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4 years ago

Can someone please help me?

Mathematics

1 answer:

KonstantinChe [14]4 years ago

5 0

Negative Integers are :

Less than zero
To the left of zero on a number line.

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PLEASE HELP FAST <br><br>WILL MARK BRAINLIST

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Answer:

1, (c) 2,(c) 3,(a)

Step-by-step explanation:

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3 years ago

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3 years ago

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What is the value of 4(3x + 5) when x = 11?

navik [9.2K]

Answer:

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Step-by-step explanation:

3(11)= 33

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3 years ago

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3 years ago

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Use the Quadratic Formula to solve x2 + 20x + 98 = 0

Lubov Fominskaja [6]

$ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\if\ \Delta \ \textless \ 0\ then\ no\ solution\\\\if\ \Delta =0\ then\ one\ solution\ x_0=\dfrac{-b}{2a}\\\\if\ \Delta \ \textgreater \ 0\ then\ two\ solutions\ x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\-----------------------------$

$x^2+20x+98=0\\a=1;\ b=20;\ c=98\\\\\Delta=20^2-4\cdot1\cdot98=400-392=8 \ \textgreater \ 0\\\sqrt\Delta=\sqrt8=\sqrt{4\cdot2}=\sqrt4\cdot\sqrt2=2\sqrt2\\\\x_1=\dfrac{-20-2\sqrt2}{2\cdot1}=\dfrac{-20-2\sqrt2}{2}=-10-\sqrt2\\\\x_2=\dfrac{-20+2\sqrt2}{2\cdot1}=\dfrac{-20+2\sqrt2}{2}=-10+\sqrt2$

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3 years ago