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3 years ago

Find the system of inequalities representing the given situation?

Mathematics

1 answer:

Diano4ka-milaya [45]3 years ago

5 0

The question was written by the student in a comment because the image contains no question at all.

Kathy sells candles for $3 and flowers for $5. She plans to sell at least 200 items and likes to earn a minimum of $2500.

Answer:

$\left\{\begin{matrix}x + y = 200\\ 3x + 5y = 2500\end{matrix}\right$

Step-by-step explanation:

System of equations

Kathy sells candles for $3 and flowers for $5. Let's set the following variables:

x = number of candles Kathy sells

y = number of flowers Kathy sells

She plans to sell 200 items, thus:

x + y = 200

She also likes to earn $2,500. The equation for this condition is:

3x + 5y = 2500

The system of equations is

$\left\{\begin{matrix}x + y = 200\\ 3x + 5y = 2500\end{matrix}\right$

You might be interested in

The equation -3 + 7 = -10 simplifies to a true statement. TrueFalse

german

Answer:

False.

Step-by-step explanation:

-3 + 7 = 7 - 3 = 4.

6 0

4 years ago

Evaluate 5xyz + 3y for x=2.5, y=12, and z=4

Reika [66]

Answer:

636

Step-by-step explanation:

5xyz+3y

5(2.5)(12)(4)+3(12)

(12.5)(48)+36

600+36

636

8 0

4 years ago

Find the quotient: 15x^4y^2 - 25x^2y^3 + 45xy / 5xy:

salantis [7]

Answer:

3x^3y − 5xy^2 + 9

Step-by-step explanation:

When there is only one denominator term, the quotient can be found by dividing the numerator terms individually.

$\dfrac{15x^4y^2 - 25x^2y^3 + 45xy}{5xy}=\dfrac{15x^4y^2}{5xy}-\dfrac{25x^2y^3}{5xy}+\dfrac{45xy}{5xy}=3x^3y-5xy^2+9$

_____

There are a couple of things to keep in mind here:

an exponent tells the number of times the factor is repeated
equal numbers of identical factors cancel from numerator and denominator, for example x^3/x = (x·x·x)/x = x·x = x^2. (One factor of x cancels.)

6 0

4 years ago

Int var1 = 0b0001; int var2 = 0b1111; int results1 = var1 & var2; int results2 = var1 | var2; int results3 = var1 ^ var2; in

Afina-wow [57]

Int var1 = 0b0001;
int var2 = 0b1111;
int results1 = var1 & var2;
int results2 = var1 | var2;
int results3 = var1 ^ var2;
int printit = results1 + results2 + results3;

what are the values for results1, results2, results3 and printit after executing the code?
notes:
1. faster responses will be obtained if your code is presented line by line (in a file) before posting.
2. please specify language, many languages use the same syntax but could have differences in interpretation
-------------------------------------------------------------------------------

Assuming Java as the language. C is similar.
& bitwise AND &
^ bitwise exclusive OR
| bitwise inclusive OR

So
results1=var1&var2=0b0001&0b1111=0b0001
results2=var1|var2=0b0001&0b1111=0b1111
results3=var1^var2=0b0001&0b1111=0b1110
printit=results1+results2+results3=0b0001+0b1111+0b1110
=0b10000+0b1110
=0b11110

Note: by default, int has 4 signed bytes, ranging from decimal -2147483648 to +2147483647

8 0

4 years ago

Write the given expression as an algebraic expression in x. tan(2 cos^-1(x))

miv72 [106K]

$\bf tan\left[ 2cos^{-1}(x) \right]\implies tan\left[ 2cos^{-1}\left( \frac{x}{1} \right) \right]
\\\\\\
\textit{if we say }cos^{-1}\left( \frac{x}{1} \right)=\theta\textit{ that means }tan\left[ 2cos^{-1}\left( \frac{x}{1} \right) \right]\iff tan(2\theta)\\\\
-----------------------------\\\\
cos^{-1}\left( \frac{x}{1} \right)=\theta\implies cos(\theta)=\cfrac{x}{1}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c}\\\\
-----------------------------\\\\
$

$\bf \textit{again, using the pythagorean theorem to get the opposite side}
\\\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\implies \pm\sqrt{1^2-x^2}=b
\\\\\\
\pm\sqrt{1-x^2}=b\\\\
-----------------------------\\\\
tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\implies tan(2\theta)=\cfrac{2\cdot \frac{\pm\sqrt{1-x^2}}{x}}{1-\left( \frac{\pm\sqrt{1-x^2}}{x} \right)^2}
\\\\\\$

$\bf tan(2\theta)=\cfrac{\frac{\pm2\sqrt{1-x^2}}{x}}{1-\frac{1-x}{x}}\implies 
tan(2\theta)=\cfrac{\frac{\pm2\sqrt{1-x^2}}{x}}{\frac{x-1+x}{x}}
\\\\\\
tan(2\theta)=\cfrac{\pm2\sqrt{1-x^2}}{x}\cdot \cfrac{x}{2x-1}
\implies 
tan(2\theta)=\cfrac{\pm 2\sqrt{1-x^2}}{2x-1}$

6 0

4 years ago