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3 years ago

How do you add integers with different signs?

2 answers:

7 0

A Because whenever you’re adding integers that have the same thing you keep the same signs in at the absolute value of each number

hjlf3 years ago

5 0

B because if you have something like 2+-3 it would equal -1 because 3-2 is 1 and -3 is the larger number

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Identify the correct equation that matches the graph.

kobusy [5.1K]

Answer:

I don't know either B or D ........

4 0

3 years ago

Read 2 more answers

Which point is on the line that passes through point Z and is perpendicular to line AB?

tankabanditka [31]

Step 1

Find the slope of the line AB

we know that

the formula to find the slope between two points is equal to

$m=\frac{(y2-y1)}{(x2-x1)}$

we have

$A(-2,4)\\B(0,-4)$

substitute in the formula

$m=\frac{(-4-4)}{(0+2)}$

$m=\frac{(-8)}{(0+2)}$

$m1=-4$

Step 2

Find the slope of the line that passes through point Z and is perpendicular to line AB

we know that

If two lines are perpendicular, then the product of their slopes is equal to minus one

$m1*m2=-1$

$m2=-\frac{1}{m1}$

we have

$m1=-4$

substitute

$m2=\frac{1}{4}$

Step 3

Find the equation of the line that passes through point Z and is perpendicular to line AB

we have

$Z(0,2)\\m2=\frac{1}{4}$

we know that

the equation of the line into point -slope form is equal to

$y-y1=m*(x-x1)$

substitute

$y-2=\frac{1}{4}*(x-0)$

$y=\frac{1}{4}x+2$

Verify which point is on the line that passes through point Z and is perpendicular to line AB

we know that

If a point is on the line, then the coordinates of the point must be satisfy the equation of the line

we are going to proceed to verify each point to determine the solution

Substitute the values of x and y of the point in the equation of the line. If the equation is true, then the point is on the line

Step 4

point $A(-4,1)$

$1=\frac{1}{4}*(-4)+2$

$1=1$ --------> the equation is true

therefore

the point $A(-4,1)$ is on the line

Step 5

point $B(1,-2)$

$-2=\frac{1}{4}*(1)+2$

$-2=\frac{9}{4}$ --------> the equation is not true

therefore

the point $B(1,-2)$ is not on the line

Step 6

point $C(2,0)$

$0=\frac{1}{4}*(2)+2$

$0=\frac{5}{2}$ --------> the equation is not true

therefore

the point $C(2,0)$ is not on the line

Step 7

point $D(4,4)$

$4=\frac{1}{4}*(4)+2$

$4=3$ --------> the equation is not true

therefore

the point $D(4,4)$ is not on the line

the answer is the option

$A(-4,1)$

4 0

3 years ago

Read 2 more answers

10x^3+40x^2+15x ------------------------------ 5x

Anna71 [15]

Answer:

Step-by-step explanation:

$\dfrac{10x^3 + 40x^2 + 15x}{5x}\\ \dfrac{10x^3}{5x} +\dfrac{40x^2}{5x}+\dfrac{15x}{5x}$

2x^2 + 8x + 3 is your final answer.

4 0

3 years ago

Given the following sets.

fenix001 [56]

It is an empty set because nothing relates the two subsets

4 0

3 years ago

Consider a function f(x) such that f(5x) =x-5/5x-1. Find f(x) and hence write down the domain of f(x).

nasty-shy [4]

Answer:

$f(x) = \frac{x - 25}{5x - 5}$

Domain: $x \ne 1$

Step-by-step explanation:

Given

$f(5x) = \frac{x - 5}{5x - 1}$

Required

Determine f(x) and its domain

To determine f(x), we replace 5x with x in f(5x)

So, we have:

$f(5x) = \frac{\frac{5x}{5} - 5}{5x - 1}$

$f(x) = \frac{\frac{x}{5} - 5}{x - 1}$

Take LCM

$f(x) = \frac{\frac{x - 25}{5}}{x - 1}$

$f(x) = \frac{x - 25}{5} * \frac{1}{x-1}$

$f(x) = \frac{x - 25}{5x - 5}$

To get the domain, we set the denominator as

$5x - 5 \ne 0$

$5x \ne 5$

$x \ne 1$

4 0

3 years ago