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2 years ago

What is the term for protection that guarantees payment to you in the event of financial loss?

1 answer:

5 0

The term for protection that guarantees payment to you in the event of financial loss is the insurance. It is <span>a practice or arrangement by which a company or government agency provides a guarantee of compensation for specified loss, damage, illness, or death in return for payment of a premium.</span>

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A tennis player hits a serve that cannot be returned 45 percent of time. out of 300 serves, how many can you predict will not be

Aleks [24]

135 ; If you are allowed to use a calculator to determine percents you can do 300 × 45% = and it gives you your answer.

8 0

2 years ago

Read 2 more answers

Alma was shopping for a T-shirt at a store that advertised "15% off on everything in the store." The regular price of the T-shir

marshall27 [118]

Answer:

$13.60 is the correct sale price

Step-by-step explanation:

If a t-shirt is 15% off then you pay 85%. This means you will pay 0.85(16) = 13.60 is the correct sale price. $14 is not the correct price.

3 0

3 years ago

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Find the value(s) of x where the tangent to the graph of y=e^5x is parallel to the

Alik [6]

Answer:

There are none.

Step-by-step explanation:

<u>No calculus involved:</u>

The line, in slope-intercept form, has equation $y=-10x+17$ , ie is always decreasing (easy to spot applying the definition)

Meanwhile, $y=e^{5x}$ is always increasing over its domain.

At no point the tangent will be decreasing.

<u>Let's use calculus</u>

We are to solve the equation $y'(x) = -10 \rightarrow 5e^{5x} = -10 \rightarrow e^{5x}=-2$ which has no real solutions.

8 0

2 years ago

X²+Y²=250 and XY=117<br> What are the values of X and Y?

yan [13]

Answer:

Step-by-step explanation:

$x^2+y^2=250\\\\x^2-2xy+y^2+2xy=250\\\\(x-y)^2=250-2xy\\\\(x-y)^2=250-2\cdot117\\\\ (x-y)^2=16\\\\x-y=4\qquad\qquad\vee\qquad \qquad x-y=-4\\\\x=4+y \qquad\qquad \vee\qquad\qquad x=-4+y\\\\(y+4)y=117\qquad\vee\qquad\quad (y-4)y=117\\\\y^2+4y-117=0\qquad\vee\qquad y^2-4y-117=0\\\\y=\dfrac{-4\pm\sqrt{4^2-4(-117)}}{2\cdot1}\qquad\vee\qquad y=\dfrac{4\pm\sqrt{4^2-4(-117)}}{2\cdot1}\\\\y=\dfrac{-4\pm\sqrt{16+468}}{2}\qquad\ \ \vee\qquad y=\dfrac{4\pm\sqrt{16+468}}{2}$

$y_1=\dfrac{-4-22}{2}\ ,\quad y_2=\dfrac{-4+22}{2}\ ,\quad y_3=\dfrac{4-22}{2}\ ,\quad y_4=\dfrac{4+22}{2}\\\\y_1=-13\ ,\qquad y_2=9\ ,\qquad\quad\qquad\ y_3=-9\ ,\qquad y_4=13\\\\x_{1,2}=4+y_{1,2}\qquad\qquad\qquad\qquad\qquad x_{3,4}=-4+y_{3,4}\\\\x_1=-9\ ,\qquad x_2=13\ ,\qquad\quad\qquad x_3=-13\ ,\qquad x_4=9$

6 0

3 years ago

PLEASE HELP ME FOR BRAINLIESTTT!!!

lyudmila [28]

Answer:

Sorry I cant help But I think that well I dont know but if you have any other questions i can help you with I can.

Step-by-step explanation:

4 0

1 year ago