All categories

3 years ago

PLEASE HELP ME!!!..

Mathematics

1 answer:

Alex17521 [72]3 years ago

4 0

Step-by-step explanation:

By Pythagoras' Theorem, we have

Hypotenuse = √[(13.9)² + (10.2)²] = 17.2. (B)

You might be interested in

How do I solve a liner equation in the form a(x-b)=cx? I have no idea plz help

Vitek1552 [10]

Distribute on the left
3x-15=7x

Subtract 3x from both sides
-15= 4x

Divide both sides by 4
-15/4= x

Final answer: x=-15/4 or -3.75

7 0

3 years ago

What is the relationship between lines a and b?

balu736 [363]

Answer:

they are parallel, aka "A"

Step-by-step explanation:

the lines (NOT line segments) never ever cross or meet eachother so they must be paralell

5 0

3 years ago

Read 2 more answers

Graph the line that contains the point (-2,-4) and has a slope of<br> 1/4

mixer [17]

Answer:

y = 1/4x - 7/2

Step-by-step explanation:

To answer this question, we have to use point-slope form:

y - y1 = m(x - x1)

Substitute the numbers into the variables:

y - (-4) = 1/4(x - (-2))

y + 4 = 1/4(x + 2)

y + 4 = 1/4x + 1/2

y = 1/4x - 7/2

To verify your answer, simply substitute the values into your equation:

-4 = 1/4(-2) - 7/2

-4 = -1/2 - 7/2

-4 = -8/2

-4 = -4

3 0

3 years ago

(30 points!) Powers of i worksheet

Gelneren [198K]

Answer:

x+7=14

x=7

(7)+7=14

Step-by-step explanation:

7 0

3 years ago

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.

8 0

3 years ago