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Which two numbers on the number line have an absolute value of 1.75? Select the location of both numbers on the number line

hram777 [196]

Solution

To find Which two numbers on the number line have sn

absolute value of 1.75 .

Absolute value means " how far a number is from

zero " .

"1.75" is 1.75 away from zero .

and " -1.75" is also 1.75 away from zero .

i.e

| 0 - 1.75 | = 1.75

| 1.75 - 0 | = 1.75

So , the absolute value of "1.75" is 1.75

similarly , absolute value of "1.75" is 1.75 .

Hence , the two numbers are -1.75 and 1.75 .

8 0

3 years ago

Read 2 more answers

Find the value of x a) 1 b) 31 c) 24 d) 14

Furkat [3]

So the right answer is 14.

Look at the attached picture

Hope it will help you

Good luck on your assignment

6 0

3 years ago

What is the surface area of this 3-D solid?

Dvinal [7]

Answer:

I think its flat sorry if it's wrong

5 0

3 years ago

Carbon-14 dating assumes that the carbon dioxide on Earth today has the same radioactive content as it did centuries ago. If thi

defon

7978 years

Step-by-step explanation:

A = A02^(-t/hl)

where hl = half-life.

Dividing both sides by A0 and taking the logarithm, we get

ln(A/A0) = -(t/hl)ln2

or solving for t,

t= -(hl)ln(A/A0)/ln2

note that A/A0 = 0.38

t = -(5715 yrs)[ln(0.38)/ln2]

= 7978 years

7 0

3 years ago

3x^2+kx=-3 What is the value of K will result in exactly one solution to the equation?

Scorpion4ik [409]

Answer:

For k = 6 or k = -6, the equation will have exactly one solution.

Step-by-step explanation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = (x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

If $\bigtriangleup = 0$ , the equation has only one solution.

In this problem, we have that:

$3x^{2} + kx + 3 = 0$

So

$a = 3, b = k, c = 3$

$\bigtriangleup = b^{2} - 4ac$

$\bigtriangleup = k^{2} - 4*3*3$

$\bigtriangleup = k^{2} - 36$

We will only have one solution if $\bigtriangleup = 0$ . So

$\bigtriangleup = 0$

$k^{2} - 36 = 0$

$k^{2} = 36$

$k = \pm \sqrt{36}$

$k = \pm 6$

For k = 6 or k = -6, the equation will have exactly one solution.

3 0

3 years ago