Will give brainlyest HELPP

Mathematics

1 answer:

aev [14]3 years ago

6 0

Answer:

16:30

Step-by-step explanatThis is the anwser

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PLZZ HELP ASAP ILL MAKE U BRAINLIEST

user100 [1]

Translation right down c=

4 0

3 years ago

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You pay 1.5% interest on your credit card bill every month. This month your

IrinaVladis [17]

Answer:

Step-by-step explanation:

Bill=3475

Interest=1.5%

Total payment made=3475+3475x1.5/100

=3475x101.5/100

=3527.125$ is the total payment (answer)

4 0

3 years ago

Lines a and b are parallel.

VashaNatasha [74]

I have added a screenshot with the complete question along with a diagram representing the scenario.

Answer:
s = 22°

Explanation:
1- getting the top right angle of line B:
We are given that:
the top right angle of line A = 158°
Since lines A and B are parallel, therefore, the top right angle of line A and the top right angle of line B are corresponding angles which means that they are equal
This means that:
Top right angle of line B = 158°

2- getting the value of s:
Now, taking a look at line B, we can note that:
angle s and the top right angle form a straight angle. This means that the sum of these two angles is 180°
Therefore:
180 = s + 158
s = 180 - 158
s = 22°

Hope this helps :)

8 0

3 years ago

What is the solution to 5x+9-3x=18+15 A.x=8 B.x=12 C.x=21 D.x=3

Vlad [161]

Your answer will be B) x=12

1)5x-3x
2) move 9 to the other side (change signs)
3)18+15-9
4) 2x= 24 (divide 2 from both sides
5)x=12

hope this helps

7 0

3 years ago

Read 2 more answers

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