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KatRina [158]
1 year ago
5

Can someone solve this for me?

Mathematics
1 answer:
Helga [31]1 year ago
5 0

well, we know it's a rectangle, so that means the sides JK = IL and JI = KL, so

\stackrel{JK}{3x+21}~~ = ~~\stackrel{IL}{6y}\implies 3(x+7)=6y\implies x+7=\cfrac{6y}{3} \\\\\\ x+7=2y\implies \boxed{x=2y-7} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{JI}{6y-6}~~ = ~~\stackrel{KL}{2x+20}\implies 6(y-1)=2(x+10)\implies \cfrac{6(y-1)}{2}=x+10 \\\\\\ 3(y-1)=x+10\implies 3y-3=x+10\implies \stackrel{\textit{substituting from the 1st equation}}{3y-3=(2y-7)+10} \\\\\\ 3y-3=2y+3\implies y-3=3\implies \blacksquare~~ y=6 ~~\blacksquare ~\hfill \blacksquare~~ \stackrel{2(6)~~ - ~~7}{x=5} ~~\blacksquare

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Given the following system of equations: 2x + 6y = 12 4x + 3y = 15 Which action creates an equivalent system that will eliminate
ozzi

Answer:

Multiply the second equation by −2 to get −8x − 6y = −30.

Step-by-step explanation:

{2x + 6y = 12

{4x + 3y = 15

{2x + 6y = 12

{−8x − 6y = −30 >> New Equation

* Doing this will give you <em>additive</em><em> </em><em>inverses</em><em> </em>of −6y and 6y, which result in 0, so they are both ELIMINATED.

** [3, 1] is your solution.

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Find the derivative of f(x) = 6x + 7 at x = 9
Ket [755]

Answer:

f'(9) = 6

Step-by-step explanation:

f(x) = 6x+7

The derivative is

f'(x) = 6

Evaluated at  x=9

f'(9) = 6

3 0
3 years ago
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a language arts test is worth 100 points there is a total of 26 questions . there are spelling word questions that ae worth 2 po
natka813 [3]

10 question of spelling words and 16 questions of vocabulary are present

<em><u>Solution:</u></em>

Let "x" be the number of spelling word questions

Let "y" be the number of vocabulary word question

<em><u>There is a total of 26 questions. Therefore, we get</u></em>

number of spelling word questions + number of vocabulary word question = 26

x + y = 26 --------- eqn 1

<em><u>There are spelling word questions that worth 2 points each and vocabulary word question worth 5 points each</u></em>

The language arts test is worth 100 points

Therefore, we frame a equation as:

number of spelling word questions x 2 + number of vocabulary word question x 5 = 100

x \times 2 + y \times 5 = 100

2x + 5y = 100 --------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

From eqn 1,

x = 26 - y ------- eqn 3

<em><u>Substitute eqn 3 in eqn 2</u></em>

2(26 - y) + 5y = 100

52 - 2y + 5y = 100

3y = 100 - 52

3y = 48

<h3>y = 16</h3>

<em><u>Substitute y = 16 in eqn 3</u></em>

x = 26 - 16

<h3>x = 10</h3>

Thus 10 question of spelling words and 16 questions of vocabulary are present

4 0
3 years ago
Can someone help me with this problem://
fiasKO [112]

You can simply collect terms, subtract the constant and divide by the x-coefficient. It is generally considered easier to do those steps if you eliminate fractions first (multiply by 12).

Multiply by 12

... 4(x -1) +3(x +5) = 6

... 4x -4 +3x +15 = 6 . . . . . eliminate parentheses

... 7x +11 = 6 . . . . . . . . . . . .collect terms

... 7x = -5 . . . . . . . . . . . . . . subtract the constant 11

... x = -5/7 . . . . . . . . . . . . . divide by the x-coefficient

_ _ _ _ _ _ _

Here it is the other way.

... x(1/3 +1/4) +(-1/3 +5/4) = 1/2

... (7/12)x + 11/12 = 1/2 . . add the fractions to finish collecting terms

... x + 11/7 = 6/7 . . . . . . . multiply by 12/7

... x = -5/7 . . . . . . . . . . . subtract 11/7

At the third step here, you could subtract 11/12 before doing the multiply. You get the same answer, but you have to do the extra conversion of 1/2=6/12.

8 0
3 years ago
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