Which statement represents the expression 2q+3 ?

Answer the following questions CORRECTLY I will know if this is wrong. I WILL REPORT ANY INCORRECT ANSWERS!

liberstina [14]

Answer by JKismyhusbandbae: 9

(5 + 12 + 8 + 12 + 11 + 8 + 5 + 5 + 7 + 12)/10=9

To find the mean, add up all the lengths and divide by 10.

4 0

2 years ago

Solve the system of equations. - - 10x + 3y = 5 x = y - 4 2 y =

Hitman42 [59]

Is the answer -1/6?!

3 0

2 years ago

I know the selected answer is correct but I'm not too sure how to get that answer.

Kryger [21]

$\tt{ Hey \: there , \: Mr.Panda \: ! }$ ;)

♨ $\large{ \tt{ E \: X \: P \: L \: A \: N \: A \: T \: I\: O \: N}}:$

⤻ Before solving the given question , you should know the answer of these questions :

✺How do you find the hypotenuse , perpendicular and base when the angle ( $\theta \: , \alpha \: ,\beta$ ) is given ?

⇾ The longest side , which is the opposite side of right angle is the hypotenuse ( h ). There are two other sides , the opposite and the adjacent. The naming of these sides depends upon which angle is involved. The opposite is the side opposite the angle involved and it is called the perpendicular ( p ) . The adjacent us the side next to the angle involved ( buy not the hypotenuse ) and it is called the base ( b ).

☄ $\large{ \tt{REMEMBER}} :$

$\bf{ \sin \theta = \frac{opposite}{hypotenuse} = \frac{perpendicular}{hypotenuse} }$

$\bf{ \cos\theta = \frac{adjacent}{hypotenuse} = \frac{base}{hypotenuse} }$

$\bf{ \tan \theta = \frac{opposite}{adjacent} = \frac{perpendicular}{base} }$

In the above cases , $\theta$ is taken as the angle of reference.

♪ Our Q/A part ends up here! Let's start solving the question :

❈ $\large{ \tt{GIVEN}} :$

Perpendicular ( p ) = ? , Hypotenuse ( h ) = 18 & base ( b ) = 16

✧ $\large{ \tt{TO \: FIND} : }$

Value of tan $\theta$

✎ $\large{ \tt{SOLUTION}} :$

Firstly , Finding the value of perpendicular ( p ) using Pythagoras theorem :

❃ $\boxed{ \sf{ {h}^{2} = {p}^{2} + {b}^{2} }}$ [ Pythagoras theorem ]

$\large{ ⇢ \sf{p}^{2} + {b}^{2} = {h}^{2} }$

$\large{⇢ \sf{ {p}^{2} = {h}^{2} - {b}^{2} }}$

$\large{ ⇢\sf{ {p}^{2} = {18}^{2} - {16}^{2} }}$

$\large{⇢ \sf{ {p}^{2} = 324 - 256}}$

$\large{⇢ \sf{ {p}^{2} = 68}}$

$\large{⇢ \sf{p = \sqrt{68}}}$

$\large{ ⇢\sf{p = \boxed{ \tt{2 \sqrt{17}}} }}$

Okey, We found out the perpendicular i.e $\tt{2 \sqrt{17}}$ . Now , We know :

❊ $\large{ \sf{ \tan \theta} = \frac{perpendicular}{base} }$

$\large {\tt{↬ \: tan \theta = \frac{2 \sqrt{17} }{16}}}$

$\large{ \tt{ ↬ tan \theta = \frac{ \cancel{2} \: \sqrt{17} }{ \cancel{16} \: \: 8} }}$

$\large{ \tt{ ↬ \boxed{ \tt{tan \theta = \frac{ \sqrt{17} }{8}}}}}$

⟿ $\boxed{ \boxed{ \tt{OUR\: FINAL \: ANSWER : \boxed{ \underline{ \bf{ \frac{ \sqrt{17} }{8}}}}}}}$

۵ Yay! We're done!

♕ $\large\tt{RULE \: OF \:SUCCESS }:$

Never lose hope & keep on working ! ✔

ツ Hope I helped!

☃ Have a wonderful day / evening! ☼

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3 0

2 years ago

Read 2 more answers

The minimum and maximum distances from a focus to a point on an ellipse occur when that point on the ellipse is an endpoint of t

Sauron [17]

The answer is True.

Explanation:
Let a = major axis
Let b = minor axis
Let c = focal length.

Consider the right focus, located a distance c from the center of the ellipse (at the origin).
From the right focus to the right point on the major axis is equal to a-c. This is the minimum distance.
From the right focus to the left point on the major axis is equal to a+c. This is the maximum distance.

7 0

3 years ago

Read 2 more answers

If someone can explain this I will greatly appreciate it. I don’t really understand it.

erma4kov [3.2K]

Answer:

The second answer choice.

Step-by-step explanation:

6 0

2 years ago