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scoray [572]
3 years ago
9

Use the identity below to complete the tasks:

Mathematics
2 answers:
quester [9]3 years ago
7 0

Answer:

a=[2q^{2}r]

b=[3s^{2}t]

Step-by-step explanation:

we have

8q^{6}r^{3}+27s^{6}t^{3}

we know that

8q^{6}r^{3}=(2^{3})(q^{2})^{3}r^{3}=[2q^{2}r]^{3}

27s^{6}t^{3}=(3^{3})(s^{2})^{3}t^{3}=[3s^{2}t]^{3}

therefore

a=[2q^{2}r]

b=[3s^{2}t]

substitute

a^{3} +b^{3}=(a+b)(a^{2} -ab+b^{2})

[2q^{2}r]^{3} +[3s^{2}t]^{3}=([2q^{2}r]+[3s^{2}t])([2q^{2}r]^{2} -[2q^{2}r][3s^{2}t]+[3s^{2}t]^{2})

[2q^{2}r]^{3} +[3s^{2}t]^{3}=([2q^{2}r]+[3s^{2}t])([4q^{4}r^{2}] -6[q^{2}r][s^{2}t]+[9s^{4}t^{2}])

maw [93]3 years ago
7 0

Answer:

a= 2q^2r

b=3s^t

Step-by-step explanation:

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Suppose f is a continuous function on [-2, 2) such that
Leviafan [203]

Answer:

2. a and b only.

Step-by-step explanation:

We can check all of the given conditions to see which is true and which false.

a. f(c)=0 for some c in (-2,2).

According to the intermediate value theorem this must be true, since the extreme values of the function are f(-2)=1 and f(2)=-1, so according to the theorem, there must be one x-value for which f(x)=0 (middle value between the extreme values) if the function is continuous.

b. the graph of f(-x)+x crosses the x-axis on (-2,2)

Let's test this condition, we will substitute x for the given values on the interval so we get:

f(-(-2))+(-2)

f(2)-2

-1-1=-3  lower limit

f(-2)+2

1+2=3 higher limit

according to these results, the graph must cross the x-axis at some point so the graph can move from f(x)=-3 to f(x)=3, so this must be true.

c. f(c)<1 for all c in (-2,2)

even though this might be true for some x-values of of the interval, there are some other points where this might not be the case. You can find one of those situations when finding f(-2)=1, which is a positive value of f(c), so this must be false.

The final answer is then 2. a and b only.

4 0
3 years ago
Please answer this correctly, and you'll recieve brainliest. I need the CORRECT answer.
miv72 [106K]

Answer:

B.  

Plzzzzz give me Brainliest!  

5 0
3 years ago
Answers for these im strugglinh
Rainbow [258]

Answer:

Step-by-step explanation:

40.

Multiples of 3: 3,6,9,12, 15, …

Multiples of 5: 5, 10, 15, 20, …

LCM of 3 and 5 is 15

Multiples of 6: 6, 12, 18, 24,…

Multiples of 9: 9, 18, 27,…

LCM of 6 and 9 is 18

42.

1/2 and 2/3

denominator 2 has multiples: 2, 4, 6, 8, 10, 12 …

denominator 3 has multiples: 3, 6, 9, 12,…

LCD  of 1/2 and 2/3 is 6

equivalent fraction 1/2 = 3(1/2) = 3/6

equivalent fraction 2/3 = 2(2/3) = 4/6

43.

5/12 and 2/15

denominator 12 has multiples: 12, 24, 36, 48, 60…

denominator 15 has multiples: 15, 30, 45, 60…

LCD  of 5/12 and 2/15 is 60

equivalent fraction 5/12 = 5(5/12) = 25/60

equivalent fraction 2/15 = 4(2/15) = 8/60

44.

16/48 has the prime factorization 2*2*2*2/ 2*2*2*2*3

16/48 simplification is 1/3

45.

12/40 has the prime factorization 2*2*3/2*2*2*5

12/40 simplification is 3/ 10

46.

18/63 has the prime factorization 2*3*3/ 3*3*7

18/63 simplification is 2/7

5 0
3 years ago
Х Y
Montano1993 [528]

Answer:

y = 0.5x + 3

Step-by-step explanation:

7 0
2 years ago
SOMEONE HELP ME ON THESE QUESTIONS
dem82 [27]

1.)  The sum(addition) of 21 and 5 times(multiplication) a number f is(=) 61.

f = unknown number/variable     [So 21 plus 5f(5 times f) equals 61]

21 + 5f = 61   [21(one-time) + 5f(number x variable) = 61(total)]

2.)  Seventeen more(addition) than seven times(multiplication) a number j is(=) 87.

j = unknown number/variable    [So 17 plus 7j(7 times j) equals 87]

17 + 7j = 87  

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18 + 0.05n = 50.50  

[Company charges $18 plus five cents per call(n), and the total charge was $50.50]

4.)     s = the number of students

40 + 30s = 220

[Tutor charges $40 plus $30 per student(s), and the total charge was $220]

4 0
3 years ago
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