All categories

2 years ago

Find the slope of the the line graphed

Mathematics

1 answer:

saveliy_v [14]2 years ago

4 0

Answer:

$m=\frac{-5}{2}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Slope Formula: $m=\frac{y_2-y_1}{x_2-x_1}$

Step-by-step explanation:

Step 1: Define

Find points from graph.

Point (1, 1)

Point (3, -4)

Step 2: Find slope m

Simply plug in the 2 coordinates into the slope formula to find slope m

Substitute [SF]: $m=\frac{-4-1}{3-1}$
Subtract: $m=\frac{-5}{2}$

You might be interested in

An injury from a motor vehicle crash happens in the United States every seconds.

larisa [96]

The answer is Every 15 seconds

8 0

3 years ago

X is less than or equal to 24.

GREYUIT [131]

Answer:

yes

Step-by-step explanation:

x isnt a number

6 0

2 years ago

Read 2 more answers

Help ASAP !!!!! Which number line represents the solutions to \x + 4) = 2?

Pachacha [2.7K]

Answer:

first option

Step-by-step explanation:

Given

| x + 4 | = 2

The absolute value function always produces a positive value but the expression inside can be positive or negative, thus

x + 4 = 2 ( subtract 4 from both sides )

x = - 2

- (x + 4) = 2, distribute left side

- x - 4 = 2 ( add 4 to both sides )

- x = 6 ( multiply both sides by - 1 )

x = - 6

Thus solutions are x = - 2, x = - 6

These are represented on the graph by solid blue circles at x = - 2, x = - 6

The first option represents the solution

6 0

2 years ago

-3 + bracket -5 bracket close

vodomira [7]

-3 (-5) = 15

two negatives= a positive

when a number is next to a number in brackets in means multiply.

8 0

2 years ago

Show that if a and b are positive integers, then ab = gcd (a, b). lcm (a, b) . [Hint: Use prime factorizations of a and b and th

AleksandrR [38]

Answer:

$ab=\gcd(a,b)\cdot \text{lcm}(a,b)$

Step-by-step explanation:

Using the hint, write a and b in the following prime factorization:

$a=p_1^{x_1}p_2^{x_2}\cdots p_t^{x_t}\cdot q,$

$b=p_1^{y_1}p_2^{y_2}\cdots p_t^{y_t}\cdot r,$

where $\gcd(p_i,r)=1,\ \gcd(p_i,q)=1,\ \gcd(r,q)=1,\ \gcd(p_i,p_j)=1,$ for i ≠ j.

Then by the formulae for gcd(a,b) and lcm(a,b) we know that:

$\gcd(a,b)=p_1^{\min(x_1,y_1)}p_2^{\min (x_2,y_2)} \cdots p_t^{\min(x_t,y_t)}$

$\text{lcm}(a,b)= q\cdot r\cdot p_1^{\max(x_1,y_1)}p_2^{\max(x_2,y_2)}\cdots p_t^{\max(x_t,y_t)}$

Note that the expression $\min(x_i,y_i)+\max(x_i,y_i)=x_i+y_i$ for all i, since if the minimum is, without loss of generality, $x_i$ , then the maximum must be $y_i$ , and viceversa. Then, it is straightforward to verify that when we multiply gcd(a, b) and lcm(a, b) its prime factorization matches the prime factorization of ab, and so we can see the equaility holds:

$\gcd(a,b)\cdot \text{lcm}(a,b)=ab.$

7 0

3 years ago