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3 years ago

3# Find the missing variable (s). Round to the nearest tenth.

Mathematics

1 answer:

rewona [7]3 years ago

7 0

Answer:

Step-by-step explanation:

This triangle is already positioned to match trig functions (also known as circular functions).

Point A is origin, side AC (length x) is on X axis,

side CB (length y) is parallel to Y axis.

***** You just remember *****

cos is horizontal, on X axis,

sin is vertical, on Y axis.

To get from unit triangle with hypotenuse 1,

horizontal side cos(θ), vertical side sin(θ), you multiply all sides by same number, namely, the specified length of the hypotenuse.

***** that's all you need to know *****

Draw a unit circle centered at A, mark point Y at intersection of circle and AB, and drop a vertical line from Y crossing X axis at X.

Then radial segment AY is 1, AB is 10

horizontal segment AX is cos(32°), AC is 10AX

vertical segment XY is sin(32°), CB is 10XY.

So x = 10 cos(32°), y= 10 sin(32°).

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Answer with workings please

icang [17]

AB = 9 cm
BC = 6cm

CD = 7 cm
AE = 6 cm

3BC = AB
3ED = AE

  AB = AE
  BC ED
  ⁹/₃ = ⁶/ₓ
3 · 6 = 9 · x
18 = 9x
9 9
2 = x

ED = 2 cm

5 0

3 years ago

ASAP WHO EVER ANSWERS THIS WILL BE MARKED BRANLIEST

Kaylis [27]

Answer:

option A is the answer hope you find it helpful

3 0

3 years ago

A tank contains 100 gal of brine made by dissolving 80 Ib of salt in water. Pure water runs into the tank at the rate of 4 gal/m

ohaa [14]

Answer:

Step-by-step explanation:

Amount of brine = 100 gal

amount of salt = 80 lb

rate of water running = 4 gal / min

Let A be the amount of salt in lb in the tank at any time t.

(a) dA/dt = rate of salt in - rate of salt out

Rate of salt in = 0

rate of salt out = A x 4 / 100 = A / 25 lb/min

So, dA/dt = 0 - A/25 = - A/25 lb/min

$\frac{dA}{A}=-\frac{1}{25}dt$

$\int \frac{dA}{A}=-\frac{1}{25}\int dt$

$A(t) = C e^{-0.04t}$

When t = 0, A = 80 lb

So,

$A(t) = 80 e^{-0.04t}$

(b) Now, A = 40 lb

So,

$40 = 80 e^{-0.04t}$

$0.5 = e^{-0.04t}$

take log on both the sides

$ln 0.5 = - 0.04 t$

t = 17.3 minutes

8 0

3 years ago

Show that the following statement is an identity by transforming the left side into the right side. sin θ (sec θ + csc θ) = tan

ExtremeBDS [4]

Answer:

Step-by-step explanation:

Required to prove that:

Sin θ(Sec θ + Cosec θ)= tan θ+1

Steps:

Recall sec θ= 1/cos θ and cosec θ=1/sin θ

Substitution into the Left Hand Side gives:

Sin θ(Sec θ + Cosec θ)

= Sin θ(1/cos θ + 1/sinθ )

Expanding the Brackets

=sinθ/cos θ + sinθ/sinθ

=tanθ+1 which is the Right Hand Side as required.

Note that from trigonometry sinθ/cosθ = tan θ

8 0

4 years ago

What is the coordinates of the center of an ellipse defined by the equation 16x^2 + 25y^2 + 160x - 200y + 400 = 0 ? Please give

pickupchik [31]

16x^2 + 25y^2 + 160x - 200y + 400 = 0     Rearrange and regroup.

(16x^2 + 160x) + (25y^2 - 200y ) = 0-400.     Group the xs together and the ys together.

16(X^2 + 10x) + 25(y^2-8y) = -400.     Factorising.

We are going to use completing the square method.

Coefficient of x in the first expression = 10.

Half of it = 1/2 * 10 = 5. (Note this value)

Square it = 5^2 = 25.     (Note this value)

Coefficient of y in the second expression = -8.

Half of it = 1/2 * -8 = -4. (Note this value)

Square it = (-4)^2 = 16. (Note this value)

We are going to carry out a manipulation of completing the square with the values

25 and 16. By adding and substracting it.

16(X^2 + 10x) + 25(y^2-8y) = -400

16(X^2 + 10x + 25 -25) + 25(y^2-8y + 16 -16) = -400

Note that +25 - 25 = 0.    +16 -16 = 0. So the equation is not altered.

16(X^2 + 10x + 25) -16(25) + 25(y^2-8y + 16) -25(16) = -400

16(X^2 + 10x + 25) + 25(y^2-8y + 16) = -400 +16(25) + 25(16)    Transferring the terms -16(25) and -25(16)

to other side of equation. And 16*25 = 400

16(X^2 + 10x + 25) + 25(y^2-8y + 16) = 25(16)

16(X^2 + 10x + 25) + 25(y^2-8y + 16) = 400

We now complete the square by using the value when coefficient was halved.

16(x-5)^2 + 25(y-4)^2 = 400

Divide both sides of the equation by 400

(16(x-5)^2)/400 + (25(y-4)^2)/400 = 400/400              Note also that, 16*25 = 400.

((x-5)^2)/25 + ((y-4)^2)/16 = 1

((x-5)^2)/(5^2) + ((y-4)^2)/(4^2) = 1

Comparing to the general format of an ellipse.

((x-h)^2)/(a^2) + ((y-k)^2)/(b^2) = 1

Coordinates of the center = (h,k).

Comparing   with above   (x-5) = (x - h) , h = 5.

Comparing   with above   (y-k) = (y - k) , k = 4.

Therefore center = (h,k) = (5,4).

Sorry the answer came a little late. Cheers.

3 0

3 years ago