Without using a calculator, determine the number of real zeros of the function

In the diagram the measure of angle 3 is 105 degrees which angle must also measure 105 degrees?

valentina_108 [34]

Answer: C

Step-by-step explanation:

I am sorry if I don't correct

7 0

3 years ago

Sin tita= 0.6892.find the value Of tita correct to two decimal places<br><br>

Anvisha [2.4K]

Answer:

$\theta \approx 6.28n + 2.38, \quad n \in \mathbb{Z}$

or

$\theta \approx 6.28n + 0.76, \quad n \in \mathbb{Z}$

Considering $\theta \in (0, 2\pi]$

$\theta \approx 2.38$

or

$\theta \approx 0.76$

Step-by-step explanation:

$\sin(\theta)=0.6892$

We have:

$\sin (x)=a \Longrightarrow x=\arcsin (a)+2\pi n \text{ or } x=\pi -\arcsin (a)+2\pi n \text{ as } n\in \mathbb{Z}$

Therefore,

$\theta= \arcsin (0.6892)+2\pi n, \quad n \in \mathbb{Z}$

or

$\theta = \pi -\arcsin (0.6892)+2\pi n, \quad n\in \mathbb{Z}$

---------------------------------

$\theta \approx 6.28n + 2.38, \quad n \in \mathbb{Z}$

or

$\theta \approx 6.28n + 0.76, \quad n \in \mathbb{Z}$

4 0

3 years ago

Twenty percent of the students at Drew's school walk to school. There are 300 students in Drew's school. How many students walk

swat32

20% of 300 equals 60

7 0

3 years ago

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HELP ASAP PLZZZZ

Tcecarenko [31]

QUESTION 1

The given system of equations is

$3d - e = 7...eqn(1)$
$d + e = 5...eqn(2)$

To solve by linear combination, we add equation (1) to equation (2) to get,

$3d + d= 7 + 5$

$4d = 12$

We divide through by 4 to obtain,

$d = \frac{12}{4}$

$d = 3$

We put d=3 into equation (2) to get,

$3+ e = 5$

$e = 5 - 3$

$e = 2$

$\boxed {The \: solution \: is \: (3, 2)}$

QUESTION 2

The given system is

4x + y = 5 ...eqn(1)

3x + y = 3 ...eqn(2)

To solve by linear combination, we subtract equation (2) from equation (1) to eliminate y from the equation.

This will give us,

$4x - 3x = 5 - 3$

This implies that,

$x = 2$

Put x=3 into equation (1) to get,

$4(2) + y = 5$

$8+ y = 5$

$y = 5 - 8$

$y = - 3$

The solution is

$(2,-3)$

QUESTION 3

We want to solve the system;

a – 2b = –2 ....eqn(1)

2a + 2b = 14...eqn(2)

by linear combination.

We need to add equation (1) to equation (2) to eliminate b.

This implies that,

$2a + a = 14 + - 2$

Simplify,

$3a = 12$

Divide both sides by 3 to get,

$a = 4$
Put a=4 into equation (2) to obtain,

$2(4) + 2b = 14$

$8 + 2b = 14$
$2b = 14 - 8$

$2b = 6$

$b = 3$

The ordered pair in the form (a, b) is

$(4,3)$

QUESTION 4

The given system of equations is

11x + 4y = 18 ...eqn(1)

3x + 4y = 2 ...eqn(2)

We subtract equation (2) from equation (1) to get,

$11x - 3x = 18 - 2$

$8x = 16$

$x = 2$

Put x=2 into equation (2) to obtain,

$3(2) + 4y = 2$

This implies that,

$6 + 4y = 2$

$4y = 2 - 6$

$4y = - 4$

$y=-1$

The correct answer is (2,-1).

QUESTION 5

The given system is ;

2d + e = 8...eqn1

d – e = 4...eqn2

We add the two equations to eliminate e.

This implies that,

$2d + d = 8 + 4$

$3d = 12$

We divide both sides by 3 to get,

$d = 4$

We put d=4 into equation (2) to get,

$4 - e = 4$

$- e = 4 - 4$

$- e = 0$

$e = 0$

The solution is

$(4,0)$

7 0

3 years ago

A tortoise is walking in the desert. It takes

dlinn [17]

96 meters per hour :)

7 0

3 years ago

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