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SSSSS [86.1K]
3 years ago
14

What is the solution to the expression Negative 2 minus 3 minus (negative 4)? A number line going from negative 10 to positive 1

0. –9 –3 –1 5
Mathematics
2 answers:
Gnom [1K]3 years ago
7 0

Answer:

The right answer is C

Step-by-step explanation:

It's right in edg! :)

beks73 [17]3 years ago
5 0

The solution of the expression is -1 ⇒ 3rd answer

Step-by-step explanation:

Let us revise some important rules on the number line

  • If we add number to another number we move to right
  • If we subtract a number from another number we move to left

∵ The expression is negative 2 minus 3 minus (negative 4)

∵ negative 2 = (-2)

∵ minus means (-)

(-2) minus 3 means move 3 units from (-2) to left, so you will pass -3, -4, and stop at -5

∴ (-2) - 3 = -5

∵ minus (negative 4) = - (-4)

- Remember (-)(-) = (+)

∴ minus (negative 4) = + 4

That means move 4 units to the right from -5, then you will pass -4, -3, -2, and stop at -1

∴ -5 + 4 = -1

∴ The expression (-2) - 3 - (-4) = -1

The solution of the expression is -1

Learn more:

You can learn more about the directed numbers in brainly.com/question/10364988

#LearnwithBrainly

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when 6 is subtracted from the square of a number, the result is 5 times the number. Find the negative solution.
sveta [45]

When 6 is subtracted from the square of a number, the result is 5 times the number, then the negative solution is -1

<h3><u>Solution:</u></h3>

Given that when 6 is subtracted from the square of a number, the result is 5 times the number

To find: negative solution

Let "a" be the unknown number

Let us analyse the given sentence

square of a number = a^2

6 is subtracted from the square of a number = a^2 - 6

5 times the number = 5 \times a

<em><u>So we can frame a equation as:</u></em>

6 is subtracted from the square of a number = 5 times the number

a^2 - 6 = 5 \times a\\\\a^2 -6 -5a = 0\\\\a^2 -5a -6 = 0

<em><u>Let us solve the above quadratic equation</u></em>

For a quadratic equation ax^2 + bx + c = 0 where a \neq 0

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

Here in this problem,

a^2-5 a-6=0 \text { we have } a=1 \text { and } b=-5 \text { and } c=-6

Substituting the values in above quadratic formula, we get

\begin{array}{l}{a=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(-6)}}{2 \times 1}} \\\\ {a=\frac{5 \pm \sqrt{25+16}}{2}=\frac{5 \pm \sqrt{49}}{2}} \\\\ {a=\frac{5 \pm 7}{2}}\end{array}

We have two solutions for "a"

\begin{array}{l}{a=\frac{5+7}{2} \text { and } a=\frac{5-7}{2}} \\\\ {a=\frac{12}{2} \text { and } a=\frac{-2}{2}}\end{array}

<h3>a = 6 or a = -1</h3>

We have asked negative solution. So a = -1

Thus the negative solution is -1

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andrey2020 [161]

Answer:

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Step-by-step explanation:

Please look at the figure attached to get more clear solution.

We have given:

FG||CB

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And alternate interior angles on transverse line are equal

So, ∠1=∠4

And ∠4=28^{\circ}

Hence,  ∠1=∠4=28^{\circ}

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90^{\circ}+∠2+28^{\circ}=180^{\circ}

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On simplification we get:

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Answer:

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Step-by-step explanation:

The diameter is a line that splits a circle in half, while the radius is only half of that.

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