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2 years ago

What ratios are equivalent to 12:9

Mathematics

1 answer:

VARVARA [1.3K]2 years ago

8 0

$\huge\text{Hey there!}$

$\huge\textbf{Equation:}$

$\mathsf{12:9}$

$\huge\textbf{Simplifying:}$

$\mathsf{12:9}$

$\mathsf{= 12\div3 : 9 \div 3}$

$\mathsf{= 4:3}$

$\huge\textbf{Therefore, your answer should be:}$

$\huge\boxed{\frak{4:3}}\huge\checkmark$

$\huge\text{Good luck on your assignment \& enjoy your day!}$

~ $\frak{Amphitrite1040:)}$

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Please answer this and will marked as brainlist

xeze [42]

<h3>Answer: 1</h3>

where x is nonzero

=======================================================

Explanation:

We'll use two rules here

(a^b)^c = a^(b*c) ... multiply exponents
a^b*a^c = a^(b+c) ... add exponents

------------------------------

The portion [ x^(a-b) ]^(a+b) would turn into x^[ (a-b)(a+b) ] after using the first rule shown above. That turns into x^(a^2 - b^2) after using the difference of squares rule.

Similarly, the second portion turns into x^(b^2-c^2) and the third part becomes x^(c^2-a^2)

-------------------------------

After applying rule 1 to each of the three pieces, we will have 3 bases of x with the exponents of (a^2-b^2), (b^2-c^2) and (c^2-a^2)

Add up those exponents (using rule 2 above) and we get

(a^2-b^2)+(b^2-c^2)+(c^2-a^2)

a^2-b^2+b^2-c^2+c^2-a^2

(a^2-a^2) + (-b^2+b^2) + (-c^2+c^2)

0a^2 + 0b^2 + 0c^2

0+0+0

All three exponents add to 0. As long as x is nonzero, then x^0 = 1

4 0

3 years ago

This is my warmup can you please help me out with it

Makovka662 [10]

Answer:,

the answer is (4,2)

Step-by-step explanation:

is you look at the 2 equations and plug in the points where the lines intercets you would get the answer soo examples

4+2=6

the first one checks out

the second one

4+(2*2)=8

so the parinthises go first 2 times 2 = 4 plus 4 = 8

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3 years ago

The graph of f(x) = x^2 has been shifted into the form f(x) = (x − h)^2 + k

stepan [7]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the y-axis}$

$\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}$

with that template in mind, let's see, it went to the right 2 units, and then up 3 units.

that simply means, C = -2, D = 3.

8 0

3 years ago

I need help with only one question on number 4? anyone pls?

Murljashka [212]

I think 2,-2 is right

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2 years ago

Let v⃗ 1=⎡⎣⎢033⎤⎦⎥,v⃗ 2=⎡⎣⎢1−10⎤⎦⎥,v⃗ 3=⎡⎣⎢30−3⎤⎦⎥ be eigenvectors of the matrix A which correspond to the eigenvalues λ1=−1, λ2

kaheart [24]

Answer:

- x as a linear combination :

x = -1 v1+ 0 v2+ 1 v3.

- Transpose Ax = (12, -6, -6)

Step-by-step explanation:

Given v1 = (0 3 3),v2 = (1 −1 0), v3 = (3 0 −3) be eigenvectors of the matrix A which correspond to the eigenvalues λ1 = −1, λ2 = 0, and λ3 = 1, respectively, and let x = (−2 −4 0). Express x as a linear combination of v1, v2, and v3, and find Ax .

To write x as a linear combination of v1, v2, and v3

x = -1 v1+ 0 v2+ 1 v3.

To find Ax

Write A = (0 ......3 ......3 )

...................(1 ......-1 ......0)

...................(3 ......0......-3)

Since transpose x = (-2, 4, 0)

Ax =......... (0 ......3......3 )(-2)

...................(1 ......-1 ......0)(4)

...................(3 ......0......-3)(0)

= (0×-2 + 3×4 + 3×0)

...(1×-2 + -1×4 + 0×0)

.. (3×-2 + 0×4 + -3×0)

As = (12)

....(-6)

Transpose Ax = (12, -6, -6)

7 0

3 years ago