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3 years ago

Of the 400 employees, 5% of them are park managers. How many managers does the park employe

Mathematics

2 answers:

Anvisha [2.4K]3 years ago

8 0

Answer:

The park employed 20 managers.

Step-by-step explanation:

5% of 400

(5 ÷ 100) × 400

0.05 × 400

= 20

Lapatulllka [165]3 years ago

8 0

The answer is definitely 20!

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3 years ago

39. Solve this system: 3x+y=-7 . 4x-y=14A. (–1, 10)B. (1, –10)C. (–1, –10)D. no solution

valkas [14]

Given,

The linear pair of the equation is,

$\begin{gathered} 3x+y=-7 \\ 4x-y=14 \end{gathered}$

Required

The solution of the linear equation.

Taking the equation first as,

$\begin{gathered} 3x+y=-7 \\ y=-7-3x \end{gathered}$

Substituting the value of y in second equation then,

$\begin{gathered} 4x-y=14 \\ 4x+7+3x=14 \\ 7x+7=14 \\ 7x=7 \\ x=1 \end{gathered}$

Substituting the value of x in first equation then,

$\begin{gathered} 3x+y=-7 \\ 3(1)+y=-7 \\ y=-7-3 \\ y=-10 \end{gathered}$

Hence, the solution of the system is (1, -10).

8 0

1 year ago

Need help simplifying this

diamong [38]

The simplified answer is $\frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}$ .

<u>Step-by-step explanation:</u>

$$\frac{3 y+2 x}{z+2 x}-\frac{2 y-3 x}{3 x+y}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}$

To add or subtract denominators of the fraction must be same.

If it is not the same, we must take LCM of the denominators. and so we can add the fractions.

To make the denominator same multiply the 1st term $(\frac{3x+y}{3x+y})$ and 2nd term by $(\frac{z+2x}{z+2x})$

= $\frac{(3 y+2 x)(3 x+y)}{(z+2 x)(3 x+y)}-\frac{(2 y-3 x)(z+2 x)}{(3 x+y)(z+2 x)}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}$

LCM of the denominators is 6x²+ 3xz + 2xy +yz.

Multiply the factors in the numerator.

= $\frac{\left(6 x^{2}+3 y^{2}+11 x y\right)}{(z+2 x)(3 x+y)}-\frac{\left(2 y z+4 x y-3 x z-6 x^{2}\right)}{(3 x+y)(z+2 x)}-\frac{2 z y+6 x z}{6 x^{2}+3 x z+2 x y+y z}$

Now, the denominators are same, you can subtract it.

= $\frac{\left(6 x^{2}+6 x^{2}+11 x y-4 x y-2 y z-2 y z+3 x z-6 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}$

= $\frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}$

Thus the simplified solution is $\frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}$

4 0

3 years ago