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4 years ago

A line that includes the points (7,u) and (6,6) has a slope of -8. What is the value of u?

Mathematics

2 answers:

Zielflug [23.3K]4 years ago

8 0

Answer:

(6-u)/(6-7)= (6-u)/-1= 8

6 - u = -8

-u = -14

u = 14

Tju [1.3M]4 years ago

3 0

Answer:u = - 2

Step-by-step explanation:

Slope = change in value of y on the vertical axis / change in value of x on the horizontal axis

change in the value of y = y2 - y1

Change in value of x = x2 -x1

y2 = final value of y

y 1 = initial value of y

x2 = final value of x

x1 = initial value of x

From the information given,

Slope = - 8

y2 = 6

y1 = u

x2 = 6

x1 = 7

Therefore,

- 8 = (6 - u)/(6 - 7)

- 8 = (6 - u)/ - 1

6 - u = 8

u = 6 - 8 = - 2

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The perimeter of a rectangle is 220 inches.

irina1246 [14]

The length=77 and the Width=33 because if you add 77+77+33+33 it equals 220

6 0

3 years ago

Solve for x in the equation x^2-14+31=63

Bingel [31]

To solve for a variable, you need to get it (x) by itself.

x² - 14 + 31 = 63 Add 31 to -14
x² + 17 = 63 Subtract 17 from both sides of the equation
x² = 46 Square root both sides
x = $\pm \sqrt{46}$

Check your work by plugging in $\sqrt{46}$ for x and then $- \sqrt{46}$ .

$\sqrt{46}$ ² - 14 + 31 = 63 Square $\sqrt{46}$
46 - 14 + 31 = 63 Subtract 14 from 46
32 + 31 = 63 Add
63 = 63

$- \sqrt{46}$ ² - 14 + 31 = 63 Square $- \sqrt{46}$
46 - 14 + 31 = 63 Subtract 14 from 46
32 + 31 = 63 Add
63 = 63

So, x is $\sqrt{46}$ and $- \sqrt{46}$ .

3 0

3 years ago

Determine which of the following sets of three points constitute the vertices of a right triangle: (a) 3 + 5i,2 +2i,5i; (b)2i,3

Illusion [34]

Answer:

Option (c) is correct

Step-by-step explanation:

Case (a)

A = 3 + 5i = (3, 5)

B = 2 + 2i = (2, 2)

C = 5i = (0, 5)

Use the distance formula to find the distance between two points

$AB = \sqrt{(2-3)^{2}+(2-5)^{2}}=\sqrt{10}$

$BC = \sqrt{(0-2)^{2}+(5-2)^{2}}=\sqrt{13}$

$CA = \sqrt{(0-3)^{2}+(5-5)^{2}}=\sqrt{9}$

For the triangle to be right angles triangle

$BC^{2}=AB^{2}+CA^{2}$

Here, it is not valid, so these are not the points of a right angled triangle.

Case (b)

A = 2i = (0, 2)

B = 3 + 5i = (3, 5)

C = 4 + i = (4, 1)

Use the distance formula to find the distance between two points

$AB = \sqrt{(3-0)^{2}+(5-2)^{2}}=\sqrt{18}$

$BC = \sqrt{(4-3)^{2}+(1-5)^{2}}=\sqrt{17}$

$CA = \sqrt{(4-0)^{2}+(1-2)^{2}}=\sqrt{17}$

For the triangle to be right angles triangle

$AB^{2}=BC^{2}+CA^{2}$

Here, it is not valid, so these are not the points of a right angled triangle.

Case (c)

A = 6 + 4i = (6, 4)

B = 7 + 5i = (7, 5)

C = 8 + 4i = (8, 4)

Use the distance formula to find the distance between two points

$AB = \sqrt{(7-6)^{2}+(5-4)^{2}}=\sqrt{2}$

$BC = \sqrt{(8-7)^{2}+(4-5)^{2}}=\sqrt{2}$

$CA = \sqrt{(8-6)^{2}+(4-4)^{2}}=\sqrt{4}$

For the triangle to be right angles triangle

$CA^{2}=BC^{2}+AB^{2}$

Here, it is valid, so these are the points of a right angled triangle.

7 0

3 years ago

The sum of the first 15 terms of a geometric sequence with 7 as the first term and a common ratio of -3 is __________.

Andre45 [30]

Step-by-step explanation:

the formula for the sum of the first n terms of a geometric sequence is

Sn = s1(1 - r^n)/(1 - r)

with r being the common ratio and s1 is the first term.

so,

S15 = 7×(1 - (-3)¹⁵)/(1 - -3) = 7×(1 - -14,348,907)/4 =

= 7×14,348,908/4 = 7×3,587,227 = 25,110,589

4 0

2 years ago

2a2b3 and -4a2b3 are like terms.<br><br>True<br>False

Mademuasel [1]

The answer would be true, yes these terms are like terms.

By definition, like terms are any two terms that contain the same variable which are raised to the same power.

Because both terms have two variables, a and b, and the a variable of both are raised to the second power, and the b is raised to the 3rd power, both terms are considered like terms.

8 0

3 years ago