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Georgia [21]
3 years ago
14

What's one third multiply 3

Mathematics
2 answers:
allsm [11]3 years ago
7 0
Simple! 1/3 times 3 equals 3/3, or 1.
ollegr [7]3 years ago
5 0
1/3*3/1
Multiply
Final answer: 1
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ANSWER FOR EXTRA POINTS AND BRAINLIST ⭐️⭐️⭐️⭐️⭐️‼️‼️
Finger [1]

Answer:

c

Step-by-step explanation:

I don't have anything to say but I double checked

8 0
2 years ago
Name the remainder for the quotient and describe or show how you found the remainder.
horsena [70]

Answer:

Step-by-step explanation:

The expression x^3 - x^2 - 17x -15) ÷ (x-5) shows that x+5 is one of the factor if the polynomial function. According to factor theorem, we will equate the factor to zero and find x;

x+5 = 0

x = -5

Next is to generate the remainder.

Substitute x = -5 into the polynomial function

P(x) = x^3 - x^2 - 17x -15

P(-5) = (-5)^3 - (-5)^2 - 17(-5) -15

P(-5) = -125-25+85-15

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P(-5) = -80

Hence the remainder is -80

5 0
3 years ago
4x−1<11 answer my question
Hitman42 [59]

Answer:

Step-by-step explanation:

4 0
2 years ago
Find the area of triangle ABC when b= 141 degrees a= 7 and c=8
Vlad [161]
  You basically use the formula height*base/2 to find the area of the triangle. For instance, let's say a is your chosen base, which has a length of 7. You then use the pythagorean theorem of the right triangle (which is formed by splitting the triangle in half), which is a^2+b^2=c^2, and you substitute half your base for a and the other length (8) for c, which is the hypotenuse of the triangle. Note how this is all being done to find "b", which is the height of the triangle, which will then help you substitute all of your known values into the area formula of a triangle to answer your question. I'm not sure if b=141 degrees would have an impact on this question, but I hope this helped you in some way.
7 0
3 years ago
How to solve this?<br>\int \frac { 4 - 3 x ^ { 2 } } { ( 3 x ^ { 2 } + 4 ) ^ { 2 } } d x​
ivanzaharov [21]

\Large \mathbb{SOLUTION:}

\begin{array}{l} \displaystyle \int \dfrac{4 - 3x^2}{(3x^2 + 4)^2} dx \\ \\ = \displaystyle \int \dfrac{4 - 3x^2}{x^2\left(3x + \dfrac{4}{x}\right)^2} dx \\ \\ = \displaystyle \int \dfrac{\dfrac{4}{x^2} - 3}{\left(3x + \dfrac{4}{x}\right)^2} dx \\ \\ \text{Let }u = 3x + \dfrac{4}{x} \implies du = \left(3 - \dfrac{4}{x^2}\right)\ dx \\ \\ \text{So the integral becomes}  \\ \\ = \displaystyle -\int \dfrac{du}{u^2} \\ \\ = -\dfrac{u^{-2 + 1}}{-2 + 1} + C \\ \\ = \dfrac{1}{u} + C \\ \\ = \dfrac{1}{3x + \dfrac{4}{x}} + C \\ \\ = \boxed{\dfrac{x}{3x^2 + 4} + C}\end{array}

5 0
3 years ago
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