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Gwar [14]
3 years ago
10

Which polynomial below is considered a binomial? Question 3 options:

Mathematics
1 answer:
sukhopar [10]3 years ago
3 0

For this case we have that by definition, a binomial is an algebraic expression that is formed by the sum (or difference) of two terms, also called monomials.

Examples:

ax + b\\xy ^ 2 + 3xz

So, according to the previous definition we have that the binomial is:

9a ^ 2-5

Answer:

Option B

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Find (f+g)(x) when f(x)=3/x+5 and g(x)=2/x
olga_2 [115]
What does that 3/x and 2/x mean? 2x and 3x or x^2 and x^3?
8 0
3 years ago
compute the projection of → a onto → b and the vector component of → a orthogonal to → b . give exact answers.
Nina [5.8K]

\text { Saclar projection } \frac{1}{\sqrt{3}} \text { and Vector projection } \frac{1}{3}(\hat{i}+\hat{j}+\hat{k})

We have been given two vectors $\vec{a}$ and $\vec{b}$, we are to find out the scalar and vector projection of $\vec{b}$ onto $\vec{a}$

we have $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$

The scalar projection of$\vec{b}$onto $\vec{a}$means the magnitude of the resolved component of $\vec{b}$ the direction of $\vec{a}$ and is given by

The scalar projection of $\vec{b}$onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}$

$$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\sqrt{1^2+1^1+1^2}} \\&=\frac{1^2-1^2+1^2}{\sqrt{3}}=\frac{1}{\sqrt{3}}\end{aligned}$$

The Vector projection of $\vec{b}$ onto $\vec{a}$ means the resolved component of $\vec{b}$ in the direction of $\vec{a}$ and is given by

The vector projection of $\vec{b}$ onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \cdot(\hat{i}+\hat{j}+\hat{k})$

$$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\left(\sqrt{1^2+1^1+1^2}\right)^2} \cdot(\hat{i}+\hat{j}+\hat{k}) \\&=\frac{1^2-1^2+1^2}{3} \cdot(\hat{i}+\hat{j}+\hat{k})=\frac{1}{3}(\hat{i}+\hat{j}+\hat{k})\end{aligned}$$

To learn more about scalar and vector projection visit:brainly.com/question/21925479

#SPJ4

3 0
1 year ago
How do you solve (Arc)QPT if <QZT = 120
yuradex [85]
By definition, the arc length is given by:
 arc = R * theta * ((2 * pi) / 360)
 Where,
 theta: angle in degrees
 R: radio
 We have then:
 (Arc) QPT if <QZT = 120:
 theta = 360-120 = 240 degrees
 R = 13.5 units
 Substituting values we have:
 (Arc) QPT = R * theta * ((2 * pi) / 360)
 (Arc) QPT = (13.5) * (240) * ((2 * pi) / 360)
 (Arc) QPT = 56.55 units
 Answer:
 
(Arc) QPT = 56.55 units
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3 years ago
Add.
Ann [662]

Answer:

17. +5-2-1=+2

18. -7-6-2=15

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What is the snswer to 3a+2b x 7c
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I am pretty sure your answer is this because 3+2=5 and 5*7=35 abc
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3 years ago
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