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3 years ago

Solve for y in the equation 4y+15=-5y+9 a)-6 b) -2/3 c)2/3 d)6

2 answers:

7 0

Your goal is to isolate y

To begin isolating y, perform the opposite operation by subtracting 15 on both sides of the equation

4y = -5y - 6

Perform the opposite operation by adding 5y on both sides of the equation

9y = -6

Perform the opposite operation by dividing both sides of the equation by 9

y= $y= -\frac{1}{3}$

galina1969 [7]3 years ago

3 0

First off, you need to have your ys on the same side. Here's how to do it:
4y + 15 = -5y + 9
4y + 5y = 9 - 15 (once a number goes to the other side, it changes its sign)
9y = -6
y = -6/9
y = -2/3

The correct answer is B. -2/3

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3 years ago

Mae has a number cube with 6 sides that are numbered 1 through 6. What is the probability that Mae will roll an odd number?

Stels [109]

Answer:

C) $P(E) = \frac{3}{6}$

The probability that Mae will roll an odd number

$P(E) = \frac{3}{6}$

Step-by-step explanation:

Step(i):-

Given that Mae has a number cube with 6 sides that are numbered 1 through 6.

n(S) = { 1,2,3,4,5,6,} = 6

Let 'E ' be the event of odd numbers

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n(E) = {1, 3, 5} = 3

Step(ii):-

The probability that Mae will roll an odd number

$P(E) = \frac{n(E)}{n(S)} = \frac{3}{6}$

$P(E) = \frac{3}{6}$

4 0

3 years ago

Describe the transformation of the graph of f into the graph of g as either a horizontal or vertical stretch. f(x)=sqrt(x) and g

SpyIntel [72]

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

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

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

with that template in mind, let's see $\bf f(x)=\sqrt{x}\qquad 
\begin{array}{llll}
g(x)=&\sqrt{0.5x}\\
&\quad \uparrow \\
&\quad B
\end{array}$

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