PLS PLS PLS HELP Honors 9th Grade Math!!!!

7. Estimote the following percentages of 60.<br> 10% =<br> 25% =<br> 50% =<br> 20% =

Orlov [11]

Answer:

10% of 60 = 6

25% of 60 = 15

50% of 60 = 30

20% of 60 = 12

Step-by-step explanation:

3 0

2 years ago

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Identify the domain and range

UkoKoshka [18]

Answer:

Domain: 1 ≤ x ≤ 4

Range : 1 ≤ f(x) ≤ 4

Step-by-step explanation:

The domain of a function f(x) is the limit within which the values of x varies.

Here, in the graph, it shows that the maximum value of x is 4 and the minimum value of x is 1.

Therefore, the domain of the function is 1 ≤ x ≤ 4

Again the range of a function f(x) is the limit within which the values of f(x) vary.

Here, the graph shows that the maximum value of f(x) is 4 and the minimum value of f(x) is 1.

Therefore, the range of the function f(x) is 1 ≤ f(x) ≤ 4. (Answer)

6 0

3 years ago

What percentage of 440 is 325

GuDViN [60]

The answer would be 73.86

3 0

2 years ago

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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

The system of equations may have a unique solution, an infinite number of solutions, or no solution. Use matrices to find the ge

Leno4ka [110]

Answer:

Infinite number of solutions.

Step-by-step explanation:

We are given system of equations

$5x+4y+5z=-1$

$x+y+2z=1$

$2x+y-z=-3$

Firs we find determinant of system of equations

Let a matrix A= $\left[\begin{array}{ccc}5&4&5\\1&1&2\\2&1&-1\end{array}\right]$ and B= $\left[\begin{array}{ccc}-1\\1\\-3\end{array}\right]$

$\mid A\mid=\begin{vmatrix}5&4&5\\1&1&2\\2&1&-1\end{vmatrix}$

$\mid A\mid=5(-1-2)-4(-1-4)+5(1-2)=-15+20-5=0$

Determinant of given system of equation is zero therefore, the general solution of system of equation is many solution or no solution.

We are finding rank of matrix

Apply $R_1\rightarrow R_1-4R_2$ and $R_3\rightarrow R_3-2R_2$

$\left[\begin{array}{ccc}1&0&1\\1&1&2\\0&-1&-3\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\1\\-5\end{array}\right]$

Apply $R_2\rightarrow R_2-R_1$

$\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&-1&-3\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\6\\-5\end{array}\right]$

Apply $R_3\rightarrow R_3+R_2$

$\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&0&-2\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\6\\1\end{array}\right]$

Apply $R_3\rightarrow- \frac{1}{2}$ and $R_2\rightarrow R_2-R_3$

$\left[\begin{array}{ccc}1&0&1\\0&1&0\\0&0&1\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]$

Apply $R_1\rightarrow R_1-R_3$

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$ : $\left[\begin{array}{ccc}-\frac{9}{2}\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]$

Rank of matrix A and B are equal.Therefore, matrix A has infinite number of solutions.

Therefore, rank of matrix is equal to rank of B.

4 0

3 years ago