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3 years ago

HELP ASAP. WILL GIVE BRAINLIEST.

Mathematics

1 answer:

Delvig [45]3 years ago

3 0

Answer:

thr solutions of the answer is x5,they try

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Which is the constant of proportionality for the relationship shown in the graph?

vladimir2022 [97]

Answer:

COP: 2X

Step-by-step explanation:

K= Y/X

K= 4/2

K= 8/4

K= 12/6

K=16/8

8 0

3 years ago

Factorize (x-3)(x-5)(x-7)(x-9)+15

zhannawk [14.2K]

The expanded form of the equation is x^4-24x^3+222x^2-240+960

<h3>Expansion of expression</h3>

Given the expression below;

(x-3)(x-5)(x-7)(x-9)+15

Expand

(x-3)(x-5)(x-7)(x-9) + 15

(x^2-5x-3x+15)(x^2-9x-7x+63) +15

(x^2-8x+15)(x^2-16x+63) + 15

Expand further to have;

x^4-16x^3+63x^2-8x^3+144x^2-504+15x^2-240x+945+15

x^4-24x^3+222x^2-240+960

Hence the expanded form of the equation is x^4-24x^3+222x^2-240+960

Learn more on expansion here: brainly.com/question/29114
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6 0

1 year ago

((MULTIPLE CHOICE ))

lesya692 [45]

Answer:

Neither.

Step-by-step explanation:

you are correct about the pattern, so it's not adding the same number and not multiplying by the same number. So it's neither.

6 0

3 years ago

Read 2 more answers

Translate 7 units right and units up

Fantom [35]

Answer:

point G : (7,4)

point F : (7,1)

point J : (2,4)

point H: (5,4)

Step-by-steWhen you translate you take each point and move it how many units it asks you to move. In this problen you move each point 7 units to the right and 9 units up to get your answer.

8 0

3 years ago

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