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3 years ago

Use Structure A line passes through points Aln, 4) and B(6, 8) and is parallel to y = 2x - 5. What is the value of n?

Mathematics

1 answer:

artcher [175]3 years ago

8 0

Step-by-step explanation:

From y=2x-5, gradient=2

Change in y = Gradient

Change in x

8-4 =2

6-n

4 = 2(6-n)

4 =12 -2n

4 - 12 = -2n

-8 = -2n

-8 =n

-2

n= 4

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Nadusha1986 [10]

Your answer would be D.

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4 years ago

If I could have help on this that would be great.

denis-greek [22]

Answer: Choice C) Harold

Darren is correct because he subtracted c from both sides

Julie is correct because she multiplied both sides by H, and then divided both sides by L

Brenda is correct because she multiplied both sides by r

Harold is not correct. We can see this with a counter example. If A = 7 and h = 1, then the first equation Harold wrote leads to b = 14. Now plug those same values into the second equation and we get h = 2b/A turning into 1 = 2*14/7 and that simplifies to 1 = 4, which is a false statement. This contradiction is one example why the two equations Harold wrote aren't equivalent.

6 0

3 years ago

Read 2 more answers

APPLICATIONS

n200080 [17]

Answer:

(a) (4,9), (8,14).; (b) 1.25 in/h; (c) d = 1.25h + 4 ; (d) 4 in; (e) 19 in

Step-by-step explanation:

(a) Information as coordinate pairs

4 h, 9 in ⟶ (4, 9)

8 h, 14 in ⟶ (8,14)

The coordinates of the two points are (4,9), (8,14).

(b) Slope of the line

$\begin{array}{rcl}m & = & \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\ & = & \dfrac{14 - 9}{8 - 4}\\\\& = & \dfrac{5}{4}\\\\& = & \mathbf{1.25}\end{array}$

The slope of the line is 1.25 in/h

(c) Equation of the line

$\begin{array}{rcl}d & = & mh + b\\9 & = & \dfrac{5}{4} \times 4 + b\\\\9& = & 5 + b\\b & = & \mathbf{4}\\\end{array}$

The y intercept is at (0,4)

d = 1.25h + 4

(d) Depth when snowfall began

$\begin{array}{rcl}d &= &\text{0 h } \times \dfrac{\text{1.25 in}}{\text{1 h}} \text{+ 4 in}\\\\&= &\text{0 in+ 4 in}\\& = & \textbf{4 in}\\\end{array}$

The depth was 4 in when the snowfall began.

(e) Depth after 12 h

$\begin{array}{rcl}d &= &\text{12 h } \times \dfrac{\text{1.25 in}}{\text{1h}} \text{+ 4 in}\\\\&= &\text{15 in+ 4 in}\\& = & \textbf{19 in}\\\end{array}$

The depth would be 19 in after 12 h.

The figure below is a graph of your function. It shows that the depth was 4 in at the start of the snowfall, that it increases by 5 in every 4 h, and predicts a depth of 19 in after 12 h.