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3 years ago

Which is greater 6km or 54,000 mm

2 answers:

7 0

1 km = 1,000,000 millimeters
6 km = 6 * 1,000,000 millimeters = 6,000,000 mm

6,000,000 mm is greater than 54.000 mm

Therefore, 6 km is greater than 54,000 mm,

uranmaximum [27]3 years ago

7 0

1 km is 1,000,000 mm. So 6k is the equivalent of 6,000,000 mm. So 6km > 54,000mm

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Simplify: 6^20/6^5 please help me out and answer with the options given below:) ty

AURORKA [14]

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When you simply the fraction you get 6^15 or 470184984576.

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3 years ago

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If a home goods retailer pays $35.50 for the vacuum cleaner shown here, answer the following questions. (Round dollars to the ne

Marianna [84]

The Markup percentage is mathematically given as

M=42%

<h3>What is Mark up percentage?</h3>

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1 year 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

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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)

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