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Sidana [21]
3 years ago
15

PLEASE HELP ILL MARK YOU BRAINLIEST

Mathematics
1 answer:
trapecia [35]3 years ago
7 0

Answer:

Part A: x⁴ + x³ + x²

Part B: (4·x + 2) + (-5·x + 6) = -x + 8

Step-by-step explanation:

Part A:

We note that standard form means that the terms of the polynomial are arranged in order starting from the the largest exponential of the polynomial to the smallest exponent

Therefore, we have

x⁴ + x³ + x² is a fourth-degree polynomial with three terms in standard form

It is known that the above polynomial is in standard form based on its ordered arrangement from the largest exponential to the smallest exponential

Part B:

Polynomial are closed under addition, that is when two polynomials are added only the coefficients change as the exponents and variables remain unchanged as follows;

(4·x + 2) + (-5·x + 6)

= 4·x + 2 - 5·x + 6

= 4·x - 5·x + 2 + 6

= -x + 8 which is a polynomial with the same variable and exponent.

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Select the correct answer Consider functions p and q. p(x) = log₂ (x - 1) g(x) = 2^x - 1 Which statement is true about these fun
goldenfox [79]

The x-intercept of p(x) is x = 2, the x-intercept of g(x) is x = 0, then the correct option is B.

<h3>What can we say about the x-intercepts of the given functions?</h3>

For a function f(x), the x-intercept is the value of x such that:

f(x) = 0.

Here we have:

p(x) = log₂(x - 1)

Remember that:

logₙ(1)  = 0

For any base n, then the x-intercept of p(x) is x = 2, because:

p(2) = log₂(2 - 1) = log₂(1) = 0.

The other function is:

g(x) = 2ˣ - 1

Remember that any number to the power of zero is equal to 1, then:

g(0) = 2⁰ - 1 = 1 - 1 = 0

The x-intercept of p(x) is x = 2, the x-intercept of g(x) is x = 0, then the correct option is B.

If you want to learn more about x-intercepts:

brainly.com/question/3951754

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8 0
2 years ago
Easy Geometry Riddle
Stella [2.4K]

Answer:

A postage stamp

Step-by-step explanation:

A postage stamp sits in the corner of an envelope, yet the envelope that it's on can travel around the world.

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3 0
3 years ago
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What is 2 3/10 as a percent?
kobusy [5.1K]
2 3/10 = 2.3
To get this number as a percent, we need to multiply it by 100: 2.3 is a whole number, and can be represented by fraction 2.3/1
A percentage is represented by a fraction over 100. So to transform 2.3/1 to a percentage, just multiply both numerator and denominator by 100, and you'll get 230/100.
So 2 3/10 is 230% as a percent.

Hope this Helps! :)
4 0
3 years ago
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Please answer this question, i request
Jet001 [13]

{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}

\star  \:  \tt \cot  \theta = \dfrac{7}{8}

{\large{\textsf{\textbf{\underline{\underline{To \: Evaluate :}}}}}}

\star \:  \tt \dfrac{(1  +  \sin \theta)(1 - \sin \theta) }{(1 +  \cos \theta) (1  -  \cos \theta) }

{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}

Consider a \triangle ABC right angled at C and \sf \angle \: B = \theta

Then,

‣ Base [B] = BC

‣ Perpendicular [P] = AC

‣ Hypotenuse [H] = AB

\therefore \tt \cot  \theta   =  \dfrac{Base}{ Perpendicular}  =  \dfrac{BC}{AC} = \dfrac{7}{8}

Let,

Base = 7k and Perpendicular = 8k, where k is any positive integer

In \triangle ABC, H² = B² + P² by Pythagoras theorem

\longrightarrow \tt {AB}^{2}  =   {BC}^{2}  +   {AC}^{2}

\longrightarrow \tt {AB}^{2}  =   {(7k)}^{2}  +   {(8k)}^{2}

\longrightarrow \tt {AB}^{2}  =   49{k}^{2}  +   64{k}^{2}

\longrightarrow \tt {AB}^{2}  =   113{k}^{2}

\longrightarrow \tt AB  =   \sqrt{113  {k}^{2} }

\longrightarrow \tt AB = \red{  \sqrt{113}  \:  k}

Calculating Sin \sf \theta

\longrightarrow  \tt \sin \theta = \dfrac{Perpendicular}{Hypotenuse}

\longrightarrow  \tt \sin \theta = \dfrac{AC}{AB}

\longrightarrow  \tt \sin \theta = \dfrac{8 \cancel{k}}{ \sqrt{113} \: \cancel{ k } }

