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Svet_ta [14]
3 years ago
6

Can some one please help and please hurry​

Mathematics
1 answer:
fenix001 [56]3 years ago
8 0
Hi don’t know but have a wonderful day
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Find the volume of the right triangle retangular prism
umka21 [38]
The correct answer is C, hope it’s right :)
8 0
2 years ago
Simplify. (√5+4) (√5 - 2)​
son4ous [18]

Answer:

2√5 - 3

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Algebra I</u>

  • Terms/Coefficients
  • Expand by FOIL

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

(√5 + 4)(√5 - 2)

<u>Step 2: Simplify</u>

  1. Expand [FOIL]:                                                                                                  (√5)² - 2√5 + 4√5 - 8
  2. Combine like terms:                                                                                           (√5)² + 2√5 - 8
  3. Evaluate exponents:                                                                                        5 + 2√5 - 8
  4. Combine like terms:                                                                                         2√5 - 3
6 0
2 years ago
Read 2 more answers
Pre-Calc: Find all the zeros of the function.
uysha [10]

The zeros of the polynomial function are y = 4/5, y = -4/5 and y = ±4/5√i and the polynomial as a product of the linear factors is f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

<h3>What are polynomial expressions?</h3>

Polynomial expressions are mathematical statements that are represented by variables, coefficients and operators

<h3>How to determine the zeros of the polynomial?</h3>

The polynomial equation is given as

f(y) = 625y^4 - 256

Express the terms as an exponent of 4

So, we have

f(y) = (5y)^4 - 4^4

Express the terms as an exponent of 2

So, we have

f(y) = (25y^2)^2 - 16^2

Apply the difference of two squares

So, we have

f(y) = (25y^2 - 16)(25y^2 + 16)

Apply the difference of two squares

So, we have

f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

Set the equation to 0

So, we have

(5y - 4)(5y + 4)(25y^2 + 16) = 0

Expand the equation

So, we have

5y - 4 = 0, 5y + 4 = 0 and 25y^2 + 16 = 0

This gives

5y = 4, 5y = -4 and 25y^2 = -16

Solve the factors of the equation

So, we have

y = 4/5, y = -4/5 and y = ±4/5√i

Hence, the zeros of the polynomial function are y = 4/5, y = -4/5 and y = ±4/5√i

How to write the polynomial as a product of the linear factors?

In (a), we have

The polynomial equation is given as

f(y) = 625y^4 - 256

Express the terms as an exponent of 4

So, we have

f(y) = (5y)^4 - 4^4

Express the terms as an exponent of 2

So, we have

f(y) = (25y^2)^2 - 16^2

Apply the difference of two squares

So, we have

f(y) = (25y^2 - 16)(25y^2 + 16)

Apply the difference of two squares

So, we have

f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

Hence, the polynomial as a product of the linear factors is f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

Read more about polynomial at

brainly.com/question/17517586

#SPJ1

5 0
1 year ago
Please look at photo and ignore already selected answer
navik [9.2K]

Answer:

Question (A) firgue b is pallorgram

Question ( B) Only A is a rentengle

Question (C) is already right

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Complete the square to rewrite y-x^2-6x+14 in vertex form. then state whether the vertex is a maximum or minimum and give its co
ycow [4]

Answer:

y= x^2 -6x +(\frac{6}{2})^2 +14 -(\frac{6}{2})^2

And solving we have:

y= x^2 -6x +9 + 14 -9

y= (x-3)^2 +5

And we can write the expression like this:

y-5 = (x-3)^2

The vertex for this case would be:

V= (3,5)

And the minimum for the function would be 3 and there is no maximum value for the function

Step-by-step explanation:

For this case we have the following equation given:

y= x^2 -6x +14

We can complete the square like this:

y= x^2 -6x +(\frac{6}{2})^2 +14 -(\frac{6}{2})^2

And solving we have:

y= x^2 -6x +9 + 14 -9

y= (x-3)^2 +5

And we can write the expression like this:

y-5 = (x-3)^2

The vertex for this case would be:

V= (3,5)

And the minimum for the function would be 3 and there is no maximum value for the function

4 0
3 years ago
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