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Flauer [41]
3 years ago
14

What is 1(-1) + 1(-1) +4 =?

Mathematics
2 answers:
RideAnS [48]3 years ago
7 0
Hopefully this helps you understand
-Dominant- [34]3 years ago
5 0

Answer:

2

Step-by-step explanation

1 times -1 is -1. -1 + -1=-2. -2+-4= 2

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Urgent!! Will mark brainliest!!
horsena [70]

Answer:

1) x is negative and y is positive ⇒ last answer

2) cotФ = -12/35 ⇒ second answer

3) The right identity is cot²Ф - csc²Ф = -1 ⇒ last answer

Step-by-step explanation:

* For any point (x , y) lies on the terminal side of the angle Ф

 in standard position

* x = cosФ and y = sinФ

- If Ф in the first quadrant, then x , y are positive

∴ All trigonometry functions are positive

- If Ф in the second quadrant, then x is negative , y is positive

∴ sinФ only is positive

- If Ф in the third quadrant, then x is negative , y is negative

∴ tanФ only is positive

- If Ф in the fourth quadrant, then x is positive , y is negative

∴ cosФ only is positive

* Lets solve the problems

∵ Ф = 3π/4 ⇒ (135°)

∴ It lies on the second quadrant

∴ x is negative and y is positive

* Lets revise the reciprocal of sinФ, cosФ and tanФ

- cscФ = 1/sinФ

- secФ = 1/cosФ

- cotФ = 1/tanФ

∵ secФ = -37/12

∴ cosФ = -12/37

∵ π/2 < Ф < π

∴ Ф lies on the second quadrant

∴ cotФ is negative values

∵ tan²Ф = sec²Ф - 1

∵ secФ = -37/12

∴ tan²Ф = (-37/12)² - 1 = 1225/144 ⇒ take√ for both sides

∴ tanФ = ± 35/12

∵ cotФ = ± 12/35

∵ cotФ is negative value

∴ cotФ = -12/35

* In the standard position of the angle Ф the terminal

 of it lies on the unit circle O

- By using Pythagorean theorem

∵ x² + y² = 1

∵ x = cosФ and y = sinФ

∴ cos²Ф + sin²Ф = 1 ⇒ (1)

∴ cos²Ф = 1 - sin²Ф

∴ sin²Ф = 1 - cos²Ф

* Divide (1) by cos²Ф

∴ cos²Ф/cos²Ф + sin²Ф/cos²Ф = 1/cos²Ф

* Remember sin²Ф/cos²Ф = tan²Ф and 1/cos²Ф = sec²Ф

∴ 1 + tan²Ф = sec²Ф ⇒ (2) ⇒ subtract 1 from both sides

∴ tan²Ф = sec²Ф - 1 ⇒ subtract sec²Ф from both sides

∴ tan²Ф - sec²Ф = -1

* Divide (1) by sin²Ф

∴ cos²Ф/sin²Ф + sin²Ф/si²Ф = 1/sin²Ф

* Remember cos²Ф/sin²Ф = cot²Ф and 1/sin²Ф = csc²Ф

∴ cot²Ф + 1 = csc²Ф ⇒ (3) ⇒ subtract 1 from both sides

∴ cot²Ф = csc²Ф - 1 ⇒ subtract csc²Ф from both sides

∴ cot²Ф - csc²Ф = -1

* The right identity is cot²Ф - csc²Ф = -1

3 0
3 years ago
Carlos is tiling a kitchen counter that is 12 feet by 3 feet.The counter has a rectangular hole 3 feet by 2 feet cut in it for a
alina1380 [7]

Answer:

Carlos has to tile 30 sq ft


3 0
3 years ago
Compare the graph of f (x) with the graph of k (x) = 2 (x-8)2
Bingel [31]

Answer:

one is not linear

Step-by-step explanation:

7 0
3 years ago
Find the distance between the point (3,1) and the line with the equation y=2x-5
True [87]

Answer:

B:  0

Step-by-step explanation:

Note that (3, 1) satisfies the equation y = 2x - 5:

1 = 2(3) - 5

1  =  6 -5  = 1

Therefore, the distance between this point and this line is zero.

B is the correct answer.

3 0
3 years ago
Y''+y'+y=0, y(0)=1, y'(0)=0
mars1129 [50]

Answer:

y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form ay''+by'+cy=0.

Given: y''+y'+y=0

Let y=e^{rt} be it's solution.

We get,

\left ( r^2+r+1 \right )e^{rt}=0

Since e^{rt}\neq 0, r^2+r+1=0

{ we know that for equation ax^2+bx+c=0, roots are of form x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} }

We get,

y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}

For two complex roots r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta, the general solution is of form y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )

i.e y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

Applying conditions y(0)=1 on e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right ), c_1=1

So, equation becomes y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

On differentiating with respect to t, we get

y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )

Applying condition: y'(0)=0, we get 0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}

Therefore,

y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

3 0
3 years ago
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