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3 years ago

Give an example of a number whose square root and cube root are equal.

Mathematics

1 answer:

RoseWind [281]3 years ago

4 0

The only one I can thing of would be the Number 1

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Describe the transformations necessary to transform f(x) into g(x)

Aleks [24]

Using translation concepts, the transformations are given as follows:

a) The function is horizontally compressed by a factor of 3 and shifted down one unit.

b) The function is shifted right 3 units and vertically stretched by a factor of 2.

<h3>What is a translation?</h3>

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction in it's definition.

Item a:

The transformations are:

x -> 3x, hence the function is horizontally compressed by a factor of 3.

y -> y - 1, hence the function is shifted down one unit.

Item b:

The transformations are:

x -> x - 3, hence the function is shifted right 3 units.

y -> 2y, hence the function is vertically stretched by a factor of 2.

More can be learned about translation concepts at brainly.com/question/4521517

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2 years ago

The drama club is washing. cars for a fundraiser. If the rate continues, how many cars will they was in 4 hours?

solong [7]

You need to be more specific

5 0

3 years ago

24. Laisha paid for her and a friend to go to the movies. She

sammy [17]

Answer:

$26.91

Step-by-step explanation:

9.50 +13.93 =23.43

50.34 - 23.43 = 26.91

4 0

3 years ago

Expand (2x+2)^6<br> 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

Please see the picture .......................

aivan3 [116]

Answer:

Explanation: dont got one

8 0

3 years ago

Read 2 more answers