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STALIN [3.7K]
3 years ago
13

Please answer whats on the screenshot for brainliest

Mathematics
1 answer:
belka [17]3 years ago
3 0

Answer:

the third

Step-by-step explanation:

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How are the greater than, less than, and equal to symbols useful?
Snowcat [4.5K]
They are called relation symbols
they tell how something relates to something else

greater than tells ou that the the result must be more than, but not eqal to that numberd so example, we have to have something that flys higher than that bridge (100ft) but it cannot fly 100 or else it will crash

less than tells you that you must be less than, but not euqal to. example
 this thing must be less than 7 feet tall to fit under the doorway, it cannot be 7 feet or else it wont fit

equal to is an exact amount, nothing more or less
we have to have a screw that is this big, nothing more or less



less and greater than is nice since if you found a solution and you wanted to notate it, example
if the naswer is any number less than 10, it would be nicer to write x<10 instead of listing all numbers less than 10 since they could get very close example 9.999999999 forever
greater is the same, any number bigger than 30, you woul dhave to include 30.0000000001 and smaller which is hard, but with the x>30, it is easier

the equals just shows equality, this is equal to this, period
3 0
3 years ago
Read 2 more answers
Choose the correct reasoning of when a projectile reaches its highest point. A. A projectile reaches its highest point when its
alexira [117]

Answer: A projectile reaches its highest point when its vertical velocity component is zero.

Step-by-step explanation:

In projectile motion, objects moves in parabolic path.

Projectile motion is a type of motion in which an object moves in a bilaterally symmetrical, parabolic path. Projection motion happens when the force is applied at the start of the trajectory (where the object follows) which follows an interference which is gravity. Gravity takes over and afterwards the object accelerates downwards. The acceleration is constant so the vertical velocity (Vy) varies linearly. The maximum height of the projectile which is the highest height is reached when the vertical velocity (Vy) is equal to zero.

After the projectile reaches its maximum height which is its highest point, it begins to drop downwards.

5 0
3 years ago
Use the drop down menus to correctly complete the statement.
Lubov Fominskaja [6]

If the diagonals of a quadrilateral bisect all the angles, then it's a rhombus, converse of a property. If the diagonals of a quadrilateral are perpendicular bisectors of each other, then it's a rhombus, converse of a property.

6 0
2 years ago
Verify the following <br> i.[ -3/4]3 = -27/64<br> ii. [-2/3]6 = 64/ 729
Juliette [100K]

Answer:

The relationships are correct and valid

Step-by-step explanation:

The correct question is as follows;

Verify the following

i.[ -3/4]^3 = -27/64

ii. [-2/3]^6 = 64/ 729

We have the solution as follows;

i) we have;

(-3/4)^3

That means ;

-3^3 = -3^3 = -27

4^3 = 64

so;

(-3/4)^3 = -27/64

ii) (-2/3)^6

(-2)^6 = 64

3^6 = 729

Thus;

(-2/3)^6 = 64/729

4 0
3 years ago
If 13cos theta -5=0 find sin theta +cos theta / sin theta -cos theta​
Ivahew [28]

Step-by-step explanation:

<h3>Need to FinD :</h3>

  • We have to find the value of (sinθ + cosθ)/(sinθ - cosθ), when 13 cosθ - 5 = 0.

\red{\frak{Given}} \begin{cases} & \sf {13\ cos \theta\ -\ 5\ =\ 0\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \big\lgroup Can\ also\ be\ written\ as \big\rgroup} \\ & \sf {cos \theta\ =\ {\footnotesize{\dfrac{5}{13}}}} \end{cases}

Here, we're asked to find out the value of (sinθ + cosθ)/(sinθ - cosθ), when 13 cosθ - 5 = 0. In order to find the solution we're gonna use trigonometric ratios to find the value of sinθ and cosθ. Let us consider, a right angled triangle, say PQR.

Where,

  • PQ = Opposite side
  • QR = Adjacent side
  • RP = Hypotenuse
  • ∠Q = 90°
  • ∠C = θ

As we know that, 13 cosθ - 5 = 0 which is stated in the question. So, it can also be written as cosθ = 5/13. As per the cosine ratio, we know that,

\rightarrow {\underline{\boxed{\red{\sf{cos \theta\ =\ \dfrac{Adjacent\ side}{Hypotenuse}}}}}}

Since, we know that,

  • cosθ = 5/13
  • QR (Adjacent side) = 5
  • RP (Hypotenuse) = 13

So, we will find the PQ (Opposite side) in order to estimate the value of sinθ. So, by using the Pythagoras Theorem, we will find the PQ.

Therefore,

\red \bigstar {\underline{\underline{\pmb{\sf{According\ to\ Question:-}}}}}

\rule{200}{3}

\sf \dashrightarrow {(PQ)^2\ +\ (QR)^2\ =\ (RP)^2} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ +\ (5)^2\ =\ (13)^2} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ +\ 25\ =\ 169} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ =\ 169\ -\ 25} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ =\ 144} \\ \\ \\ \sf \dashrightarrow {PQ\ =\ \sqrt{144}} \\ \\ \\ \dashrightarrow {\underbrace{\boxed{\pink{\frak{PQ\ (Opposite\ side)\ =\ 12}}}}_{\sf \blue{\tiny{Required\ value}}}}

∴ Hence, the value of PQ (Opposite side) is 12. Now, in order to determine it's value, we will use the sine ratio.

\rightarrow {\underline{\boxed{\red{\sf{sin \theta\ =\ \dfrac{Opposite\ side}{Hypotenuse}}}}}}

Where,

  • Opposite side = 12
  • Hypotenuse = 13

Therefore,

\sf \rightarrow {sin \theta\ =\ \dfrac{12}{13}}

Now, we have the values of sinθ and cosθ, that are 12/13 and 5/13 respectively. Now, finally we will find out the value of the following.

\rightarrow {\underline{\boxed{\red{\sf{\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}}}}}}

  • By substituting the values, we get,

\rule{200}{3}

\sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ {\footnotesize{\dfrac{\Big( \dfrac{12}{13}\ +\ \dfrac{5}{13} \Big)}{\Big( \dfrac{12}{13}\ -\ \dfrac{5}{13} \Big)}}}} \\ \\ \\ \sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ {\footnotesize{\dfrac{\dfrac{17}{13}}{\dfrac{7}{13}}}}} \\ \\ \\ \sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ \dfrac{17}{13} \times \dfrac{13}{7}} \\ \\ \\ \sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ \dfrac{17}{\cancel{13}} \times \dfrac{\cancel{13}}{7}} \\ \\ \\ \dashrightarrow {\underbrace{\boxed{\pink{\frak{\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ \dfrac{17}{7}}}}}_{\sf \blue{\tiny{Required\ value}}}}

∴ Hence, the required answer is 17/7.

6 0
2 years ago
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