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3 years ago

Please help if you’re good at math need an answer asap!!

Mathematics

1 answer:

mars1129 [50]3 years ago

8 0

Answer:

Zero real roots:

f(x) = x^2-x+1

f(x) = x^2+2x+3

One real root:

Two real roots:

f(x)=x^2+2x+1

f(x)=x^2-3x+2

Step-by-step explanation:

These were determined by using the quadratic formula.

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Use the given conditions to write an equation for the line in point-slope form and general form. Passing through (8 comma negat

miskamm [114]

Answer:

The answer to your question is y = -5x + 32 point-slope form

5x + y - 32 = 0 general form

Step-by-step explanation:

Data

(8, -8)

⊥ x - 5y - 6 = 0

Process

1.- Get the slope of the line given

x - 5y - 6 = 0

-5y = -x + 6

y = -x/-5 + 6/-5

y = x/5 - 6/5

slope = 1/5

slope of the new line -5, because the lines are perpendicular

2.- Get the equation of the new line

y - y1 = m(x - x1)

y + 8 = -5(x - 8)

y + 8 = -5x + 40

y = -5x + 40 - 8

y = -5x + 32 point-slope form

Equal to zero to find the general form

5x + y - 32 = 0 general form

5 0

3 years ago

Find the explicit formula for the given sequence.

Art [367]

Answer:

Option a.) an = -4(-3)^(n-1)

4 0

2 years ago

What is the value of cd in the equation 32cd=11cd-42?

USPshnik [31]

Answer:

32cd-11cd=-42

21cd=-42

cd=-42/21

cd=(-2)

the answer is d

7 0

3 years ago

-6x – 4y = -14 4x – 12y = -20 a. (1, 2) b. (-2, 1) c. (-2,4) d. (-2,-4)

dangina [55]

It’s probably A. (1,2) hope this helps..

5 0

3 years ago

Read 2 more answers

Explain how to multiply the following whole numbers 21 x 14

Lesechka [4]

Answer:

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

________

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

Step-by-step explanation:

Given

$21\:\times \:14$

Line up the numbers

$\begin{matrix}\space\space&2&1\\ \times \:&1&4\end{matrix}$

Multiply the top number by the bottom number one digit at a time starting with the ones digit left(from right to left right)

Multiply the top number by the bolded digit of the bottom number

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

Multiply the bold numbers: 1×4=4

$\frac{\begin{matrix}\space\space&2&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&\space\space&4\end{matrix}}$

Multiply the bold numbers: 2×4=8

$\frac{\begin{matrix}\space\space&\textbf{2}&1\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}$

Multiply the top number by the bolded digit of the bottom number

$\frac{\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}$

Multiply the bold numbers: 1×1=1

$\frac{\begin{matrix}\space\space&\space\space&2&\textbf{1}\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&\space\space&1&\space\space\end{matrix}}$

Multiply the bold numbers: 2×1=2

$\frac{\begin{matrix}\space\space&\space\space&\textbf{2}&1\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&2&1&\space\space\end{matrix}}$

Add the rows to get the answer. For simplicity, fill in trailing zeros.

$\frac{\begin{matrix}\space\space&\space\space&2&1\\ \space\space&\times \:&1&4\end{matrix}}{\begin{matrix}\space\space&0&8&4\\ \space\space&2&1&0\end{matrix}}$

adding portion

$\begin{matrix}\space\space&0&8&4\\ +&2&1&0\end{matrix}$

Add the digits of the right-most column: 4+0=4

$\frac{\begin{matrix}\space\space&0&8&\textbf{4}\\ +&2&1&\textbf{0}\end{matrix}}{\begin{matrix}\space\space&\space\space&\space\space&\textbf{4}\end{matrix}}$

Add the digits of the right-most column: 8+1=9

$\frac{\begin{matrix}\space\space&0&\textbf{8}&4\\ +&2&\textbf{1}&0\end{matrix}}{\begin{matrix}\space\space&\space\space&\textbf{9}&4\end{matrix}}$

Add the digits of the right-most column: 0+2=2

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

Therefore,

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

________

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

6 0

3 years ago