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Basile [38]
3 years ago
5

Simplify

20%20%7D%7B%5Cfrac%7B1%7D%7B3%7Dof2%5Cfrac%7B2%7D%7B5%7D%20%20%7D" id="TexFormula1" title="\frac{2\frac{1}{5}-1\frac{1}{2}+1\frac{7}{10} }{\frac{1}{3}of2\frac{2}{5} }" alt="\frac{2\frac{1}{5}-1\frac{1}{2}+1\frac{7}{10} }{\frac{1}{3}of2\frac{2}{5} }" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Lemur [1.5K]3 years ago
4 0
Answer:
3

You need to first subtract everything on the numerator by changing the fractions in the numerator. Each fraction should have a similar denominator. You should end up getting 24/10, simplifying it down to 12/5. After that you multiply the fractions in the denominator part, due to the key word “of” indicating that you need to multiply. Once you do that you have your numerator and denominator simplified down from each equation, you then simplify down further.
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Find the solution(s) for x in the equation below. x^2 + 10x + 21= 0
Law Incorporation [45]

The solutions for ‘x’ in the given equation are – 3  and  - 7

<u>Step-by-step explanation:</u>

Given equation:

                   x^{2} + 10 x + 21 = 0

To find the ‘x’ value, try to factor, because in this case it works, it's fast. By using factor method, we get

                    (x + 3) (x + 7) = 0  (adding both value we get 10 and multiply as 21 as in equation and check with signs also while factoring)

                     x = - 3, -7

Verify above values by multiply both terms,

                     (x + 3) (x + 7) = 0

                x^{2} + 7 x + 3 x + 21 = 0

                x^{2} + 10 x + 21 = 0 (so values obtained from factor method are correct)

Or, can use quadratic formula, for a x^{2} + b x + c=0, the solutions are given by:

                      x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

In the given equation, a = 1, b = 10, c = 21, apply these in above formula

                     x=\frac{-10 \pm \sqrt{10^{2}-(4 \times 1 \times 21)}}{2(1)}=\frac{-10 \pm \sqrt{100-84}}{2}

                     x=\frac{-10 \pm \sqrt{16}}{2}=\frac{-10 \pm 4}{2}

So,

When  x=\frac{-10+4}{2}=\frac{-6}{2}=-3

When x=\frac{-10-4}{2}=\frac{-14}{2}=-7

Hence, the values for ‘x’ are - 3 and - 7

7 0
3 years ago
You are given the following system.x=2y+42x-3y=11 What method would you use to solve this system?
qwelly [4]

Answer:

substitution

Step-by-step explanation:

We assume the two equations are ...

  • x = 2y+4
  • 2x -3y = 11

You are given an expression for x. Using that in the second equation will result in an immediate solution for y.

... 2(2y+4) -3y = 11 . . . . . substituting 2y+4 for x

.. y +8 = 11

... y = 3

... x = 2·3 +4 = 10

The solution is (x, y) = (10, 3)

7 0
4 years ago
Monroe is gift wrapping a box that has the shape of a right rectangular prism. The box has a length of 20 cm, a width of 7 cm, a
Allisa [31]
I would assume you would do base*width*weight to get your area.
6 0
3 years ago
How do i solve thissss (mainly the question on the bottom)
Ymorist [56]

a)

well done

b)

well, a function for a relationship means, the x-coordinates must not repeat in a set, namely that for every "y" there must be a unique "x" coordinate, no X-REPEATS.

so for example in a relation that say is { (3, 4) , (5, 4) , (7, 4) , (10, 11) }, we do have y-coordinates repeated, but for a function that doesn't matter, our x-coordinates are not repeated thus, it's a function.

in this case, let's see the relationship set, just a few points { (3,5) , (3,6) , (3,8) , (6,10)}, well darn, we have 3 repeated three times in the x-coordinate slots, therefore is not a function, just a relation.

6 0
3 years ago
A sixth grader weighs 90 pounds, which is 120% of what he weighed in fourth grade. How much did he weigh in fourth grade?
Klio2033 [76]

Answer:

75lb

Step-by-step explanation:

3 0
3 years ago
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