Anyone no 8th grade math if so please help my sister out

Mathematics

2 answers:

qaws [65]3 years ago

7 0

Answer:

2/5

Step-by-step explanation:

you want to start with the point on the left and then count 2 up to the line that the other point is on than you count over 5 to the point on the right

then you put rise over run which makes it 2/5

Brilliant_brown [7]3 years ago

6 0

Answer:

2/5

Step-by-step explanation:

rise over run formula

Therefore the slope is increasing by two fifths units.

2/5 is your answer. ❤

Brainliest please!? ✌

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Step-by-step explanation:

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3 years ago

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Change the subject of the formula L = v 4kt - p to k.

Romashka [77]

Answer:

$\boxed{k = \frac{L^2 + p}{4t}}$

General Formulas and Concepts:

Algebra I

Basic Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Step-by-step explanation:

Step 1: Define

Identify given.

$\displaystyle L = \sqrt{4kt - p}$

Step 2: Solve for k
We can use equality properties to help us rewrite the equation to get k as our subject:

Let's first square both sides:

$\displaystyle\begin{aligned}L = \sqrt{4kt - p} & \rightarrow L^2 = \big( \sqrt{4kt - p} \big) ^2 \\& \rightarrow L^2 = 4kt - p \\\end{aligned}$

Next, add p to both sides:

$\displaystyle\begin{aligned}L = \sqrt{4kt - p} & \rightarrow L^2 = \big( \sqrt{4kt - p} \big) ^2 \\& \rightarrow L^2 = 4kt - p \\& \rightarrow L^2 + p = 4kt \\\end{aligned}$

Next, divide 4t by both sides:

$\displaystyle\begin{aligned}L = \sqrt{4kt - p} & \rightarrow L^2 = \big( \sqrt{4kt - p} \big) ^2 \\& \rightarrow L^2 = 4kt - p \\& \rightarrow L^2 + p = 4kt \\& \rightarrow \frac{L^2 + p}{4t} = k \\\end{aligned}$