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3 years ago

Solve the compound inequality 4 (x-5) <12 or x+2>9

Mathematics

1 answer:

arsen [322]3 years ago

8 0

Answer would be -19 and your welcome :)

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Which best describes the dna of a cloned animal

pychu [463]

I think it has the same DNA as its parent

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2 years ago

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How can you use order numbers that are written as fractions, decimals, and percents?

galina1969 [7]

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2 years ago

Carter works in a department store selling clothing. He makes a guaranteed salary of $400 per week, but is paid a commision on t

pentagon [3]

Answer:

$837.5

Step-by-step explanation:

Step one:

given data

He makes a guaranteed salary of $400 per week

Commision on top of his base salary equal to 25% of his total sales for the week

Required

His total salary when he made sales of 25%

Step two

let us solve for 25% of $1750 made

=25/100*$1750

=0.25*$1750

=$437.5

Hence his total payment for the week will be

=$400+$437.5

=$837.5

Also, if the sales is $x his pay would be

let y be his pay

y=0.25x+400

8 0

2 years ago

46225 find squre root

Margarita [4]

Square root means 46225^1/2 so the answer is 65

4 0

2 years ago

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Write the equation of a hyperbola with vertices (0, -4) and (0, 4) and foci (0, -5) and (0, 5).

andre [41]

Check the picture below. So, more or less looks like so.

notice, the center is clearly at the origin, and notice how long the "a" component is, also, bear in mind that, is opening towards the y-axis, that means the fraction with the "y" variable is the positive one.

Also notice, the "c" distance from the center to either foci, is just 5 units.

$\bf \textit{hyperbolas, vertical traverse axis }\\\\
\cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
\qquad 
\begin{cases}
center\ ({{ h}},{{ k}})\\
vertices\ ({{ h}}, {{ k}}\pm a)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{{{ a }}^2+{{ b }}^2}
\end{cases}\\\\
-------------------------------\\\\$

$\bf \begin{cases}
h=0\\
k=0\\
a=4\\
c=5
\end{cases}\implies \cfrac{(y-{{ 0}})^2}{{{ 4}}^2}-\cfrac{(x-{{ 0}})^2}{{{ b}}^2}=1\implies \cfrac{y^2}{16}-\cfrac{x^2}{b^2}=1
\\\\\\
c=\sqrt{a^2+b^2}\implies c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\\\\\\
\sqrt{5^2-4^2}=b\implies \boxed{3=b}
\\\\\\
\cfrac{y^2}{16}-\cfrac{x^2}{3^2}=1\implies \boxed{\cfrac{y^2}{16}-\cfrac{x^2}{9}=1}$

7 0

3 years ago