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3 years ago

Which sets of side lengths could from a triangle? choose all that apply.

Mathematics

2 answers:

Maurinko [17]3 years ago

7 0

Answer:

Step-by-step explanation:

andreev551 [17]3 years ago

5 0

Answer:

d and e

Step-by-step explanation:

for them to be possible side lengths of a triangle the two shorter sides added together have to be greater than the longest side and an answer D and e 6 + 4 is greater than 9 and 5 + 5 is greater than 9 to so those are possible side lengths

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PLEASE ANSWER I REALLY NEED IT ASAP (30points) ZOOM IN TO SEE IT BETTER

arsen [322]

Answer:

x = 1 ; x = 8

Step-by-step explanation:

6x² - 5x + 10 = 5x² + 4x + 2

x² - 9x + 8 = 0

x² - 8x - x + 8 = 0

x(x - 8) - (x -8) = 0

(x - 8)(x - 1) = 0

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4 0

3 years ago

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HELP <br> Please don’t answer if you don’t know it.

inysia [295]

Answer:

d=5

Step-by-step explanation:

4 0

3 years ago

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Solve for the variable x+9=12

labwork [276]

Answer:

Step-by-step explanation:

subtract 9 from both sides

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3 years ago

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If x is 1 what is the value of y?9x - 5y = 29

mylen [45]

Given

The equation,

9x - 5y = 29.

To find the value of y when x is 1.

Explanation:

It is given that,

$9x-5y=29$

That implies,

When x=1,

$\begin{gathered} 9(1)-5y=29 \\ 9-5y=29 \\ -5y=29-9 \\ -5y=20 \\ y=\frac{20}{-5} \\ y=-4 \end{gathered}$

Hence, the value of y is -4.

4 0

1 year ago

Solve 5x^2 - 7x + 2 = 0 by completing the square.

Reika [66]

Answer: The correct option is (c) $\dfrac{49}{100}.$

Step-by-step explanation: We are given to solve the following quadratic equation by the method of completing the square:

$5x^2-7x+2=0~~~~~~~~~~~~~~~~~~~(i)$

Also, we are to find the constant added on both sides to form the perfect square trinomial.

We have from equation (i) that

$5x^2-7x+2=0\\\\\Rightarrow x^2-\dfrac{7}{5}x+\dfrac{2}{5}=0\\\\\\\Rightarrow x^2-2\times x\times \dfrac{7}{10}+\left(\dfrac{7}{10}\right)^2+\dfrac{2}{5}=\left(\dfrac{7}{10}\right)^2\\\\\\\Rightarrow \left(x-\dfrac{7}{10}\right)^2=\dfrac{49}{100}-\dfrac{2}{5}\\\\\\\Rightarrow \left(x-\dfrac{7}{10}\right)^2=\dfrac{49-40}{100}\\\\\\\Rightarrow \left(x-\dfrac{7}{10}\right)^2=\dfrac{9}{100}\\\\\\\Rightarrow x-\dfrac{7}{10}=\pm\dfrac{3}{10}\\\\\\\Rightarrow x=\pm\dfrac{3}{10}+\dfrac{7}{10}.$

So,

$x=\dfrac{3}{10}+\dfrac{7}{10},~~~~~~~~x=-\dfrac{3}{10}+\dfrac{7}{10}\\\\\\\Rightarrow x=\dfrac{10}{10},~~~~~~~~\Rightarrow x=\dfrac{-3+7}{10}\\\\\\\Rightarrow x=1,~-\dfrac{2}{5}.$

Thus, the required solution is $x=1,~-\dfrac{2}{5}.$ and the value of the constant added is $\dfrac{49}{100}.$

Option (c) is correct.

3 0

3 years ago

Which sets of side lengths could from a triangle? choose all that apply.​

Which sets of side lengths could from a triangle? choose all that apply.