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3 years ago

Simplify : (4+5√3)^2

Mathematics

1 answer:

algol [13]3 years ago

4 0

Answer:

91 + 40 $\sqrt{3}$

Step-by-step explanation:

(4 + 5 $\sqrt{3}$ )² = (4 + 5 $\sqrt{3}$ ) (4 + 5 $\sqrt{3}$ )

= 4( 4 + 5 $\sqrt{3}$ )+ 5 $\sqrt{3}$ (4 + 5 $\sqrt{3}$ )

= 16 + 20 $\sqrt{3}$ + 20 $\sqrt{3}$ + 25 $\sqrt{9}$

= 16 + 40 $\sqrt{3}$ + 25· 3

= 16 + 40 $\sqrt{3}$ + 75

= 91 + 40 $\sqrt3}$

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The events "male” and "buys lunch” are not independent because

kvv77 [185]

The events "male” and "buys lunch” are not independent because P(male | buys lunch) = 0.4 and P(male) = 0.3.

<h3>Complete question</h3>

James surveyed people at school and asked whether they bring their lunch to school or buy their lunch at school more often. The results are shown below.

Bring lunch: 46 males, 254 females

Buy lunch: 176 males, 264 females

The events "male" and "buys lunch" are not independent because

<h3>How to determine the probability?</h3>

The number of male students is:

Male = 46 + 176

Male = 222

The number of female students is:

Female = 254 + 264

Female = 518

The total student is:

Total = 222 + 518

Total = 740

Next, calculate the probability of selecting a male student

P(Male) = 222/740

P(Male) = 0.3

Of all the 440 that buy lunch, 176 are male,

So, we have:

P(male | buy lunch) = 176 / 440

P(male | buy lunch) = 0.4

Because

P(male | buy lunch) and P(male) are not equal.

Then, the events "male” and "buys lunch” are not independent

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3 0

2 years ago

a car was valued at $41,000 in the year 2009 by 2013 the car value has depreciated to 19,000 if the car value continues to by th

lubasha [3.4K]

Answer:

$6,376.92

Step-by-step explanation:

-Let d be the rate of depreciation per year.

-Therefore, the value after n years can be expressed as:

$A=P(1-d)^n\\\\A=Value \ after \ n \ years\\P=Initial \ Value\\d=Rate \ of \ depreciation\\n=Time \ in \ years$

#We substitute for the years 2009-2013 to solve for d:

$A=P(1-d)^n\\\\19000=41000(1-d)^4\\\\0.475=(1-d)^4\\\\d=1-0.475^{0.25}\\\\d=0.1698$

#We then use the calculated depreciation rate above to solve for A after 10 yrs:

$A=P(1-d)^n\\\\=41000(1-0.1698)^{10}\\\\=\$6,376.92$

Hence, the value of the car after 10 yrs is $6,376.92

3 0

3 years ago

Write the equation of a line with slope -3 containing (-6,2)

Kitty [74]

Answer:

the equation for the line would be y=-1/3x

5 0

4 years ago

Trevon is making payments on a car that costs $26,555. He makes 36 equal payments.If he rounds the equal payments up to the near

scoundrel [369]

Trevon will overpay around $13

Further explanation:

One time payment will be calculated by dividing the total amount by 36 and then the answer will be rounded off to nearest dollar

So,

Given

$Cost\ of\ car=\ 26555\\Number\ of\ payments=36\\Amount\ of\ one\ payment = \frac{26555}{36}\\=737.64$

One time payment is 737.64. Rounding off to nearest dollar gives us 738.

If 738 is paid 36 times,

$Total\ amount\ paid=36*738=26568$

The amount that will be overpaid is:

$=26568-26555=13$

Trevon will overpay around $13

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6 0

3 years ago

98 - 5q for q = 7 help

Molodets [167]

you have to multiply the 5 and the 7 and you get 35 then subtract 98 and 35 and you get 63

3 0

4 years ago

Read 2 more answers