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2 years ago

Taking a test will mark braniniliest

Mathematics

2 answers:

Vlad [161]2 years ago

8 0

Answer:

whats the question?

erastovalidia [21]2 years ago

5 0

Answer:

Five hundred and three, you multiply the 5 by the 10 to the second power and get 500 because 10 to the second power is 10x10 which is 100, then add the three

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Can someone answer this??

kifflom [539]

Answer:

I think x will stay the same

5 0

2 years ago

Solve the following proportion for X

mixer [17]

Answer: 3.7 (to nearest tenth)

Working:
7/17 = x/9
63 = 17x
x = 63/17
x = 3.7 (to nearest tenth)

8 0

3 years ago

Which of the following gives a valid reason for using the given solution method to solve the system of equations shown? Equation

alukav5142 [94]

Answer:

* Elimination; a coefficient in Equation I is an integer multiple of a coefficient in Equation II.

* Elimination; a coefficient in Equation II is an integer multiple of a coefficient in Equation I.

Step-by-step explanation:

Equation I: 4x − 5y = 4

Equation II: 2x + 3y = 2

These equation can only be solved by Elimination method

Where to Eliminate x :

We Multiply Equation I by a coefficient of x in Equation II and Equation II by the coefficient of x in Equation I

Hence:

Equation I: 4x − 5y = 4 × 2

Equation II: 2x + 3y = 2 × 4

8x - 10y = 20

8x +12y = 6

Therefore, the valid reason using the given solution method to solve the system of equations shown is:

* Elimination; a coefficient in Equation I is an integer multiple of a coefficient in Equation II.

* Elimination; a coefficient in Equation II is an integer multiple of a coefficient in Equation I.

4 0

3 years ago

Suppose f (x) = x^2. Find the graph of f(x+2)

NISA [10]

$\bf ~~~~~~~~~~~~\textit{function transformations}
\\\\\\
% templates
f(x)={{ A}}({{ B}}x+{{ C}})+{{ D}}
\\\\
~~~~y={{ A}}({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)={{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\\\
f(x)={{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\\\
f(x)={{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\\\\
--------------------$

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

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

with that template in mind,

$\bf f(x)=x^2\qquad \quad f(x+2)=(x+2)^2\implies f(x+2)=\stackrel{A}{1}(\stackrel{B}{1}x\stackrel{C}{+2})^2\stackrel{D}{+0}$

C = 2 B = 1 C/B = 2/1 or +2, horizontal left shift of 2 units

f(x) shifted left by 2 units is f(x+2).

8 0

2 years ago

‼️ PLEASEEE HELPPPP ‼️

V125BC [204]

Answer: answer is in the photo

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers