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3 years ago

What’s the slope of the line?

Mathematics

2 answers:

Alexeev081 [22]3 years ago

6 0

Answer:

4/3

Step-by-step explanation:

You use Rise over Run.

It runs 3 blocks to the left (+).

It rises 4 blocks up (+).

Mariulka [41]3 years ago

3 0

Answer:

is the change in y over the change in x

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Which expression is equivalent to 10i/2-i a. –5i – 2 b. –4i – 2 c. 4i – 2 d. 5i – 2

scoray [572]

The answer to the question is
C. 4i-2

4 0

4 years ago

Read 2 more answers

A scientist conducted an experiment. Due to chemical reaction, the temperature of a compound rose 1/10 degree every 1/4 hour for

klio [65]

Answer:

The rate is $\frac{2}{5}$ degree per hour.

Step-by-step explanation:

Given : A scientist conducted an experiment. Due to chemical reaction, the temperature of a compound rose $\frac{1}{10}$ degree every $\frac{1}{4}$ hour for a limited time.

To find : What was the rate, in degrees per hour that the temperature of the compound rose ?

Solution :

In every $\frac{1}{4}$ hour the temperature of a compound rose $\frac{1}{10}$ degree

To get the rate, in degrees per hour,

In every 1 hour the temperature of a compound rose is

$r=\dfrac{\frac{1}{10}}{\frac{1}{4}}$ degree

$r=\dfrac{4}{10}$ degree/hr

$r=\dfrac{2}{5}$ degree/hr

Therefore, the rate is $\frac{2}{5}$ degree per hour.

5 0

3 years ago

ACTIVITY 2 (19) Mr Duma recently inherited a rectangular plot, part of the estate left by his late father. The plot with the fol

seraphim [82]

The sum of the lengths of three sides of the rectangle, gives the length

of the fencing, while one third of the rectangle area is for the pavement.

Responses:

Project A: The formula for the length of the fencing is, L = 4·x - 1

Project B: Length of the fancy wall = 2·x + 1

$Project \ C:Area \ of \ the \ paving = \underline{ \dfrac{2\cdot x^2 }{3} - \dfrac{x }{3} - \dfrac{1}{3}}$

<h3>Which method can be used to find the length and area of paving from the given equations?</h3>

Project A: Let QP and SR represent the longest sides of the rectangle, we have;

PQ = SR = 2·x + 1

Given parameters are;

Length of the rectangular plot = 2·x + 1

Width of the rectangular plot = x - 1

Vertices of the rectangular plot are; QPSR

Project A: Let QP and SR represent the longest sides of the rectangle, we have;

PQ = SR = 2·x + 1

Which gives;

SP = QR = x - 1

The length of the fencing, L = SP + PQ + QR = x - 1 + 2·x + 1 + x - 1 = 4·x - 1

The formula for the length of the fencing is, L = 4·x - 1

Project B: The front side is SR

Therefore;

Length of the fancy wall = 2·x + 1

Project C:

Area, A = Length × Width

Area of the plot, A = (2·x + 1) × (x - 1) = 2·x² - x - 1

$Area \ of \ the \ paving = \dfrac{1}{3} \times \left(2 \cdot x^2 - x - 1\right) = \underline{ \dfrac{2\cdot x^2 }{3} - \dfrac{x }{3} - \dfrac{1}{3}}$