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jeka94
1 year ago
7

Please do #4 It’s it the one on the bottom

Mathematics
1 answer:
spin [16.1K]1 year ago
3 0

We need to graph this equation:

16x+2y=300

Its solutions are the points through which it graph passes. Since it's a linear equation its graph is a straight line and we only need two of its points to draw it. But before graphing let's re-write the equation. We can substract 16x from both sides:

\begin{gathered} 16x+2y=300 \\ 16x+2y-16x=300-16x \\ 2y=300-16x \end{gathered}

And we divide both sides by 2:

\begin{gathered} \frac{2y}{2}=\frac{300-16x}{2}=\frac{300}{2}-\frac{16x}{2} \\ y=150-8x \end{gathered}

So now with this equation if we pick two random x values we'll get their corresponding y values. This way we'll find two points that are part of the graph which is the line that passes through both. We can begin with x=0:

y=150-8\cdot0=150

So the first point is (0,150). Then we can take x=10 and we get:

y=150-8\cdot10=150-80=70

So the second point is (10,70). Then the graph is the line that passes through points (0,150) and (10,70). In order to represent it

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We are given

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\frac{8}{sin(45)}=\frac{BC}{sin(60)}

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6 0
3 years ago
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Answer:

I'm guessing it's 70°

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4 0
2 years ago
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rewrite the polynomial 2y^2+ 6y^3-11-17y^4+8y^5 in the standard form also find its degree and coefficient of y^4
OleMash [197]

Answer:

The standard form is  8 y ⁵ - 17 y⁴ + 6 y³ +2 y² - 11

The degree of given polynomial is '5'

the co-efficient of  y⁴  is '-17'

Step-by-step explanation:

Given standard form 2 y²+ 6 y³-11-17 y⁴+8 y⁵

<em>The form ax² + b x + c is called the standard form of the quadratic expression of 'x'.This is second degree standard form of polynomial.</em>

<em>The form ax⁵ + b x⁴ + c x³ +d x² +ex +f  is called the standard form of the quadratic expression of 'x'.This is fifth degree standard form of polynomial</em>

now Given polynomial is  2 y²+ 6 y³-11-17 y⁴+8 y⁵

The standard form is

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<u><em>Conclusion</em></u>:-

<em>The degree of given polynomial is '5'</em>

<em>The co-efficient of  y⁴  is '-17'</em>

<em>   </em>

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Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
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