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3 years ago

Please help! (ignore the answer I chose)

Mathematics

1 answer:

Andrei [34K]3 years ago

5 0

The numerator factors as (t-8)(t+4), so the whole thing is (t-8)(t+4)/(t-8). Now we can cancel the t-8 AS LONG as t isn't equal to 8, otherwise, we are canceling a zero from both numerator and denominator which is invalid. What is left is t+4.

So your choice was right.

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a certain state the license plate has two digits, then two letters, and then two digits. How many different license plates are p

xxTIMURxx [149]

Step-by-step explanation:

Choosing 4 digits out of 10 = 10C4 = 210.

Arranging the 4 digits in 4 places = 4!

Total number of ways to arrange digits

= 210 * 4! = 5040.

Choosing 2 letters out of 26 = 26C2 = 325

Arranging the 2 letters in 2 places = 2!

Total number of ways to arrange letters

= 325 * 2! = 650.

Total number of possible license plates

= 5040 * 650 = 3,276,000.

7 0

3 years ago

Read 2 more answers

How do you find a vector that is orthogonal to 5i + 12j ?

Rashid [163]

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\
slope=\cfrac{a}{{{ b}}}\qquad negative\implies -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\
-------------------------------\\\\$

$\bf \boxed{5i+12j}\implies 
\begin{array}{rllll}
\ \textless \ 5&,&12\ \textgreater \ \\
x&&y
\end{array}\quad slope=\cfrac{y}{x}\implies \cfrac{12}{5}
\\\\\\
slope=\cfrac{12}{{{ 5}}}\qquad negative\implies -\cfrac{12}{{{ 5}}}\qquad reciprocal\implies - \cfrac{{{ 5}}}{12}
\\\\\\
\ \textless \ 12, -5\ \textgreater \ \ or\ \ \textless \ -12,5\ \textgreater \ \implies \boxed{12i-5j\ or\ -12i+5j}$

if we were to place <5, 12> in standard position, so it'd be originating from 0,0, then the rise is 12 and the run is 5.

so any other vector that has a negative reciprocal slope to it, will then be perpendicular or "orthogonal" to it.

so... for example a parallel to <-12, 5> is say hmmm < -144, 60>, if you simplify that fraction, you'd end up with <-12, 5>, since all we did was multiply both coordinates by 12.

or using a unit vector for those above, then

$\bf \textit{unit vector}\qquad \cfrac{\ \textless \ a,b\ \textgreater \ }{||\ \textless \ a,b\ \textgreater \ ||}\implies \cfrac{\ \textless \ a,b\ \textgreater \ }{\sqrt{a^2+b^2}}\implies \cfrac{a}{\sqrt{a^2+b^2}},\cfrac{b}{\sqrt{a^2+b^2}}
\\\\\\
\cfrac{12,-5}{\sqrt{12^2+5^2}}\implies \cfrac{12,-5}{13}\implies \boxed{\cfrac{12}{13}\ ,\ \cfrac{-5}{13}}
\\\\\\
\cfrac{-12,5}{\sqrt{12^2+5^2}}\implies \cfrac{-12,5}{13}\implies \boxed{\cfrac{-12}{13}\ ,\ \cfrac{5}{13}}$

4 0

3 years ago

Lins father is paying for a 20 dollar meal. He has a 15% - off coupon for the meal. After the discount, a 7% sales tax is applie

fredd [130]

Answer 18.19

20x0.85=17
17x(1.07)=18.19

8 0

3 years ago

In the base-ten number system, ten digits {0, 1, 2, 3, 4, 5, 7, 8, 9) are used for number

SVETLANKA909090 [29]

Answer:

..............I think its wrong .......

6 0

2 years ago

How many possible committees of 4 people can be chosen from 15 men and 20 women so that at least two women should be on each com

Mnenie [13.5K]

If there has to be 2 men and 2 women, we know that we must take a group of 2 men out of the group of 15 men and a group of 2 women out of the group of 20 women. Therefore, we have:

(15 choose 2) x (20 choose 2)

(15 choose 2) = 105

(20 choose 2) = 190

190*105 = 19950

Therefore, there are 19950 ways to have a group of 4 with 2 men and 2women.

If there has to be 1 man and 3 women, we know that we must take a group of 1 man out of the group of 15 men and a group of 3 women out of the group of 20 women. Therefore, we have:

(15 choose 1) x (20 choose 3)

(15 choose 1) = 15

(20 choose 3) = 1140

15*1140 = 17100

Therefore, there are 17100 ways to have a group of 4 with 3 women and 1 man.

We now find the total outcomes of having a group with 4 women.

We know this is the same as saying (20 choose 4) = 4845

Therefore, there are 4845 ways to have a group of 4 with 4 women.

We now add the outcomes of 2 women, 3 women, and 4 women and get the total ways that a committee can have at least 2 women.

19950 + 17100 + 4845 = 41895 ways that there will be at least 2 women in the committee

5 0

3 years ago