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3 years ago

11

Type an equation for the line shown in the graph

1 answer:

Andreas93 [3]3 years ago

7 0

Y=3/2x+0

The y-intercept is 0 as the line intercepts at the origin.

-6/-4= 3/2

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If i roll 2 dice what is the probability that neither of them are prime nor even?

Eduardwww [97]

Answer:

There are 3 prime numbers shown on the die: 2, 3 and 5. The probability of showing a prime number on a single die is 1/2, hence the probability of not showing a prime number is also 1/2. Total possible outcomes when two dice are rolled = 6*6 = 36. Total possible outcomes when two dice are rolled = 6*6 = 36.

4 0

3 years ago

2u^2 + v when u=3 and v=7

Mariana [72]

Answer: 25

Step-by-step explanation:

There are two variables which is u and v and is given that u is 3 and v is 7 so input that into the expression and solve.

2(3)^2 + 7 Solve the exponent first

2(9) + 7 Now multiply then add

18 + 7 = 25

5 0

3 years ago

What is the product?

Marina86 [1]

Answer:

$\frac{2}{k+2}$

Step-by-step explanation:

Simplify the expression and you will get this.

hope this helps

3 0

3 years ago

Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

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

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

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

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

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

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

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

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

4 0

3 years ago

A sequence is defined by the following: an = 4n – 1 What is a1?

Alexandra [31]

an = 4n – 1

a1 means n=1

a1 = 4(1) -1

a1 = 4-1

a1 =3

Answer: 3

6 0

3 years ago

Read 2 more answers