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2 years ago

For the function f(x)=1/x which of these could be a value of F(x) when x is close to zero

Mathematics

2 answers:

alukav5142 [94]2 years ago

7 0

Answer:

The function in quadrant I, where y is increasing upward.

Step-by-step explanation:

The smaller x becomes, the larger f(x) becomes. The answer is that f(x) or y approaches infinity as x becomes smaller and smaller. This helps to understand why division by zero is undefined.

Here you go! Hope this helps!! :-)

Sindrei [870]2 years ago

4 0

Answer:

-10,000

Step-by-step explanation:

apex ;)

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Find four consecutive multiples of 3 such that the product of the first, third, and fourth is 1080

Dovator [93]

Answer:

<u>Therefore, we have 6, 9, 12, and 15, that are the first, second, third and fourth consecutive multiples of 3, such the product of 6, 12 and 15 is 1,080.</u>

Step-by-step explanation:

Let's find the factors of 1,080, as follows:

1,080 Dividing by 2 (1,080/2)

540 Dividing by 2 (540/2)

270 Dividing by 2 (270/2)

135 Dividing by 3 (135/3)

45 Dividing by 3 (45/3)

15 Dividing by 3 (15/3)

5 Dividing by 5 (5/5)

In consequence, we have:

5 * 3 * 3 * 3 * 2 * 2 * 2

Let's find out what multiples of 3 there are:

5 * 3 = 15 is a multiple of 3,

And 3 * 3 * 2 * 2 * 2 is remaining.

3 * 2 = 6 is also a multiple of 3,

And 3 * 2 * 2 is remaining

3 * 2 * 2 = 12 is also a multiple of 3.

<u>Therefore, we have 6, 9, 12, and 15, that are the first, second, third and fourth consecutive multiples of 3, such the product of 6, 12 and 15 is 1,080.</u>

7 0

2 years ago

WILL GIVE BRAINLIEST: if the roots of the equation x^3 +px^2+qx+r = 0 are 2, 3, and 4, find the values of p, q, and r.

klasskru [66]

8+4p+2q+r=0
27+9p+3q+r=0
64+16p+4q+r=0
P=-9
Q=26
R=-24

5 0

2 years ago

Expand the function f(x)=(3x+2)^4 with the binomial theorem, not factored

lozanna [386]

Step-by-step explanation:

Enter a problem...

Algebra Examples

Popular Problems

Algebra

Expand using the Binomial Theorem (3x+2)^4

(3x+2)4(3x+2)4

Use the binomial expansion theorem to find each term. The binomial theorem states (a+b)n=n∑k=0nCk⋅(an−kbk)(a+b)n=∑k=0n⁡nCk⋅(an-kbk).

4∑k=04!(4−k)!k!⋅(3x)4−k⋅(2)k∑k=04⁡4!(4-k)!k!⋅(3x)4-k⋅(2)k

Expand the summation.

4!(4−0)!0!⋅(3x)4−0⋅(2)0+4!(4−1)!1!⋅(3x)4−1⋅(2)+4!(4−2)!2!⋅(3x)4−2⋅(2)2+4!(4−3)!3!⋅(3x)4−3⋅(2)3+4!(4−4)!4!⋅(3x)4−4⋅(2)44!(4-0)!0!⋅(3x)4-0⋅(2)0+4!(4-1)!1!⋅(3x)4-1⋅(2)+4!(4-2)!2!⋅(3x)4-2⋅(2)2+4!(4-3)!3!⋅(3x)4-3⋅(2)3+4!(4-4)!4!⋅(3x)4-4⋅(2)4

Simplify the exponents for each term of the expansion.

1⋅(3x)4⋅(2)0+4

Hope this helps!

8 0

2 years ago

Read 2 more answers

How do you do this? Distributive property plz help (num 3)

Advocard [28]

-9-(12-3x)=-9-1*(12)-1*(-3)=-9-12+3x=-21+3x=3(x-7)

4 0

2 years ago

Form a quadratic equation with roots (1+√3) and (1-√3)