\longrightarrow  \tt \sin \theta =  \purple{  \dfrac{8}{ \sqrt{113} } }

Calculating Cos \sf \theta

\longrightarrow  \tt \cos \theta = \dfrac{Base}{Hypotenuse}

\longrightarrow  \tt \cos \theta =  \dfrac{BC}{ AB}

\longrightarrow  \tt \cos \theta =  \dfrac{7 \cancel{k}}{ \sqrt{113} \:  \cancel{k } }

\longrightarrow  \tt \cos \theta =  \purple{ \dfrac{7}{ \sqrt{113} } }

<u>Solving the given expression</u><u> </u><u>:</u><u>-</u><u> </u>

\longrightarrow \:  \tt \dfrac{(1  +  \sin \theta)(1 - \sin \theta) }{(1 +  \cos \theta) (1  -  \cos \theta) }

Putting,

• Sin \sf \theta = \dfrac{8}{ \sqrt{113} }

• Cos \sf \theta = \dfrac{7}{ \sqrt{113} }

\longrightarrow \:  \tt \dfrac{ \bigg(1 +  \dfrac{8}{ \sqrt{133}} \bigg) \bigg(1 - \dfrac{8}{ \sqrt{133}} \bigg) }{\bigg(1 +  \dfrac{7}{ \sqrt{133}} \bigg) \bigg(1 - \dfrac{7}{ \sqrt{133}} \bigg)}

<u>Using</u><u> </u><u>(</u><u>a</u><u> </u><u>+</u><u> </u><u>b</u><u> </u><u>)</u><u> </u><u>(</u><u>a</u><u> </u><u>-</u><u> </u><u>b</u><u> </u><u>)</u><u> </u><u>=</u><u> </u><u>a²</u><u> </u><u>-</u><u> </u><u>b²</u>

\longrightarrow \:  \tt  \dfrac{ { \bigg(1 \bigg)}^{2}  -  { \bigg(  \dfrac{8}{ \sqrt{133} } \bigg)}^{2}   }{ { \bigg(1 \bigg)}^{2}  -  { \bigg(  \dfrac{7}{ \sqrt{133} } \bigg)}^{2}  }

\longrightarrow \:  \tt   \dfrac{1 -  \dfrac{64}{113} }{ 1 - \dfrac{49}{113} }

\longrightarrow \:  \tt   \dfrac{ \dfrac{113 - 64}{113} }{  \dfrac{113 - 49}{113} }

\longrightarrow \:  \tt { \dfrac  { \dfrac{49}{113} }{  \dfrac{64}{113} } }

\longrightarrow \:  \tt   { \dfrac{49}{113} }÷{  \dfrac{64}{113} }

\longrightarrow \:  \tt    \dfrac{49}{ \cancel{113}} \times     \dfrac{ \cancel{113}}{64}

\longrightarrow \:  \tt   \dfrac{49}{64}

\qquad  \:  \therefore  \:  \tt \dfrac{(1  +  \sin \theta)(1 - \sin \theta) }{(1 +  \cos \theta) (1  -  \cos \theta) }  =   \pink{\dfrac{49}{64} }

\begin{gathered} {\underline{\rule{300pt}{4pt}}} \end{gathered}

{\large{\textsf{\textbf{\underline{\underline{We \: know :}}}}}}

✧ Basic Formulas of Trigonometry is given by :-

\begin{gathered}\begin{gathered}\boxed { \begin{array}{c c} \\ \bigstar \:  \sf{ In \:a \:Right \:Angled \: Triangle :}  \\ \\ \sf {\star Sin \theta = \dfrac{Perpendicular}{Hypotenuse}} \\\\ \sf{ \star \cos \theta = \dfrac{ Base }{Hypotenuse}}\\\\ \sf{\star \tan \theta = \dfrac{Perpendicular}{Base}}\\\\ \sf{\star \cosec \theta = \dfrac{Hypotenuse}{Perpendicular}} \\\\ \sf{\star \sec \theta = \dfrac{Hypotenuse}{Base}}\\\\ \sf{\star \cot \theta = \dfrac{Base}{Perpendicular}} \end{array}}\\\end{gathered} \end{gathered}

{\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}

✧ Figure in attachment

\begin{gathered} {\underline{\rule{200pt}{1pt}}} \end{gathered}

3 0
2 years ago
The sum of three times a number n and 2
mario62 [17]
It would be 6n because 2times 3 is 6 and bring down the n
8 0
2 years ago
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