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3 years ago

Sketch the graphs - 2y – 3=x

Mathematics

1 answer:

Dominik [7]3 years ago

3 0

Answer:

this should be correct try mathaway but instead of away make it way. it won't let me say the actual name

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What are the important things to remember or be careful about when adding and subtracting polynomials?

prisoha [69]

Be careful you don't add unlike terms together (e.g. x^2 and x cannot be added or subtracted together).
Be careful you don't mistake the 2 in x^2 as another number. Same for x^3, x^4. etc.

8 0

4 years ago

Is the following inequality true : 4 <_ 4 ?

marissa [1.9K]

Answer:

Yes

Step-by-step explanation:

This is true because it is asking you if this is less than or EQUAL TO. 4 = 4, making it true.

8 0

2 years ago

What is 2x - 3 = -x + 3

BARSIC [14]

Simplifying 2x + -3 = -1x + 3
Reorder the terms: -3 + 2x = -1x + 3
Reorder the terms: -3 + 2x = 3 + -1x Solving -3 + 2x = 3 + -1x
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right. Add 'x' to each side of the equation. -3 + 2x + x = 3 + -1x + x
Combine like terms: 2x + x = 3x -3 + 3x = 3 + -1x + x
Combine like terms: -1x + x = 0 -3 + 3x = 3 + 0 -3 + 3x = 3
Add '3' to each side of the equation. -3 + 3 + 3x = 3 + 3
Combine like terms: -3 + 3 = 0 0 + 3x = 3 + 3 3x = 3 + 3
Combine like terms: 3 + 3 = 6 3x = 6
Divide each side by '3'. x = 2
Simplifying x = 2

answer x=2

Hope I Helped:D

7 0

3 years ago

Read 2 more answers

A market research company wishes to know how many energy drinks adults drink each week. They want to construct a 80% confidence

Diano4ka-milaya [45]

Answer:

The confidence interval = (7.8 , 8.0)

Step-by-step explanation:

Confidence Interval formula =

Mean ± z × Standard deviation/√n

Mean = 7.9

Standard deviation = 0.9

n = number of samples = 164

z = z score of an 80% confidence interval = 1.282

Confidence Interval = 7.9 ± 1.282 × 0.9/√164

= 7.9 ± 0.0900966432

Confidence Interval

= 7.9 - 0.0900966432

= 7.8099033568

Approximately to 1 decimal place = 7.8

7.9 + 0.0900966432

= 7.9900966432

Approximately to 1 decimal place = 8.0

Therefore, the confidence interval = (7.8 , 8.0)

7 0

3 years ago

Given two points P(sinθ+2, tanθ-2) and Q(4sin²θ+4sinθcosθ+2acosθ, 3sinθ-2cosθ+a). Find constant "a" and the corresponding value

vodomira [7]

Answer:

$\rm\displaystyle \displaystyle \displaystyle θ= {60}^{ \circ} , {300}^{ \circ}$

$\rm \displaystyle a = - \frac{ \sqrt{3} }{2} - 1, \frac{\sqrt{3}}{2} - 1$

Step-by-step explanation:

we are given two coincident points

$\displaystyle P( \sin(θ)+2, \tan(θ)-2) \: \text{and } \\ \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)$

since they are coincident points

$\rm \displaystyle P( \sin(θ)+2, \tan(θ)-2) = \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ )\cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)$

By order pair we obtain:

$\begin{cases} \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ) = \sin( \theta) + 2 \\ \\ \displaystyle 3 \sin( \theta) - 2 \cos( \theta) + a = \tan( \theta) - 2\end{cases}$

now we end up with a simultaneous equation as we have two variables

to figure out the simultaneous equation we can consider using substitution method

to do so, make a the subject of the equation.therefore from the second equation we acquire:

$\begin{cases} \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sinθ \cos(θ)+2a \cos(θ )= \sin( \theta) + 2 \\ \\ \boxed{\displaystyle a = \tan( \theta) - 2 - 3 \sin( \theta) + 2 \cos( \theta) } \end{cases}$

now substitute:

$\rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2 \cos(θ) \{\tan( \theta) - 2 - 3 \sin( \theta) + 2 \cos( \theta) \}= \sin( \theta) + 2$

distribute:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ)+4 \sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) - 6 \sin( \theta) \cos( \theta) + 4 \cos ^{2} ( \theta) = \sin( \theta) + 2$

collect like terms:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + 4 \cos ^{2} ( \theta) = \sin( \theta) + 2$

rearrange:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) + 4 \cos ^{2} ( \theta) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + = \sin( \theta) + 2$

by Pythagorean theorem we obtain:

$\rm\displaystyle \displaystyle 4 - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) = \sin( \theta) + 2$

cancel 4 from both sides:

$\rm\displaystyle \displaystyle - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) = \sin( \theta) - 2$

move right hand side expression to left hand side and change its sign:

$\rm\displaystyle \displaystyle - 2\sin(θ) \cos(θ)+\sin(θ ) - 4\cos( \theta) + 2 = 0$

factor out sin:

$\rm\displaystyle \displaystyle \sin (θ) (- 2 \cos(θ)+1) - 4\cos( \theta) + 2 = 0$

factor out 2:

$\rm\displaystyle \displaystyle \sin (θ) (- 2 \cos(θ)+1) + 2(- 2\cos( \theta) + 1 ) = 0$

group:

$\rm\displaystyle \displaystyle ( \sin (θ) + 2)(- 2 \cos(θ)+1) = 0$

factor out -1:

$\rm\displaystyle \displaystyle - ( \sin (θ) + 2)(2 \cos(θ) - 1) = 0$

divide both sides by -1:

$\rm\displaystyle \displaystyle ( \sin (θ) + 2)(2 \cos(θ) - 1) = 0$

by Zero product property we acquire:

$\begin{cases}\rm\displaystyle \displaystyle \sin (θ) + 2 = 0 \\ \displaystyle2 \cos(θ) - 1= 0 \end{cases}$

cancel 2 from the first equation and add 1 to the second equation since -1≤sinθ≤1 the first equation is false for any value of theta

$\begin{cases}\rm\displaystyle \displaystyle \sin (θ) \neq - 2 \\ \displaystyle2 \cos(θ) = 1\end{cases}$

divide both sides by 2:

$\rm\displaystyle \displaystyle \displaystyle \cos(θ) = \frac{1}{2}$

by unit circle we get:

$\rm\displaystyle \displaystyle \displaystyle θ= {60}^{ \circ} , {300}^{ \circ}$

so when θ is 60° a is:

$\rm \displaystyle a = \tan( {60}^{ \circ} ) - 2 - 3 \sin( {60}^{ \circ} ) + 2 \cos( {60}^{ \circ} )$

recall unit circle:

$\rm \displaystyle a = \sqrt{3} - 2 - \frac{ 3\sqrt{3} }{2} + 2 \cdot \frac{1}{2}$

simplify which yields:

$\rm \displaystyle a = - \frac{ \sqrt{3} }{2} - 1$

when θ is 300°

$\rm \displaystyle a = \tan( {300}^{ \circ} ) - 2 - 3 \sin( {300}^{ \circ} ) + 2 \cos( {300}^{ \circ} )$

remember unit circle:

$\rm \displaystyle a = - \sqrt{3} - 2 + \frac{3\sqrt{ 3} }{2} + 2 \cdot \frac{1}{2}$

simplify which yields:

$\rm \displaystyle a = \frac{ \sqrt{3} }{2} - 1$

and we are done!

disclaimer: also refer the attachment I did it first before answering the question

5 0

3 years ago

Sketch the graphs - 2y – 3=x​

Sketch the graphs - 2y – 3=x