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3 years ago

Which number is divisible by 3? 41 138 137 136

Mathematics

2 answers:

kondaur [170]3 years ago

5 0

Answer:

The number that is divisible by 3 is 138.

138 divided by 3 is 46.

Hope I helped!

SVEN [57.7K]3 years ago

4 0

Answer:

138

Step-by-step explanation:

A quick way to check for divisibility by 3 is to

add the digits together.
If that sum is divisible by 3 with no remainder then the original number is divisible by 3 with no remainder.

41: 4 + 1 = 5. 41 is not divisible by 3 There is a remainder.

138: 1 + 3 + 8 = 12 and 12 is divisible by 3 so 138 is. 138 /3 = 46

137: 1 + 3 + 7 = 11 and 11 is not divisible by 3 so 137 isn't either.

136: 1 + 3 + 6= 10 and 10 is not divisible by 3 so 136 isn't.

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If f(x)=x^3-12x^2+35x-24f(x)=x 3 −12x 2 +35x−24 and f(8)=0f(8)=0, then find all of the zeros of f(x)f(x) algebraically.

Neporo4naja [7]

Answer:

The zeros of f(x) are: (x - 1), (x - 3) and (x - 8)

Step-by-step explanation:

Given

$f(x)=x^3-12x^2+35x-24$

$f(8) = 0$

Required

Find all zeros of the f(x)

If $f(8) = 0$ then:

$x = 8$

And $x - 8$ is a factor

Divide f(x) by x - 8

$\frac{f(x)}{x - 8} = \frac{x^3-12x^2+35x-24}{x - 8}$

Expand the numerator

$\frac{f(x)}{x - 8} = \frac{x^3 - 4x^2 -8x^2 + 3x + 32x - 24}{x - 8}$

Rewrite as:

$\frac{f(x)}{x - 8} = \frac{x^3 - 4x^2 + 3x - 8x^2 +32x - 24}{x - 8}$

Factorize

$\frac{f(x)}{x - 8} = \frac{(x^2 - 4x + 3)(x - 8)}{x - 8}$

Expand

$\frac{f(x)}{x - 8} = \frac{(x^2 -x - 3x + 3)(x - 8)}{x - 8}$

Factorize

$\frac{f(x)}{x - 8} = \frac{(x - 1)(x - 3)(x - 8)}{x - 8}$

$\frac{f(x)}{x - 8} = (x - 1)(x - 3)$

Multiply both sides by x - 8

$f(x) = (x - 1)(x - 3)(x - 8)$

<em>Hence, the zeros of f(x) are: (x - 1), (x - 3) and (x - 8)</em>

7 0

3 years ago

Please help me with 1-4 and 6 and 7

Tom [10]

I could only get number one which was; 1.75

3 0

3 years ago

If f(x) = (x + 9 and g(x) = --6, which describes the range of (f + g)(x)?

S_A_V [24]

Given:

Consider the below figure attached with this question.

To find:

The value of $(f+g)(x)$ .

Solution:

The given functions are:

$f(x)=|x|+9$

$g(x)=-6$

Now,

$(f+g)(x)=f(x)+g(x)$

$(f+g)(x)=|x|+9+(-6)$

$(f+g)(x)=|x|+9-6$

$(f+g)(x)=|x|+3$

We know that,

$|x|\geq 0$

Adding 3 on both sides, we get

$|x|+3\geq 0+3$

$(f+g)(x)\geq 3$

So, $(f+g)(x)\geq 3$ for all values of x.

Hence, the correct option is A.

6 0

3 years ago

5a - 2b -3 + 2b - 6a simplify the expression and then write an equivalent expression. Please anyone help me it's my first questi

svet-max [94.6K]

Step-by-step explanation:

simplify the expression into like terms

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this will make

-a-3

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3 years ago

Read 2 more answers

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Shtirlitz [24]

Answer:

attached below is the solution

Step-by-step explanation:

|x| < 7

= -7 < x < 7

| y | < 3

= -3< y < 3

attached below is the shaded region in the xy-plane

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3 years ago