All categories

3 years ago

What’s equivalent to -60\22

Mathematics

1 answer:

Mila [183]3 years ago

5 0

Answer:

-30/11

Step-by-step explanation:

You might be interested in

Simplify the following Expression by combining like terms: y+ 3x + 4y + 2x + 3y + 3x

lys-0071 [83]

Answer:

8x+8y

Step-by-step explanation:

y + 3x + 4y + 2x + 3y +3x

With this we group x with x and y with y and in ascending order so x comes before y

3x+2x+3x

And the y with y

y+4y+3y

Which is;

3x+2x+3x+y+4y+3y

8x+8y

8 0

2 years ago

If 1/64= 4^2 s -1.162s+2 what is the value of s?

antoniya [11.8K]

Answer: Hello brainly student, The answer for this question is that s=1

Step-by-step explanation: Hope this helps you :)

8 0

2 years ago

What is 9 1\8 + 9 3\8

Murljashka [212]

Answer:

18.5 or 18 1/2

Step-by-step explanation:

9 plus 9 is 18. 1/8 plus 3/8 is 4/8 or 1/2. so 18 1/2

7 0

2 years ago

Read 2 more answers

(1+tan^2 A)cot A/cosec^2 A= tanA

Rufina [12.5K]

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identities

1 + tan²A = sec²A , secA = $\frac{1}{cosA}$ , cosecA = $\frac{1}{sinA}$ , cotA = $\frac{cosA}{sinA}$ , tanA = $\frac{sinA}{cosA}$

Consider the left side

$\frac{(1+tan^2A)cotA}{cosec^2A}$

= $\frac{sec^2AcotA}{cosec^2A}$

= $\frac{\frac{1}{cos^2A}(\frac{cosA}{sinA}) }{cosec^2A}$

= $\frac{\frac{1}{cosAsinA} }{\frac{1}{sin^2A} }$

= $\frac{1}{cosAsinA}$ × sin²A

= $\frac{sinA}{cosA}$

= tanA = right side , thus proven

4 0

2 years ago

If cos Θ = square root 2 over 2 and 3 pi over 2 < Θ < 2π, what are the values of sin Θ and tan Θ?

KIM [24]

Answer:

The answer is

$sin(\theta)=-\frac{\sqrt{2}}{2}$

$tan(\theta)=-1$

Step-by-step explanation:

we know that

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

$sin^{2}(\theta)+cos^{2}(\theta)=1$

In this problem we have

$cos(\theta)=\frac{\sqrt{2}}{2}$

$\frac{3\pi}{2}$

The angle $\theta$ belong to the third or fourth quadrant

The value of $sin(\theta)$ is negative

Step 1

Find the value of $sin(\theta)$

Remember

$sin^{2}(\theta)+cos^{2}(\theta)=1$

we have

$cos(\theta)=\frac{\sqrt{2}}{2}$

substitute

$sin^{2}(\theta)+(\frac{\sqrt{2}}{2})^{2}=1$

$sin^{2}(\theta)=1-\frac{1}{2}$

$sin^{2}(\theta)=\frac{1}{2}$

$sin(\theta)=-\frac{\sqrt{2}}{2}$ ------> remember that the value is negative

Step 2

Find the value of $tan(\theta)$

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

we have

$sin(\theta)=-\frac{\sqrt{2}}{2}$

$cos(\theta)=\frac{\sqrt{2}}{2}$

substitute

$tan(\theta)=\frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}$

$tan(\theta)=-1$

8 0

3 years ago

Read 2 more answers