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3 years ago

10

How would I solve these equations?

2 answers:

dmitriy555 [2]3 years ago

5 0

51. X = 1/5, -6 (negative six)

52. X = 0, -3/2

53. X = 0, 1/5

54. Z = 0, -7/4

kozerog [31]3 years ago

4 0

51) 5x-1=0 and x+6=0
5x=1, x=1/5 and x=-6
2 solutions x=1/5 and x=-6

52)x(2x+3)=0
2x=0 and 2x+3 =0
x=0 and x=-3/2=-1.5

53) x(5x-1)=0
x=0 and 5x-1=0
x=0 and x=1/5

54) z(4z+7) =0
z=0 and 4z+7=0
z=0 and z= - 7/4 = -1.75
z=0 and z=-1.75

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Can someone help me pls

VLD [36.1K]

Answer:

-12

Step-by-step explanation:

Since a = -6,we have to multiply 2 and -6,and you get -12 bc we have to put the sign of the bigger digit which is -6 in this problem,so you have to write -12

4 0

2 years ago

6 - 3x = 5x - 10x + 8

Jet001 [13]

Answer:

x=1

Step-by-step explanation:

6-3x=5x-10x+8=

move all terms to the left:

6-3x-(5x-10x+8)=0

add all the numbers together, and all the variables

-3x-(-5x+8)+6=0

get rid of parentheses

-3x+5x-8+6=0

add all the numbers together, and all the variables

2x-2=0

move all terms containing x to the left, all other terms to the right

2x=2

x=2/2

x=1

3 0

3 years ago

Consider two people being randomly selected. (For simplicity, ignore leap years.)

inna [77]

Answer:

$(a) = \frac{144}{133225} \\\\(b) = \frac{1}{365}$

Step-by-step explanation:

Part (a) the probability that two people have a birthday on the 9th of any month.

Neglecting leap year, there are 365 days in a year.

There are 12 possible 9th in months that make a year calendar.

If two people have birthday on 9th; P(1st person) and P(2nd person).

$=\frac{12}{365} X\frac{12}{365} = \frac{144}{133225}$

Part (b) the probability that two people have a birthday on the same day of the same month

P(2 people selected have birthday on the same day of same month) + P(2 people selected not having birthday on same day of same month) = 1

P(2 people selected not having birthday on same day of same month):

$= \frac{365}{365} X \frac{364}{365} =\frac{364}{365}$

P(2 people selected have birthday on the same day of same month) $= 1-\frac{364}{365} \\\\= \frac{1}{365}$

7 0

3 years ago

On a coordinate plane, rhombus W X Y Z is shown. Point W is at (7, 2), point X is at (5, negative 1), point Y is at (3, 2), and

Otrada [13]

Answer:

$P = 4\sqrt{13}$

Step-by-step explanation:

Given

$W = (7, 2)$

$X = (5, -1)$

$Y = (3, 2)$

$Z =(5, 5)$

Required

The perimeter

To do this, we first calculate the side lengths using distance formula

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2$

So, we have:

$WX = \sqrt{(5- 7)^2 + (-1 - 2)^2$

$WX = \sqrt{13}$

$XY = \sqrt{(3-5)^2 + (2--1)^2}$

$XY = \sqrt{13}$

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

$YZ = \sqrt{13}$

$ZW = \sqrt{(7 - 5)^2 + (2 - 5)^2}$

$ZW = \sqrt{13}$

The perimeter is:

$P = WX + XY + YZ + ZW$

$P = \sqrt{13}+\sqrt{13}+\sqrt{13}+\sqrt{13}$

$P = 4\sqrt{13}$

5 0

3 years ago

Read 2 more answers

What is an equation of the line that passes through the points (5, 2) and (-5, -6)?

lawyer [7]

Answer:

y=4/5x + 2

Step-by-step explanation:

y=mx+b

find slope which is 8/10=4/5

choose a y and plug into problem

8 0

3 years ago

Read 2 more answers