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3 years ago

HELP PLEASE!!!!!!!! :( :( :(

Mathematics

2 answers:

belka [17]3 years ago

6 0

Answer:

θ ≈ 31.1°

Step-by-step explanation:

The mnemonic SOH CAH TOA reminds you of the relations between triangle sides and trig functions of angles. Here, you have the side opposite the angle, and you have the hypotenuse. So, the relevant relation is ...

Sin = Opposite/Hypotenuse

sin(θ) = 2/√15

The inverse sine function is used to find the angle value from it sine:

θ = arcsin(2/√15) ≈ 31.0909°

The angle θ is about 31.1°.

Andrei [34K]3 years ago

3 0

Answer:

θ ≈ 31.1°

Step-by-step explanation:

The mnemonic SOH CAH TOA reminds you of the relations between triangle sides and trig functions of angles. Here, you have the side opposite the angle, and you have the hypotenuse. So, the relevant relation is ...

Sin = Opposite/Hypotenuse

sin(θ) = 2/√15

The inverse sine function is used to find the angle value from it sine:

θ = arcsin(2/√15) ≈ 31.0909°

The angle θ is about 31.1°.

_____

<em>Comment on inverse trig functions</em>

When you use your calculator for inverse trig functions, be sure it presents angles in the units you want. You may have to go to some effort to ensure that the angle is expressed in <em>degrees</em> (not radians).

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Alon was curious if segments \overline{JK} JK start overline, J, K, end overline and \overline{NM} NM start overline, N, M, end

BartSMP [9]

The answer is A because Alon could have rotated the figure and discovered that they are congruent

6 0

4 years ago

the initial vertical velocity of a rocket shot straight up is 42 meters per second. How long does it take for the rocket to retu

Snowcat [4.5K]

It takes 4.3 seconds for the rocket to return to earth.

The equation is:
$0=-9.8t^2+v_0t+h_0$
where -9.8m/sec² is the acceleration due to gravity, v₀ is the initial velocity, and h₀ is the initial height. We will go from the assumption that the rocket is launched from the ground, so h₀=0, and we are told that the initial velocity, v₀, is 42. This gives us:

$0=-9.8t^2+42t$

We will use the quadratic formula to solve this. The quadratic formula is:
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Plugging in our information we have:
$t=\frac{-42\pm \sqrt{42^2-4(-9.8)(0)}}{2(-9.8)}
\\
\\=\frac{-42\pm \sqrt{1764-0}}{-19.6}
\\
\\=\frac{-42\pm \sqrt{1764}}{-19.6}
\\
\\=\frac{-42\pm 42}{-19.6}=\frac{-42-42}{-19.6} \text{ or } \frac{-42+42}{-19.6}
\\
\\=\frac{-84}{-19.6}\text{ or }\frac{0}{-19.6}
\\
\\=4.3\text{ or }0$

x=0 is when the rocket is launched; x=4.3 is when the rocket lands.

8 0

4 years ago

Vern bought 6 candles for $ 8.50 each he bought a holder for the candles for 12.75 what is the total amount Vern spent on the ca

Alexus [3.1K]

8.50x6=51
51+12.75=63.75

So he spent $63.75

3 0

4 years ago

What is the y-coordinate of the point that divides the

natima [27]

Answer:

y-coordinate = 0

Step-by-step explanation:

Consider the below diagram attached with this question.

Section formula:

If a point divides a line segment in m:n whose end points are $(x_1,y_1)$ and $(x_2,y_2)$ , then the coordinates of that point are

$(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})$

From the below graph it is clear that the coordinates of end points are J(1,-10) and K(7,2). A point divides the line JK is 5:1.

Using section formula, the coordinates of that point are

$(\frac{(5)(7)+(1)(1)}{5+1},\frac{(5)(2)+(1)(-10)}{5+1})$

$(\frac{35+1}{6},\frac{10-10}{6})$

$(\frac{36}{6},\frac{0}{6})$

$(6,0)$

Therefore, the y-coordinate of the point that divides the directed line segment from J to k into a ratio of 5:1 is 0.

8 0

3 years ago

Read 2 more answers

In the △ABC, the height AN = 24 in, BN = 18 in, AC = 40 in. Find AB and BC. Consider all possible cases. Case 1 : N∈BC, AB = __,

aivan3 [116]

Answer:

Case 1:

$AB = 30$

$BC = 50$

Case 2:

$AB = 15.9$

$BC = 36.7$

Case 3: Not possible

Step-by-step explanation:

Given

See attachment for illustration of each case

Required

Find AB and BC

Case 1:

Using Pythagoras theorem in ANB, we have:

$AB^2 = AN^2 + BN^2$

This gives:

$AB^2 = 24^2 + 18^2$

$AB^2 = 576 + 324$

$AB^2 = 900$

Take square roots of both sides

$AB = \sqrt{900$

$AB = 30$

To calculate BC, we consider ANC, where:

$AC^2 = AN^2 + NC^2$

$40^2 = 24^2 + NC^2$

$1600 = 576 + NC^2$

Collect like terms

$NC^2 = 1600 - 576$

$NC^2 = 1024$

Take square roots

$NC = \sqrt{1024$

$NC = 32$

So:

$BC = NC + BN$

$BC = 32 + 18$

$BC = 50$

Case 2:

Using Pythagoras theorem in ANB, we have:

$AN^2 = AB^2 + BN^2$

This gives:

$24^2 = AB^2 + 18^2$

$576 = AB^2 + 324$

Collect like terms

$AB^2 = 576 - 324$

$AB^2 = 252$

Take square roots of both sides

$AB = \sqrt{252$

$AB = 15.9$

To calculate BC, we consider ABC, where:

$AC^2 = AB^2 + BC^2$

$40^2 = 252 + BC^2$

$1600 = 252 + BC^2$

Collect like terms

$BC^2 = 1600 - 252$

$BC^2 = 1348$

Take square roots

$BC = \sqrt{1348$

$BC = 36.7$

Case 3:

This is not possible because in ANC

The hypotenuse AN (24) is less than AC (40)

3 0

3 years ago

HELP PLEASE!!!!!!!! :( :( :(​

HELP PLEASE!!!!!!!! :( :( :(