Answer:
6 and 9
Step-by-step explanation:
set up an equation and solve for x
(x+3) * x = 54
easiest way to to try some numbers.
if you know your multiplication tables...
what two number multiplied together are 54
be sure the numbers you are thinking of meet the rules
is one 3 more than the other
do they multiply together to get 54
Answer:
-19
Step-by-step explanation:
0.3h+1.2=0.1h-2.6
.2h+1.2=-2.6
.2h=-3.8
h=-19
Let x represent amount invested in the higher-yielding account.
We have been given that a man puts twice as much in the lower-yielding account because it is less risky. So amount invested in the lower-yielding account would be 
.
We are also told that his annual interest is $6600 dollars. We know that annual interest for one year will be principal amount times interest rate.
, where,
I = Amount of interest,
P = Principal amount,
r = Annual interest rate in decimal form,
t = Time in years.
We are told that interest rates are 6% and 10%.
Amount of interest earned from lower-yielding account: 
.
Amount of interest earned from higher-yielding account: 
.
Let us solve for x.
Therefore, the man invested $30,000 at 10%.
Amount invested in the lower-yielding account would be 
.
Therefore, the man invested $60,000 at 6%.
First picture:
2 and 7 are alternate exterior angles, which when added together need to equal 180 degrees.
Angle 7 = 180 - 146 = 34
Answer is B.34
Second picture:
Angle 8 = 48
Angle 8 + angle 7 = 180
Angle 7 = 180 - 49 = 131
Angle 5 and angle 7 are vertical angle, so angle 5 = angle 7 = 131
Angle 6 is a vertical angle to angle 8 so = 49
Angle 1 = angle 5 = 131
Angle 2 = angle 6 = 49
Angle 3 = angle 7 = 131
Angle 4 = angle 8 = 49
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.
