Ask question

Ask question

All categories

4 years ago

15

Where do you put the number 2.78 on a number line

1 answer:

Genrish500 [490]4 years ago

3 0

In between 2 and 3. Closer to 3 than 2. A little bit after 2.5

You might be interested in

Khan academy need help with this

rewona [7]

Step-by-step explanation:

6 * 1 = 6

this means the circle can fit with 6 arcs each one of them has a length of 1.

so,

360° / 6 = 60° (Central angle for each arc)

8 0

3 years ago

Which line plot displays a data set with an outlier?<br><br><br> 20 Points + Brainiest

11111nata11111 [884]

The answer would be the the top right choice. an outlier is a point that falls outside the data and that choice is the correct one.

3 0

4 years ago

Read 2 more answers

1. 20-3 + 12 + 6 x 2

chubhunter [2.5K]

1. 41
2. 50
3. 10
4. 75
5. i’m not really sure

6 0

3 years ago

Solve by quadratic equation

Ymorist [56]

<h2>Question :</h2>

$\tt \dfrac{x+2}{x-2} + \dfrac{x-2}{x+2} = \dfrac{5}{6}$

<h2>Answer :</h2>

$\large \underline{\boxed{\bf{x = \dfrac{\pm 2\sqrt{119}}{7}}}}$

<h2>Explanation :</h2>

$\tt : \implies \dfrac{x+2}{x-2} + \dfrac{x-2}{x+2} = \dfrac{5}{6}$

$\tt : \implies \dfrac{(x+2)(x+2) + (x-2)(x-2)}{(x-2)(x+2)} = \dfrac{5}{6}$

$\tt : \implies \dfrac{(x+2)^{2} + (x-2)^{2}}{(x-2)(x+2)} = \dfrac{5}{6}$

<u>Now, we know that</u> :

$\large \underline{\boxed{\bf{(a+b)^{2} = a^{2} + b^{2}+ 2ab}}}$
$\large \underline{\boxed{\bf{(a-b)^{2} = a^{2} + b^{2} - 2ab}}}$
$\large \underline{\boxed{\bf{(a+b)(a-b) = a^{2} - b^{2}}}}$

$\tt : \implies \dfrac{x^{2}+2^{2}+ 2 \times x \times 2 + x^{2}+2^{2} - 2 \times x \times 2 }{x^{2}-2^{2}} = \dfrac{5}{6}$

$\tt : \implies \dfrac{x^{2}+ 4 + \cancel{4x} + x^{2}+ 4 - \cancel{4x}}{x^{2}-4} = \dfrac{5}{6}$

$\tt : \implies \dfrac{x^{2} + x^{2} + 4 + 4}{x^{2}-4} = \dfrac{5}{6}$

$\tt : \implies \dfrac{2x^{2} + 8}{x^{2}-4} = \dfrac{5}{6}$

<u>By cross multiply</u> :

$\tt : \implies (2x^{2} + 8)6= 5(x^{2}-4)$

$\tt : \implies 12x^{2} + 48 = 5x^{2}-20$

$\tt : \implies 12x^{2} + 48 - 5x^{2} + 20 = 0$

$\tt : \implies 7x^{2} + 68 = 0$

$\tt : \implies 7x^{2} + 0x + 68 = 0$

<u>Now, by comparing with ax² + bx + c = 0, we have</u> :

a = 7
b = 0
c = 68

<u>By using quadratic formula</u> :

$\large \underline{\boxed{\bf{x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}}}}$

$\tt : \implies x = \dfrac{-(0) \pm \sqrt{(0)^{2} - 4(7)(68)}}{2(7)}$

$\tt : \implies x = \dfrac{0 \pm \sqrt{0 - 1904}}{14}$

$\tt : \implies x = \dfrac{\pm \sqrt{- 1904}}{14}$

$\tt : \implies x = \dfrac{\pm \sqrt{2\times 2\times 2\times 2\times 7\times 17}}{14}$

$\tt : \implies x = \dfrac{\pm \cancel{2} \times 2\sqrt{7\times 17}}{\cancel{14}}$

$\tt : \implies x = \dfrac{\pm2\sqrt{119}}{7}$

$\large \underline{\boxed{\bf{x = \dfrac{\pm 2\sqrt{119}}{7}}}}$

Hence value of $\bf x =\dfrac{\pm 2\sqrt{119}}{7}$

5 0

3 years ago

Read 2 more answers

What is the equation of the quadratic function that has vertex (-3, 1) and passes through the point (-2, 4)?

Marat540 [252]

Vertex form: $y=a(x-h)^2+k$ , with (h,k) as the vertex.

For this, we will be using vertex form. Firstly, plug the vertex into the vertex form equation:

$y=a(x-(-3))^2+1\\y=a(x+3)^2+1$

Next, we need to solve for a. Plug in (-2,4) into the x and y coordinates to solve for a as such:

$4=a(-2+3)^2+1\\4=a(1)^2+1\\4=a+1\\3=a$

Putting our equation together,<u> it's $y=3(x+3)^2+1$ </u>

<u>*Additional section*</u>

Standard form: $y=ax^2+bx+c$

Converting to standard form as such:

$y=3(x+3)^2+1\\y=3(x^2+6x+9)+1\\y=3x^2+18x+27+1\\y=3x^2+18x+28$

8 0

3 years ago