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2 years ago

Find the value of f(–2) for the function f(x) = 4x + 10.

Mathematics

1 answer:

Fiesta28 [93]2 years ago

3 0

Answer:

f(-2)=2

Step-by-step explanation:

f(-2)=4x+10

f=(4)(-2)+10

f=-8+10

f=2

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You are graphing Rectangle ABCD in the coordinate plane. The following are three of the vertices of the

egoroff_w [7]

Are you looking for D on the graph? if so, then the final coordinate to complete the rectangle is (-2,-8)

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3 years ago

An urn contains six red balls numbered 1, 2, 3, 4, 5, 6, two white balls numbered 7, 8, and two black balls numbered 9, 10. A ba

Svetradugi [14.3K]

Answer:

2/10 = 1/5

Step-by-step explanation:

To figure out the probability of something, we can take

(number of outcomes of that something) / (number of total outcomes)

Here, we are trying to find the probability that the ball is white. The number of outcomes that are possible with the ball being white is 2, as there are two white balls and you can only pick one. You can pick either of the two white balls, but there is no way to pick one of them two times, pick two of them at once, or pick any other ball and have it be white.

The number of total outcomes is 10. There are 10 balls, and you can only pick one ball at a time. There are only 10 options to choose from.

Therefore, we can plug our numbers into the formula above and get 2/10 = 1/5 as our probability

8 0

2 years ago

PLZZ HELP

Digiron [165]

54 because there are 12 inches in a foot 12 divided by 3 is 4 and 6 divided by 3 is 2. Add the 2 and 4 and multiply that by 9

4 0

2 years ago

Expand (2x+2)^6 How would you find the answer using the binomial theorem?

Yanka [14]

Answer:

Step-by-step explanation:

$\displaystyle\\\sum\limits _{k=0}^n\frac{n!}{k!*(n-k)!}a^{n-k}b^k .\\\\k=0\\\frac{n!}{0!*(n-0)!}a^{n-0}b^0=C_n^0a^n*1=C_n^0a^n.\\\\ k=1\\\frac{n!}{1!*(n-1)!} a^{n-1}b^1=C_n^1a^{n-1}b^1.\\\\k=2\\\frac{n!}{2!*(n-2)!} a^{n-2}b^2=C_n^2a^{n-2}b^2.\\\\k=n\\\frac{n!}{n!*(n-n)!} a^{n-n}b^n=C_n^na^0b^n=C_n^nb^n.\\\\C_n^0a^n+C_n^1a^{n-1}b^1+C_n^2a^{n-2}b^2+...+C_n^nb^n=(a+b)^n.$

$\displaystyle\\(2x+2)^6=\frac{6!}{(6-0)!*0!} (2x)^62^0+\frac{6!}{(6-1)!*1!} (2x)^{6-1}2^1+\frac{6!}{(6-2)!*2!}(2x)^{6-2}2^2+\\\\ +\frac{6!}{(6-3)!*3!} (2a)^{6-3}2^3+\frac{6!}{(6-4)*4!} (2x)^{6-4}b^4+\frac{6!}{(6-5)!*5!}(2x)^{6-5} b^5+\frac{6!}{(6-6)!*6!}(2x)^{6-6}b^6. \\\\$

$(2x+2)^6=\frac{6!}{6!*1} 2^6*x^6*1+\frac{5!*6}{5!*1}2^5*x^5*2+\\\\+\frac{4!*5*6}{4!*1*2}2^4*x^4*2^2+ \frac{3!*4*5*6}{3!*1*2*3} 2^3*x^3*2^3+\frac{4!*5*6}{2!*4!}2^2*x^2*2^4+\\\\+\frac{5!*6}{1!*5!} 2^1*x^1*2^5+\frac{6!}{0!*6!} x^02^6\\\\(2x+2)^6=64x^6+384x^5+960x^4+1280x^3+960x^2+384x+64.$

8 0

1 year ago

Find the compound interest compounded annually on rupees 2700 rate 6 ⅔% per annum and time is 2 years

dalvyx [7]

Answer:

plz mark me as brainliest ...TOOK ME LONG TIME TO TYPE

answer = CI = 3108 RUPEES

Step-by-step explanation:

i am doing the method in which u find the simple interest of first year then second year.

SI FOR 1ST YEAR= P X R X T / 100

SI = 2700 X 20 X 2 / 3 X 100 ( RATE OF INTEREST IS 20 / 3)

SI = 108000/300

SI = 360

SI FOR SECOND YEAR =

P = 2700 + 360= 3060

SI = 3060 X 20 X 2 / 300

SI= 122400 / 300

SI = 408

COMPOUND INTEREST (CI) = PRINCIPLE + SI OF 2ND YEAR

CI = 2700 + 408

CI = 3108 RUPEES

or u can solve be the method

CI = amount - principle

Amount= principle x (change in ratio) raised to time

***************************************

6 0

2 years ago