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2 years ago

Wrote the slope-intercept form of the equation of each line.

Mathematics

1 answer:

spayn [35]2 years ago

6 0

It will be .
y=3/2 + 4

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For which interval does this quadratic inequality hold true? 6x − x2 − 8 > 0

xz_007 [3.2K]

$6x-x^2-8 \ \textgreater \ 0 \\
-x^2+6x-8\ \textgreater \ 0 \\
-x^2+2x+4x-8\ \textgreater \ 0 \\
-x(x-2)+4(x-2)\ \textgreater \ 0 \\
(-x+4)(x-2)\ \textgreater \ 0 \\ \\
\hbox{the zeros:} \\
(-x+4)(x-2)=0 \\
-x+4=0 \ \lor \ x-2=0 \\
x=4 \ \lor \ x=2$

The coefficient of x is negative, so the parabola opens downwards. The values greater than 0 are between the zeros.
The solution set for the inequality is:
$x \in (2,4)$

The answer is B.

8 0

3 years ago

Read 2 more answers

The diagram below shows part of the graph of y=g(x) .

son4ous [18]

9514 1404 393

Answer:

see attached

Step-by-step explanation:

The related function is reflected across the x-axis and translated up 3 units. Effectively, it is reflected across the line y = 3/2. The graph of the related function is shown in blue in the attachment.

6 0

2 years ago

The number of earthquake tremors in a 12-month period appears to be distributed as a Poisson random variable with a mean of 6. A

vova2212 [387]

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

Download docx

3 0

2 years ago

What number has the same value as 50 tens

Tju [1.3M]

we know that

One ten is equal to $10$

therefore

To find $50$ tens

you must times $50$ by $10$

$50*10=500$

therefore

the answer is

$500$

3 0

2 years ago

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Find cotθ, cosθ, and secθ, where θ is the angle shown in the figure.

Darina [25.2K]

Answer:

$\cos( \theta) = \frac{5}{8} \\ {8}^{2} = {5}^{2} + {opp}^{2} \\ {opp}^{2} = 64 - 25 = 39 \\ opp = \sqrt{39} \\ \sin(\theta) = \frac{\sqrt{39}}{8} \\ \tan(\theta) = \frac{ \sqrt{39} }{5} \\ \cot(\theta) = \frac{1}{\tan(\theta)} = \frac{1}{\frac{ \sqrt{39} }{5}} \\ \cot(\theta) = \frac{5}{ \sqrt{39} } \\ \csc( \theta) = \frac{1}{\sin(\theta)} = \frac{1}{\frac{\sqrt{39}}{8}} \\ \csc( \theta) = \frac{8}{ \sqrt{39} } \\ \sec( \theta) = \frac{1}{\cos( \theta) } = \frac{1}{ \frac{5}{8} } \\ \sec( \theta) = \frac{8}{5}$

5 0

2 years ago