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3 years ago

Find the slope given these two points. (-1, 4) and (4, -3)

Mathematics

1 answer:

Paraphin [41]3 years ago

3 0

Answer:

-7/5

Step-by-step explanation:

Since we have 2 points, we can find the slope using

m = ( y2-y1)/(x2-x1)

= ( -3 -4)/( 4 - -1)

( -3-4) /( 4+1)

-7/5

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The correlation coefficient for the data is 0.9976. Which statement is true about the data in the table?

Amanda [17]

Since it's close to 1 the data will fall very close to a straight line. The relationship between x and y will be the slope of the line.

Hope this helps...

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3 years ago

Polygon ABCD is translated to create polygon A’B’C’D’. Point A is located at (1,5) and point A’ is located at (-2,3). Which expr

Tatiana [17]

Answer:

The translation is 3 units left and 2 units down

Step-by-step explanation:

we know that

Point A is located at (1,5) and point A’ is located at (-2,3)

The rule of the translation is

(x,y) -----> (x+a,y+b)

(1,5) ----> (-2,3)

Find the values of a and b

1+a=-2 ----> a=-2-1 ----> a=-3

5+b=3 ----> b=3-5 ----> b=-2

The rule of the translation is

(x,y) -----> (x-3,y-2)

That means -----> The translation is 3 units left and 2 units down

8 0

3 years ago

Read 2 more answers

if a cookie jar has 24 cookies how many cookies are left in the jar after eating some.How many cookies are in the jar after you

Delvig [45]

Answer:

15&12

Step-by-step explanation:

24-9=15

15-3=12

7 0

3 years ago

Read 2 more answers

Factor the equation please help

LiRa [457]

Answer:

For A

${ \rm{6 {x}^{2}(x + 1) - 2x(x + 1) }}$

• The equation above has a common bracket "(x + 1)". So, let's first factorise out that bracket:

$\dashrightarrow \: { \rm{(x + 1) \{6 {x}^{2} - 2x \} }}$

• now in the second bracket, the common factor is x and 2.

$\dashrightarrow \: { \rm{(x + 1) \{2x(3x - 1) \}}}$

• Final answer;

${ \boxed{ \boxed{ \rm{ \dashrightarrow \: 2x(x + 1)(3x - 1) \: \: }}}}$

For B

${ \rm{3(x - 1)(x + 2) + ( {x}^{2} - x)(x + 2) }}$

• In the equation, the common bracket is (x + 2).

So let's first factorise it out:

$\dashrightarrow \: { \rm{(x + 2) \{3(x - 1) + ( {x}^{2} - x) \}}} \\ \\ { \rm{(x + 2) \{3(x - 1) + x(x - 1) \}}}$

• In the second major bracket, the common bracket is (x - 1). so factorise it out:

${ \rm{(x + 2) (x - 1) \{3 + x \}}}$

• Final answer:

${ \boxed{ \boxed{ \rm{ \dashrightarrow \: (x + 2)(x + 3)(x - 1) \: \: }}}}$

8 0

2 years ago

Pre-Calc help:

stepladder [879]

Answer:

$8x^2+8y^2-79x-32y+95=0$

Step-by-step explanation:

Let the equation be : $x^2+y^2+2gx+2fy+c=0$

The point (1,1) must satisfy this equation.

$1^2+1^2+2g*1+2f*1+c=0\\2g+2f+c=-2---(1)$

Similarly, the point (1,3) must satisfy:

$1^2+3^2+2g*1+2f*3+c=0\\1+9+2g+6f+c=0\\2g+6f+c=-10---(2)$

Also for the point (9,2), we have:

$9^2+2^2+2g*9+2f*2+c=0\\81+4+18g+4f+c=0\\18g+4f+c=-85---(3)$

Solving the equations (1), (2) and (3) simultaneously, we get:

$g=-\frac{-79}{16},f=-2,c=\frac{95}{8}$

We Substitute this values to get:

$x^2+y^2+2(\frac{-79}{16})x+2(-2)y+\frac{95}{8}=0$

$x^2+y^2-(\frac{79}{8})x-4y+\frac{95}{8}=0$

Multiply through by 8 to get:

$8x^2+8y^2-79x-32y+95=0$

8 0

3 years ago