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maria [59]
3 years ago
13

Find two consecutive odd numbers such that the sum of three-sevenths of the first number and one-third of the second number is e

qual to thirty-eight.
29 and 31
39 and 41
49 and 51
Mathematics
2 answers:
bekas [8.4K]3 years ago
8 0
The correct answer is 49 and 51
zmey [24]3 years ago
3 0

Answer:

49 and 51

Step-by-step explanation:

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The right to be heard and the right to redress. 
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WILL GIVE BRAINLIEST AND 30 POINTS<br><br> What are 2 expressions that are equivalent to 5(2j+3+j)
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5(2j+3+j)
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= 10j + 15 + 5j connect like terms j
= 15j + 15
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I need help on this please
iren [92.7K]

Answer:

m∠ACB = 82°

Step-by-step explanation:

m∠ACB = 180° - m∠ACD

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Please please help me with question 6,, I can’t figure it out.
Norma-Jean [14]

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Step-by-step explanation:

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3 years ago
Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth.
marishachu [46]

Answer:

Thus, the two root of the given quadratic equation  x^2-6=-x  is 2 and -3 .

Step-by-step explanation:

Consider, the given Quadratic equation, x^2-6=-x

This can be written as ,  x^2+x-6=0

We have to solve using quadratic formula,

For a given quadratic equation ax^2+bx+c=0 we can find roots using,

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}  ...........(1)

Where,  \sqrt{b^2-4ac} is the discriminant.

Here, a = 1 , b = 1 , c = -6

Substitute in (1) , we get,

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

\Rightarrow x=\frac{-(1)\pm\sqrt{(1)^2-4\cdot 1 \cdot (-6)}}{2 \cdot 1}

\Rightarrow x=\frac{-1\pm\sqrt{25}}{2}

\Rightarrow x=\frac{-1\pm 5}{2}

\Rightarrow x_1=\frac{-1+5}{2} and \Rightarrow x_2=\frac{-1-5}{2}

\Rightarrow x_1=\frac{4}{2} and \Rightarrow x_2=\frac{-6}{2}

\Rightarrow x_1=2 and \Rightarrow x_2=-3

Thus, the two root of the given quadratic equation x^2-6=-x is 2 and -3 .

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3 years ago
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