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marshall27 [118]

3 years ago

5

A telemarketer sells subscriptions for magazines and receives a commission of 20 % on his sales. How much would his commission b

e if he sells $375 worth subscription?

1 answer:

erastovalidia [21]3 years ago

4 0

The commission of telemarketer will be $75.

Step-by-step explanation:

Given,

Worth of subscription sales = $375

Commission rate = 20%

Amount of commission = 20% of worth of subscription sales

Amount of commission = $\frac{20}{100}*375$

Amount of commission = $0.2*375$

Amount of commission = $75

The commission of telemarketer will be $75.

Keywords: percentage, multiplication

Learn more about percentages at:

#LearnwithBrainly

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In triangle xyz, m∠z > m∠x m∠y. which must be true about △xyz? m∠x m∠z < 90° m∠y > 90° ∠x and∠y are complementary m∠x m

Anon25 [30]

m∠X + m∠Y < 90° is true.

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So, ∠X + ∠Y + ∠Z = 180°

Given, m∠Z > m∠X + m∠Y.

This implies that ∠Z is greater than the sum of the angles ∠X and ∠Y.

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1 year ago

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Step-by-step explanation:

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2 years ago

For what value of k is there one solution to the given quadratic equation (k+1)x²+4kx+2=0.

andriy [413]

Answer:

If $k = 1$ or $k = -\frac{1}{2}$ , there is only one solution to the given quadratic equation.

Step-by-step explanation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = (x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

The signal of $\bigtriangleup$ determines how many real roots an equation has:

$\bigtriangleup > 0$ : Two real and different solutions

$\bigtriangleup = 0$ : One real solution

$\bigtriangleup < 0$ : No real solutions

In this problem, we have the following second order polynomial:

$(k+1)x^{2} + 4kx + 2 = 0$ .

This means that $a = k+1; b = 4k; c = 2$

It has one solution if

$\bigtriangleup = 0$

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We can simplify by 8

$2k^{2} - k - 1 = 0$

The solution is:

$k = 1$ or $k = -\frac{1}{2}$

So, if $k = 1$ or $k = -\frac{1}{2}$ , there is only one solution to the given quadratic equation.

4 0

3 years ago