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Molodets [167]
2 years ago
9

What is the product of 4 2/7 and 11 1/4 ?

Mathematics
1 answer:
kolbaska11 [484]2 years ago
5 0

Answer:

(4 2/7) x (11 1/4) = 48 3/14

Step-by-step explanation:

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Let f(x) = x^3 + 2x^2 + 3x + 1 and g(x)= 4x-5 Find f(x) x g(x)
Brums [2.3K]

▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ { \huge \mathfrak{Answer}}▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪

Given :

f(x) =  {x}^{3}  + 2 {x}^{2}  + 3x + 1

g(x) = 4x - 5

let's solve for f(x) × g(x) :

( {x}^{3}  + 2x {}^{2}  + 3x + 1)(4x - 5)

4x {}^{4}  - 5 {x}^{3}  + 8 {x}^{3  }  - 10 {x}^{2}  + 12 {x}^{2}  - 15x + 4x - 5

4 {x}^{4}  + 3 {x}^{3}  + 2 {x}^{2}   - 11x - 5

therefore, correct choice is : D

8 0
3 years ago
If f(x) = 0, what is x?<br><br> 0 only<br> –6 only<br> –2, 1, or 3 only<br> –6, –2, 1, or 3 only
Angelina_Jolie [31]

Answer:

answer will be 0

plz mark me as brainliest

8 0
3 years ago
Solve the following quadratic equation by factoring:<br> <img src="https://tex.z-dn.net/?f=y%5E%7B2%7D" id="TexFormula1" title="
Furkat [3]

Answer:

y = 8 or y = 0

Step-by-step explanation:

Solve for y over the real numbers:

y^2 - 8 y = 0

Factor y from the left hand side:

y (y - 8) = 0

Split into two equations:

y - 8 = 0 or y = 0

Add 8 to both sides:

Answer: y = 8 or y = 0

4 0
3 years ago
7 Write an equation for the table of values. Explain how you got your answer.
e-lub [12.9K]

Answer:

y= 3x+1

Step-by-step explanation:

(You can describe the steps for your explanation. See below for the attachment)

5 0
2 years ago
What are the zeros of the function<br><img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%202x%20%5E%7B2%7D%20-8x%20-%2010" id="Tex
allsm [11]
Hello there!

To find the zeroes or the roots of a function, you just need to set the function equal to 0 then solve for x

f(x) = 2x² - 8x - 10

Set function equal to 0

2x² - 8x - 10 = 0

Now we can factorize the left side

2(x + 1)(x - 5) = 0

Then set factors equal to 0

x + 1 = 0 or x - 5 = 0

x = 0 - 1 or x = 0 + 5

x = -1 or x = 5

Thus,

The roots are -1 and 5

Let me know if you have any questions. As always, it is my pleasure to help students like you~!
7 0
3 years ago
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