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Alchen [17]
3 years ago
13

Solve the system of linear equations by elimination. 4x+3y= -5 -x+3y= -10

Mathematics
1 answer:
Georgia [21]3 years ago
5 0

4x + 3y =  - 5 \\ x - 3y = 10 \\  \\ 5x = 5 \\ x = 1 \\  \\ 4 + 3y =  - 5 \\  \:  \: 3y =  - 9 \\ y =  - 3 \\  \\ (1 \:  - 3)
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#8 Please answer the question
galben [10]

Answer:

22°

Step-by-step explanation:

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<b=22°

6 0
3 years ago
Read 2 more answers
Bonjour pourriez vous m'aider avec cette équation du second degré? j'ai du mal à comprendre le concept. merci
olga_2 [115]

If a quadratic equation has solutions, the standard procedure will always work: given an equation like

ax^2+bx+c=0

The solutions (if any) are given by

x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

In your case, you have

a=-9,\ b=0,\ c=1

So the formula becomes

x_{1,2} = \dfrac{\pm\sqrt{36}}{18}= \pm\dfrac{6}{18}=\pm\dfrac{1}{3}

Anyway, this is quite a special case, because you're missing the linear term (since b=0)

This means that you can solve equations like these more easily: rearrange the equation to make it look like ax^2=c. In your case, it becomes

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Divide both sides by 9:

x^2=\dfrac{1}{9}

Now, you know that the square of a number equals 1/9. By definition, it means that this number is the square root of 1/9. Nevertheless, both the square root and its opposite are solutions of the equation, because the minus sign will cancel out when squaring.

So, in general, you have

ax^2=c \iff x^2 = \dfrac{c}{a} \iff x=\pm\sqrt{\dfrac{c}{a}

which of course makes sense, if you're using real numbers, only if c/a>0.

In your case, this becomes

9x^2=1 \iff x^2 = \dfrac{1}{9} \iff x=\pm\sqrt{\dfrac{1}{9}} = \pm\dfrac{1}{3}

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Answer:

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Step-by-step explanation:

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3 years ago
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Elza [17]
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Hope this helps
3 0
4 years ago
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