9 \frac{20}{12} - 6 \frac{9}{12}$

Now, that should be way easier to subtract!

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Y = -3x +10 3x +y = 5 substitution method

spayn [35]

Answer:

no solutions

Step-by-step explanation:

y = -3x +10

3x +y = 5

Substitute the first equation into the second equation

Let every instance of y in the second equation be replaced with -3x+10

3x+ (-3x+10) = 5

Combine like terms

10 =5

This is never true so there are no solutions

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3 years ago

Write the following phrase as an expression "the product of 6 and n"

Kamila [148]

Answer:

Step-by-step explanation:

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2 years ago

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faltersainse [42]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

Evaluate the following integral using trigonometric substitution.

wariber [46]

Answer:

Step-by-step explanation:

1. Given the integral function $\int\limits {\sqrt{a^{2} -x^{2} } } \, dx$ , using trigonometric substitution, the substitution that will be most helpful in this case is substituting x as $asin \theta$ i.e $x = a sin\theta$ .

All integrals in the form $\int\limits {\sqrt{a^{2} -x^{2} } } \, dx$ are always evaluated using the substitute given where 'a' is any constant.

From the given integral, $\int\limits {7\sqrt{49-x^{2} } } \, dx = \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx$ where a = 7 in this case.

The substitute will therefore be $x = 7 sin\theta$

2.) Given $x = 7 sin\theta$

$\frac{dx}{d \theta} = 7cos \theta$

cross multiplying

$dx = 7cos\theta d\theta$

3.) Rewriting the given integral using the substiution will result into;

$\int\limits {7\sqrt{49-x^{2} } } \, dx \\= \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -(7sin\theta)^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -49sin^{2}\theta } } \, dx\\= \int\limits {7\sqrt{49(1-sin^{2}\theta)} } } \, dx\\= \int\limits {7\sqrt{49(cos^{2}\theta)} } } \, dx\\since\ dx = 7cos\theta d\theta\\= \int\limits {7\sqrt{49(cos^{2}\theta)} } } \, 7cos\theta d\theta\\= \int\limits {7\{7(cos\theta)} }}} \, 7cos\theta d\theta\\$

$= \int\limits343 cos^{2} \theta \, d\theta$

8 0

4 years ago