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2 years ago

What is the slope of (-2,2) (-4,-4)

Mathematics

2 answers:

Flauer [41]2 years ago

7 0

Answer:

Step-by-step explanation:

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

(-4-2)/(-4--2)

-6/-2

Andrei [34K]2 years ago

7 0

Answer:

to find slope we use the following methods

Step-by-step explanation:

given

X1 -2

X2= -4

Y1= 2

Y2= -4

slope=y2-Y1 ÷X2-X1

= -4-2÷ -4-(-2)

=-6 ÷-4+2

= -6÷-2

= 3

slop of (-2,2) (-4,-4) = 3

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Find x so that the points (x,x+1), (x+2,x+3) and (x+3,2x+4) form a right-angled triangle.

azamat

Let a, b, and c be vectors each starting at the origin and terminating at the points (x, x + 1), (x + 2, x + 3), and (x + 3, 2x + 4), respectively.

Then the vectors a - b, a - c, and b - c are vectors that point in directions parallel to each of the legs formed by the triangle with these points as its vertices.

If this triangle is to contain a right angle, then exactly one of these pairs of vectors must be orthogonal. In other words, one of the following must be true:

(a - b) • (a - c) = 0

or

(a - b) • (b - c) = 0

or

(a - c) • (b - c) = 0

We have

a - b = (x, x + 1) - (x + 2, x + 3) = (-2, -2)

a - c = (x, x + 1) - (x + 3, 2x + 4) = (-3, -x - 3)

b - c = (x + 2, x + 3) - (x + 3, 2x + 4) = (-1, -x - 1)

Case 1: If (a - b) • (a - c) = 0, then

(-2, -2) • (-3, -x - 3) = (-2)×(-3) + (-2)×(-x - 3) = 2x + 12 = 0 ==> x = -6

which would make a - c = (-3, 3) and b - c = (-1, 5), and their dot product is not zero. Then the triangles vertices are at the points (-6, -5), (-4, -3), and (-3, -8).

Case 2: If (a - b) • (b - c) = 0, then

(-2, -2) • (-1, -x - 1) = (-2)×(-1) + (-2)×(-x - 1) = 2x + 4 = 0 ==> x = -2

which would make a - c = (-3, -1) and b = (-1, 1), and their dot product is also not zero. The vertices are the points (-2, -1), (0, 1), and (1, 0).

Case 3: If (a - c) • (b - c) = 0, then

(-3, -x - 3) • (-1, -x - 1) = (-3)×(-1) + (-x - 3)×(-x - 1) = x ² + 4x + 6 = 0

but the solutions to x here are non-real, so we throw out this case.

So there are two possible values of x that make a right triangle, x = -6 and x = -2.

3 0

3 years ago

Can someone please explain to me how to solve this equation?

Nikolay [14]

Answer:

first of all erase what u have written

Step-by-step explanation:

25-3x-5=2x+16+x
20-6x=16
6x=4
x=2/3

6 0

3 years ago

155 divided by 17. .......

monitta

155 divided by 17 =
9.1176

7 0

3 years ago

Read 2 more answers

For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

C. $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

From the above expression, the power of (5) is 2

Express 2 as 6 - 4

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

Check the list of options for the expression on the left hand side

The correct answer is $(5-\frac{1}{2})^6$

3 0

3 years ago

Simultaneous equations 2x+3y=13 4x-y=-2

Rufina [12.5K]

Answer:

(0.5, 4 )

Step-by-step explanation:

Given the 2 equations

2x + 3y = 13 → (1)

4x - y = - 2 → (2)

Multiplying (2) by 3 and adding to (1) will eliminate the term in y

12x - 3y = - 6 → (3)

Add (1) and (3) term by term to eliminate y

14x = 7 ( divide both sides by 14 )

x = 0.5

Substitute x = 0.5 into either of the 2 equations and evaluate for y

Substituting into (1)

2(0.5) + 3y = 13

1 + 3y = 13 ( subtract 1 from both sides )

3y = 12 ( divide both sides by 3 )

y = 4

Solution is (0.5, 4 )

3 0

3 years ago

What is the slope of ￼(-2,2) (-4,-4)

What is the slope of (-2,2) (-4,-4)