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3 years ago

Factor completely 2x2 + 2x − 24.

Mathematics

2 answers:

nalin [4]3 years ago

7 0

2x^2 + 2x - 24

= 2x^2 + 8x - 6x -24

= 2x(x + 4) -6(x + 4)

= (x+4)(2x-6)

Lina20 [59]3 years ago

5 0

Answer:

Step-by-step explanation:

The trick here is to notice that all 3 terms can be div. by 2:

2(x^2 + x - 12)

Note that -12 factors into -3 * 4 or -4 * 3. Thus,

x^2 + x - 12 = (x-4)(x+3), and so 2x2 + 2x − 24 = 2(x-4)(x+3).

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Y= -0.25x + 4.7<br> Y=4.9x - 1.64

Andreas93 [3]

Answer:

The answer would be (1.23, 4.39)

Step-by-step explanation:

Because they are both equal to y, we can set the equations equal to each other and then solve.

4.9x - 1.64 = -0.25x + 4.7

5.15x - 1.64 = 4.7

5.15x = 6.34

x = 1.23

Now that we have the value for x, we can plug into either equation and find y.

y = -0.25x + 4.7

y = -0.25(1.23) + 4.7

y = -.31 + 4.7

y = 4.39

3 0

3 years ago

Help me with example 2 and 3

PolarNik [594]

Thanks! You got the concept, so use the formula from the top of the page to help you. Use the information provided in the examples and set up the formula with the correct numbers. Then, solve. Then you will find out how much you need to owe.

5 0

3 years ago

For the function given state the period f(t) =6sin(3t-pi/6)-1

lapo4ka [179]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}

\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{function period or frequency}\\
\qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
\qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}$

now, with that template in mind, let's take a peek at yours

$\bf \begin{array}{lllcclll}
f(t)=&6sin(&3t&-\frac{\pi }{6})&-1\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
period\qquad \cfrac{2\pi }{B}\iff\cfrac{2\pi }{3}$

8 0

3 years ago

What is the answer to 53=−19z+5+7z

nikklg [1K]

53= -19z+5+7z
53= -12z+5 (combine -19z and 7z)
48= -12z. ( subract 5 on both sides)
-4=z. (divide by -12 both sides)

8 0

3 years ago

Read 2 more answers

0.007407 to 1 significant figure

OLEGan [10]

Answer:

0.007