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tino4ka555 [31]
3 years ago
11

‼️⚠️ANSWER QUICK PLEASE⚠️‼️

Mathematics
2 answers:
Aneli [31]3 years ago
7 0

Answer:

THE ANSWER IS C

Step-by-step explanation:

kumpel [21]3 years ago
5 0
X(yz) is the answer

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What is the value of the remainder if 10x4 – 6x3 + 5x2 – x + 1 is divided by x – 3?
natita [175]

Answer:

Remainder is 691.

Step-by-step explanation:

Given function is 10x^4-6x^3+5x^2-x+1.

Now we need to find remainder if we divide given function 10x^4-6x^3+5x^2-x+1 by (x-3)

(x-3) means plug x=3 into 10x^4-6x^3+5x^2-x+1 to find remainder.

10x^4-6x^3+5x^2-x+1

=10(3)^4-6(3)^3+5(3)^2-(3)+1

=10(81)-6(27)+5(9)-(3)+1

=810-162+45-3+1

=856-162-3

=856-165

=691

Hence remainder is 691.

8 0
3 years ago
Read 2 more answers
How many 1/2-inch binders will fit on a shelf that is 30 inches long?
malfutka [58]
15 divide 30 by 2 and you have your answer

4 0
3 years ago
Find f^-1(x) for f(x) = 1/x^3
kow [346]

Answer: f^{-1}(x) = \frac{\sqrt[3]{x^{2}}}{x}

<u>Step-by-step explanation:</u>

y = \frac{1}{x^{3}}

Inverse is when you swap the x's and y's and then solve for "y":

x = \frac{1}{y^{3}}

y^{3} = \frac{1}{x}

y = \frac{1}{\sqrt[3]{x}}

y = \frac{1}{\sqrt[3]{x}}*(\frac{\sqrt[3]{x}}{\sqrt[3]{x}})^{2}

y = \frac{\sqrt[3]{x^{2}}}{x}

4 0
3 years ago
(5x^2-4X + 7) - (4x + 2)
DanielleElmas [232]
5x’2-4x+7-4x-2
We add the two 4
5x’2-8x+7-2
5x’2-8x+5
The answer is
5x’2 -8x +5
3 0
2 years ago
Which of the following has a solution set of {x | x = 0}?
notka56 [123]

Answer:

(b)  (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) =  {x | x = 0}

Step-by-step explanation:

Here, the given expression is : {x | x = 0}

So, the ONLY element in the given set = {0}

Now, take each option and solve the given expression:

(a)  x + 1 < -1

Adding -1 BOTH sides, we get:

x + 1 -1 < -1  -1

or, x < - 2 ⇒ x = { -∞ , .... , -4,-3}

Also,   x + 1 < 1

Adding -1 BOTH sides, we get:

x + 1 -1 < 1  -1

or, x <0 ⇒ x = { -∞ , .... , -4,-3,-2,-1}

So, (x + 1 < -1) ∩ (x + 1 < 1) =  { -∞ , .... , -4,-3}∩ { -∞ , .... , -4,-3,-2,-1}  

=  { -∞ , .... , -4,-3}

⇒ (x + 1 < -1) ∩ (x + 1 < 1) ≠ {0}

Similarly, solving

(b) (x + 1 ≤ 1) ∩ (x + 1 ≥ 1)

(x + 1 ≤ 1) =  x≤ 0 =   { -∞ , .... , -4,-3,-2,-1, 0}

(x + 1 ≥ 1) =  x ≥ 0 =  {0,1,2,3,... ∞}

⇒ (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) = {0}

(b)(x + 1 < 1) ∩ (x + 1 > 1)

(x + 1 < 1) =  x <  0 =   { -∞ , .... , -4,-3,-2,-1}

(x + 1 > 1) =  x  > 0 =  { 1,2,3,... ∞}

⇒ (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) = Ф

Hence,  (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) =  {x | x = 0}

8 0
3 years ago
Read 2 more answers
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