42. the ______ is the end of a vector
a) terminal point
b) initial point
c) ray
d) glide point
43-45 solve the proportion
43) $\frac{3}{4} = \frac{x}{24}$
44) $\frac{x + 2}{x} = \frac{7}{6}$
45) $\frac{2}{5} = \frac{x}{x + 15}$
46-47 solve the ratio
46) $\frac{x}{8} = \frac{8}{16}$
47) $\frac{x}{3} = \frac{12}{x}$
48) right triangle with right angle at c, bc = 9, ac = 40, ab = 41. find $\sin a$, $\cos a$, $\tan a$

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Accepted Answer

### Question 42
# Brief Explanations:
A vector has an initial point (start) and a terminal point (end). A ray is a different geometric object, and "glide point" is not a standard vector term. So the terminal point is the end of a vector.
# Answer:
A. TERMINAL POINT

### Question 43
# Explanation:
## Step1: Cross - multiply the proportion
Given $\frac{3}{4}=\frac{x}{24}$, cross - multiplying gives $4x = 3\times24$.
## Step2: Solve for x
Calculate $3\times24 = 72$, then $x=\frac{72}{4}=18$.
# Answer:
$x = 18$

### Question 44
# Explanation:
## Step1: Cross - multiply the proportion
Given $\frac{x + 2}{x}=\frac{7}{6}$, cross - multiplying gives $6(x + 2)=7x$.
## Step2: Expand and solve for x
Expand the left - hand side: $6x+12 = 7x$. Subtract $6x$ from both sides: $x = 12$.
# Answer:
$x = 12$

### Question 45
# Explanation:
## Step1: Cross - multiply the proportion
Given $\frac{2}{5}=\frac{x}{x + 15}$, cross - multiplying gives $2(x + 15)=5x$.
## Step2: Expand and solve for x
Expand the left - hand side: $2x+30 = 5x$. Subtract $2x$ from both sides: $3x = 30$, then $x = 10$.
# Answer:
$x = 10$

### Question 46
# Explanation:
## Step1: Cross - multiply the ratio
Given $\frac{x}{8}=\frac{8}{16}$, cross - multiplying gives $16x=8\times8$.
## Step2: Solve for x
Calculate $8\times8 = 64$, then $x=\frac{64}{16}=4$.
# Answer:
$x = 4$

### Question 47
# Explanation:
## Step1: Cross - multiply the ratio
Given $\frac{x}{3}=\frac{12}{x}$, cross - multiplying gives $x^{2}=3\times12$.
## Step2: Solve for x
Calculate $3\times12 = 36$, then $x=\pm6$. But since the ratio is likely defined for positive values (in a basic math context), $x = 6$ (we can also consider the domain, and if $x$ is a length or a positive quantity, $x = 6$).
# Answer:
$x = 6$ (or $x=\pm6$ depending on context)

### Question 48
# Explanation:
In a right - triangle $ABC$ with right - angle at $C$, $\sin A=\frac{\text{opposite}}{\text{hypotenuse}}$, $\cos A=\frac{\text{adjacent}}{\text{hypotenuse}}$, $\tan A=\frac{\text{opposite}}{\text{adjacent}}$.
- For $\sin A$: The side opposite to angle $A$ is $BC = 9$, and the hypotenuse $AB = 41$. So $\sin A=\frac{9}{41}$.
- For $\cos A$: The side adjacent to angle $A$ is $AC = 40$, and the hypotenuse $AB = 41$. So $\cos A=\frac{40}{41}$.
- For $\tan A$: The side opposite to angle $A$ is $BC = 9$, and the side adjacent to angle $A$ is $AC = 40$. So $\tan A=\frac{9}{40}$.
# Answer:
$\sin A=\frac{9}{41}$, $\cos A=\frac{40}{41}$, $\tan A=\frac{9}{40}$

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QUESTION IMAGE

Question

Explanation:

Question 42

Step1: Cross - multiply the proportion

Step2: Solve for x

Step1: Cross - multiply the proportion

Step2: Expand and solve for x

Answer:

Question 43