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3 years ago

Can somebody help me with this question. It’s due today. Please. Thank you!

Mathematics

1 answer:

Lady bird [3.3K]3 years ago

8 0

Answer:

Step-by-step explanation:

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HELP ME PLEASEEE! Ty:)

Jlenok [28]

I can help you if you mark me as braì LG

5 0

3 years ago

Robert and dustin agree to share the cost of buying a computer that costs $1750. if robert agrees to pay $450 more than dustin,

lawyer [7]

R + d = 1750
r = d + 450

d + 450 + d = 1750
2d + 450 = 1750
2d = 1750 - 450
2d = 1300
d = 1300/2
d = 650 <=== what Dustin pays

r = d + 450
r = 650 + 450
r = 1100 ...Robert pays 1100

7 0

4 years ago

Which of the following is a point slope equation of a line that passed through the points (10,5) and (4-7)

levacccp [35]

Answer:

Step-by-step explanation:

The point-slope form of an equation of a line:

$y-y_1=m(x-x_1)$

m - slope

(x₁, y₁) - point

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (10, 5) and (4, -7). Substitute:

$m=\dfrac{-7-5}{4-10}=\dfrac{-12}{-6}=2$

$y-5=2(x-10)$

5 0

4 years ago

Which equation represents the model shown below

Alona [7]

Answer:

-4x - 3 = -8

Step-by-step explanation:

On our left, we have -x in 4 places = 4(-x) = -4x

And on the left, we also have -1 in 3 places = 3(-1) = -3

Altogether in our left, we have: -4x +(-3)

That is: -4x - 3

On our right, we have -1 in 8 places, which is 8(-1) = -8

Therefore, the equation that represents the model would be:

-4x - 3 = -8

6 0

3 years ago

Write an equation of cosine function with amplitude 2 and a period 4pi

Arisa [49]

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

\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}\\
\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.. hmmm let's see, keeping in mind the template above

so, amplitude of 2, A=2, thus |A| = ±2, so, you can use either

and period of 4π, well, that simply means that

$\bf \cfrac{2\pi }{B}=4\pi \implies \cfrac{2\pi }{4\pi }=B\implies \cfrac{1}{2}=B\\\\
-----------------------------\\\\

\begin{array}{llll}
f(\theta)=&\pm 2cos&\left(\frac{1}{2}\theta \right)\\
&\ \uparrow &\ \uparrow \\
&A&B
\end{array}$

8 0

4 years ago