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3 years ago

Round to underline digit 3 in 32, 336

1 answer:

5 0

I dont know if you know this, but there are 3 different 3's. So how can we figure out the question if we don't know which one is underlined?

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Six more than two times a certain number is less than the number increased by twenty.

Amanda [17]

2x + 6 < x + 20
- x - x
x + 6 < 20
- 6 - 6
x < 14

Solution Set: (-∞, 14)

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3 years ago

5 2/5 x 9 2/10<br><br> Please help *multiplying fractions

Maurinko [17]

Answer:

18/5 or 3 3/5

Step-by-step explanation:

Depends if you want it simplified

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3 years ago

Factor the common factor out of 8x^3+12x

Agata [3.3K]

The answer is, 4x(2x^2+3)

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4 years ago

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PLEASE HELP!<br> Simplify:<br> 2a^n−1·3a^1−n

stealth61 [152]

The answer to your problem is
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8 0

4 years ago

Find the center that eliminates the linear terms in the translation of 4x^2 - y^2 + 24x + 4y + 28 = 0.(-3, 2)(-3,- 2)(4, 0)

baherus [9]

Step 1

Given;

$4x^2-y^2+24x+4y+28=0$

Required; To find the center that eliminates the linear terms

Step 2

$\begin{gathered} 4x^2-y^2+24x+4y=-28 \\ 4x^2+24x-y^2+4y=-28 \\ Complete\text{ the square }; \\ 4x^2+24x \\ \text{use the form ax}^2+bx\text{ +c} \\ \text{where} \\ a=4 \\ b=24 \\ c=0 \end{gathered}$ $\begin{gathered} consider\text{ the vertex }form\text{ of a }parabola \\ a(x+d)^2+e \\ d=\frac{b}{2a} \\ d=\frac{24}{2\times4} \\ d=\frac{24}{8} \\ d=3 \end{gathered}$ $\begin{gathered} Find\text{ the value of e using }e=c-\frac{b^2}{4a} \\ e=0-\frac{24^2}{4\times4} \\ e=0-\frac{576}{16}=-36 \end{gathered}$

Step 3

Substitute a,d,e into the vertex form

$\begin{gathered} a(x+d)^2+e \\ 4(x+_{}3)^2-36 \end{gathered}$ $\begin{gathered} 4(x+3)^2-36-y^2+4y=-28 \\ 4(x+3)^2-y^2+4y=\text{ -28+36} \\ \\ \end{gathered}$

Step 4

Completing the square for -y²+4y

$\begin{gathered} \text{use the form ax}^2+bx\text{ +c} \\ \text{where} \\ a=-1 \\ b=4 \\ c=0 \end{gathered}$ $\begin{gathered} consider\text{ the vertex }form\text{ of a }parabola \\ a(x+d)^2+e \\ d=\frac{b}{2a} \\ d=\text{ }\frac{4}{2\times-1} \\ d=\frac{4}{-2} \\ d=-2 \end{gathered}$ $\begin{gathered} Find\text{ the value of e using }e=c-\frac{b^2}{4a} \\ e=0-\frac{4^2}{4\times(-1)} \\ \\ e=0-\frac{16}{-4} \\ e=4 \end{gathered}$

Step 5

Substitute a,d,e into the vertex form

$\begin{gathered} a(y+d)^2+e \\ =-1(y+(-2))^2+4 \\ =-(y-2)^2+4 \end{gathered}$

Step 6

$\begin{gathered} 4(x+3)^2-y^2+4y=\text{ -28+36} \\ 4(x+3)^2-(y-2)^2+4=-28+36 \\ 4(x+3)^2-(y-2)^2=-28+36-4 \\ 4(x+3)^2-(y-2)^2=4 \\ \frac{4(x+3)^2}{4}-\frac{(y-2)^2}{4}=\frac{4}{4} \\ (x+3)^2-\frac{(y-2)^2}{2^2}=1 \end{gathered}$

Step 7

$\begin{gathered} \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1 \\ \text{This is the }form\text{ of a hyperbola.} \\ \text{From here } \\ a=1 \\ b=2 \\ k=2 \\ h=-3 \end{gathered}$

Hence the answer is (-3,2)

4 0

1 year ago