Answer:
3
Step-by-step explanation:
Solve for x:
-6 + x = -3
Add 6 to both sides of the equal sign so -6 cancels out on the left side
+6 - 6 + x = -3 + 6
x = - 3 + 6
x = 3
I hope this helps! Good luck with your schoolwork :)
For a rectangular prism, the volume is,
the area of one of the faces * the length of the other unit
so say the base is 9 * 9, then, that base times the height of 16
now, for the triangular prism, the volume is,
the area of one of the triangular faces * the length of the other unit,
so... say, find the area of one of the triangular faces
recall 1/2 b*h
and then multiply that times the other length of 20
The points M(-7,2), N(-3,4), O(-5,8)M(−7,2),N(−3,4),O(−5,8), and P(-9,6)P(−9,6) form a quadrilateral. Find the desired slopes an
pochemuha
The slope of line PO and MN is 0.5. And the slope of the line ON and MP is negative 2. Then each length of the quadrilateral is 4.47 units.
<h3>What is a quadrilateral?</h3>
It is a quadrilateral with four sides. The total interior angle is 360 degrees.
The points M(−7,2), N(−3,4), O(−5,8), and P(−9,6) form a quadrilateral.
Then the slope of each line of the quadrilateral will be
The slope of the line PO will be
PO = (8-6)/(-5+9)
PO = 2/4
PO = 0.5
The slope of the line ON will be
ON = (8-4)/(-5+3)
ON = 4/(-2)
ON = -2
The slope of the line MN will be
MN = (4-2)/(-3+7)
MN = 2/4
MN = 0.5
The slope of the line MP will be
MP = (6-2)/(-9+7)
MP = 4/(-2)
MP = -2
Then each length of the quadrilateral will be
The length of PO will be
PO = √[(-9+5)² + (6-8)²]
PO = 4.47
The length of ON will be
ON = √[(-5+3)² + (8-4)²]
ON = 4.47
The length of MN will be
MN = √[(-3+7)² + (4-2)²]
MN = 4.47
The length of MP will be
MP = √[(-9+7)² + (6-2)²]
MP = 4.47
More about the quadrilateral link is given below.
brainly.com/question/13805601
#SPJ1
Answer:
The equation of the line in slope-int form is y=2x+5
Answer:
<em>Say you have 0.8. The fraction would be 8/10. So if its the tents place the denominator is 10. hundreths would be 100 as a denominator. Last digit numerator. Was this helpful</em>
