15.
$sin 45^{circ}$ =
$cos 45^{circ}$ =
$ an 45^{circ}$ =
16.
$sin 30^{circ}$ =
$cos 30^{circ}$ =
$ an 30^{circ}$ =
17.
$sin 60^{circ}$ =
$cos 60^{circ}$ =
$ an 60^{circ}$ =
go
identify the type of function (linear, quadratic, or exponential) represented by the table or graph, and write the explicit equation for the function.
18.
x | y
0 | 3
1 | 6
2 | 12
3 | 24
4 | 48
5 | 96
unit 5, lesson 5 – ready, set, go
© 2021 open up resources and mathematics vision project
4 of 6
download for free at openup.org

Question

15.
$sin 45^{circ}$ =
$cos 45^{circ}$ =
$	an 45^{circ}$ =
16.
$sin 30^{circ}$ =
$cos 30^{circ}$ =
$	an 30^{circ}$ =
17.
$sin 60^{circ}$ =
$cos 60^{circ}$ =
$	an 60^{circ}$ =
go
identify the type of function (linear, quadratic, or exponential) represented by the table or graph, and write the explicit equation for the function.
18.
x | y
0 | 3
1 | 6
2 | 12
3 | 24
4 | 48
5 | 96
unit 5, lesson 5 – ready, set, go
© 2021 open up resources and mathematics vision project
4 of 6
download for free at openup.org

Accepted Answer

### Question 15
# Explanation:
## Step1: Recall special angle values
For $45^\circ$ in a right - isosceles triangle (where the two legs are equal, let the length of each leg be $a$, and the hypotenuse be $c=\sqrt{a^{2}+a^{2}}=\sqrt{2}a$).
The sine of an angle in a right triangle is $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, the cosine is $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$, and the tangent is $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$.
For $\theta = 45^\circ$, $\sin45^{\circ}=\frac{a}{\sqrt{2}a}=\frac{\sqrt{2}}{2}$
## Step2: Calculate $\cos45^{\circ}$
Using the formula $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$ for $\theta = 45^\circ$, since the adjacent side and the opposite side are equal (in isosceles right triangle), $\cos45^{\circ}=\frac{a}{\sqrt{2}a}=\frac{\sqrt{2}}{2}$
## Step3: Calculate $\tan45^{\circ}$
Using the formula $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$ for $\theta = 45^\circ$, since the opposite and adjacent sides are equal (length $a$), $\tan45^{\circ}=\frac{a}{a} = 1$

# Answer:
$\sin45^{\circ}=\frac{\sqrt{2}}{2}$, $\cos45^{\circ}=\frac{\sqrt{2}}{2}$, $\tan45^{\circ}=1$

### Question 16
# Explanation:
## Step1: Recall special angle values for $30^\circ$
In a $30 - 60-90$ triangle, the side opposite $30^\circ$ is the shortest side. Let the side opposite $30^\circ$ be $a$, the hypotenuse is $2a$, and the side opposite $60^\circ$ is $\sqrt{3}a$.
For $\sin30^{\circ}$, using $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, $\sin30^{\circ}=\frac{a}{2a}=\frac{1}{2}$
## Step2: Calculate $\cos30^{\circ}$
Using $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$, the adjacent side to $30^\circ$ is $\sqrt{3}a$, so $\cos30^{\circ}=\frac{\sqrt{3}a}{2a}=\frac{\sqrt{3}}{2}$
## Step3: Calculate $\tan30^{\circ}$
Using $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$, the opposite side to $30^\circ$ is $a$ and the adjacent side is $\sqrt{3}a$, so $\tan30^{\circ}=\frac{a}{\sqrt{3}a}=\frac{\sqrt{3}}{3}$

# Answer:
$\sin30^{\circ}=\frac{1}{2}$, $\cos30^{\circ}=\frac{\sqrt{3}}{2}$, $\tan30^{\circ}=\frac{\sqrt{3}}{3}$

### Question 17
# Explanation:
## Step1: Recall special angle values for $60^\circ$
In a $30 - 60-90$ triangle, the side opposite $60^\circ$ is $\sqrt{3}a$ (where $a$ is the side opposite $30^\circ$) and the hypotenuse is $2a$.
For $\sin60^{\circ}$, using $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, $\sin60^{\circ}=\frac{\sqrt{3}a}{2a}=\frac{\sqrt{3}}{2}$
## Step2: Calculate $\cos60^{\circ}$
Using $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$, the adjacent side to $60^\circ$ is $a$ (the side opposite $30^\circ$) and the hypotenuse is $2a$, so $\cos60^{\circ}=\frac{a}{2a}=\frac{1}{2}$
## Step3: Calculate $\tan60^{\circ}$
Using $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$, the opposite side to $60^\circ$ is $\sqrt{3}a$ and the adjacent side is $a$, so $\tan60^{\circ}=\frac{\sqrt{3}a}{a}=\sqrt{3}$

# Answer:
$\sin60^{\circ}=\frac{\sqrt{3}}{2}$, $\cos60^{\circ}=\frac{1}{2}$, $\tan60^{\circ}=\sqrt{3}$

### Question 18
# Explanation:
## Step1: Determine the type of function
To determine if the function is linear, quadratic, or exponential, we check the differences or ratios.
- For linear function: The first - differences (difference between consecutive $y$ - values) should be constant.
  - $6 - 3=3$, $12 - 6 = 6$, $24 - 12=12$, $48 - 24 = 24$, $96 - 48=48$. The first - differences are not constant.
- For quadratic function: The second - differences should be constant. But since the first - differences are $3,6,12,24,48$ (which are increasing by a factor), we check the ratios.
- For exponential function: The ratio of consecutive $y$ - values should be constant.
  - $\frac{6}{3}=2$, $\frac{12}{6}=2$, $\frac{24}{12}=2$, $\frac{48}{24}=2$, $\frac{96}{48}=2$. The ratio of consecutive $y$ - values is constant ($r = 2$). So the function is exponential.
## Step2: Write the explicit equation for exponential function
The general form of an exponential function is $y = ab^{x}$, where $a$ is the initial value (when $x = 0$) and $b$ is the common ratio.
When $x = 0$, $y=3$, so $a = 3$. The common ratio $b = 2$ (from the ratios of consecutive $y$ - values).
So the equation is $y=3\times(2)^{x}$

# Answer:
The function is exponential. The explicit equation is $y = 3\times2^{x}$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Question 15

Step1: Recall special angle values

Step2: Calculate \(\cos45^{\circ}\)

Step3: Calculate \(\tan45^{\circ}\)

Step1: Recall special angle values for \(30^\circ\)

Step2: Calculate \(\cos30^{\circ}\)

Step3: Calculate \(\tan30^{\circ}\)

Step1: Recall special angle values for \(60^\circ\)

Step2: Calculate \(\cos60^{\circ}\)

Step3: Calculate \(\tan60^{\circ}\)

Answer:

Question 16