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2 years ago

PLZ HELP!!! (I am testing brainly users on their skill) 50 points!

Mathematics

1 answer:

ludmilkaskok [199]2 years ago

4 0

Answer:

Step-by-step explanation

You get the answer by subtracting both sides by k and dividing both sides by (x-h)^2.

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a retailer buys a radio from the wholesaler for $75.he then marks up the price by 1/3 and sells it at a discount of 20%.what was

soldier1979 [14.2K]

He makes it up by 1/3=.333%=$75*1.333=$99.98
he then discounts it by 20%=1/5=99.98/5=$19.99
If the original profit is $99.98-$75=$24.98
and then discounted by $19.99
it would be $24.98-$19.99
=
$4.99

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3 years ago

Read 2 more answers

Can somebody prove this mathmatical induction?

Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

$\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2$

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

3 step:

Check for n=k+1 whether the statement

$\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)$

is true.

Start with the left side:

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)$

According to the 2nd step,

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

Substitute it into the $\ast$

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)$

So, you have proved the initial statement

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3 years ago

Caitlin has a movie rental card worth $85. After she rents the first movie, the card's value is $82.25. After she rents the seco

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I wish I could help you but I need help too!!

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2 years ago

If you make 300000 a year how much do you make a week

Fynjy0 [20]

300,000/yr
52 weeks/yr
300,000/52 = 5,769.2 (rounded to the nearest tenth)

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3 years ago

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The quadratic function h(t)=-16.1t^2 + 150 models a balls height, in feet, over time, in seconds, after it is dropped from a 15

zzz [600]

In order to find height from where ball is dropped, you have to find height or h(t) when time or t is zero.So plug in t=0 into your quadratic equation:h(0) = -16.1(0^2) + 150h(0) = 0 +150h(0) = 150 ft is the height from where ball is dropped. When ball hits the ground, the height is zero. So plug in h(t) = 0 and solve for t.0 = -16.1t^2 + 15016.1 t^2 = 150t^2 = 150/16.1t = sqrt(150/16.1)t = ± 3.05Since time cannot be negative, your answer is positive solution i.e. t = 3.05

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3 years ago