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Rom4ik [11]
3 years ago
7

Consider the equation 4 plus v equals 8 minus StartFraction 5 Over 3 EndFraction v.v + 4 + StartFraction 5 Over 3 EndFraction v

plus 4 equals 8 minus StartFraction 1 Over 3 EndFraction v. v = 8. What is the resulting equation after the first step in the solution?
Mathematics
2 answers:
mariarad [96]3 years ago
7 0

Answer: 2v+4=8

Step-by-step explanation:

Evgen [1.6K]3 years ago
6 0

Answer:

2v+4=8

Step-by-step explanation:

edge.

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Find the equation of the line that has a slope of a and a y-intercept of 1. ​
Stells [14]

Answer:

y=ax+1

Step-by-step explanation:

y=mx+b where m=slope and b=y-intercept

y=ax+1

4 0
2 years ago
If x and x+ 10 are a pair of adjacent angle find them ​
ikadub [295]

Answer:

85° and 95°

Step-by-step explanation:

Given that , x and x + 10 are a pair of angles .

  • If they will be pair of angles on the same line , then their sum will be 180° .

=> x + x + 10° = 180°

=> 2x = 180° -10°

=> 2x = 170°

=> x = 170°/2

=> x = 85° .

<h3><u>Hence </u><u>the </u><u>two </u><u>angles</u><u> </u><u>are </u><u>8</u><u>5</u><u>°</u><u> </u><u>and </u><u>9</u><u>5</u><u>°</u><u> </u><u>.</u></h3>
5 0
3 years ago
You need to cross a canal and want to determine the distance across the opposite side. Since you're able to take measurements on
MariettaO [177]

Answer:

36.7 ft

Step-by-step explanation:

Measurements for the sides of the canal is given as: 40 ft. and 16 ft.

You solve the above question using Pythagoras Theorem

The distance across the canal is calculated as:

√(40² - 16²)

= √(1344)

= 36.66060556 ft

Approximately = 36.7 ft

Therefore, the distance x across the canal = 36.7 ft

8 0
2 years ago
Find an equation of the plane that contains the points p(5,−1,1),q(9,1,5),and r(8,−6,0)p(5,−1,1),q(9,1,5),and r(8,−6,0).
topjm [15]
Given plane passes through:
p(5,-1,1), q(9,1,5), r(8,-6,0)

We need to find a plane that is parallel to the plane through all three points, we form the vectors of any two sides of the triangle pqr:
pq=p-q=<5-9,-1-1,1-5>=<-4,-2,-4>
pr=p-r=<5-8,-1-6,1-0>=<-3,5,1>

The vector product pq x pr gives a vector perpendicular to both pq and pr.  This vector is the normal vector of a plane passing through all three points
pq x pr
=
  i   j   k
-4 -2 -4
-3  5  1
=<-2+20,12+4,-20-6>
=<18,16,-26>

Since the length of the normal vector does not change the direction, we simplify the normal vector as
N = <9,8,-13>

The required plane must pass through all three points.
We know that the normal vector is perpendicular to the plane through the three points, so we just need to make sure the plane passes through one of the three points, say q(9,1,5).

The equation of the required plane is therefore
Π :  9(x-9)+8(y-1)-13(z-5)=0
expand and simplify, we get the equation
Π  :  9x+8y-13z=24

Check to see that the plane passes through all three points:
at p: 9(5)+8(-1)-13(1)=45-8-13=24
at q: 9(9)+8(1)-13(5)=81+9-65=24
at r: 9(8)+8(-6)-13(0)=72-48-0=24
So plane passes through all three points, as required.

3 0
3 years ago
Need help please<br> please use the image below to answer
spayn [35]

Answer:

Step-by-step explanation:

C and D answers look good

6 0
3 years ago
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