The value of x is 40.
Solution:
Given ΔVDG and ΔVNG are similar triangles.
DG = 207, NQ = 138, GQ = 60, QV = x
To find the value of x.
In two triangles are similar, then the measures of the corresponding sides are in proportional to each other.
Substitute the given values.
Do cross multiplication, we get
Divide by 207 on both sides of the equation, we get
Hence the value of x is 40.
5√18
= 5 *√9 *√2
= 5*3*<span>√2
= 15</span><span>√2</span>
Let x represent the smaller angle.
Let y represent the larger angle.
The problem states that the larger angle measures five degrees more than four times the measure of the smaller angle.
With this given information, we can create the following equation.
Also, they a supplementary, meaning that they add up to 180. We can create another equation.
Since we have two linear equations and we want to find the solution, we have a system of linear equations. Let's solve this system by using the substitution method.
Substitute
into
After substituting, you get
Now, combine the x's
Subtract both sides by 5
Divide both sides by 5.
Now, we can solve for y using the equation
since we know the value of x.
Subtract both sides by 35.
The smaller angle has a measure of 35 degrees and the larger one has a measure of 145 degrees. Have an awesome day! :)
P(Q|R) = P(Q&R)/P(R)
= (3/37)/(7/37)
= 3/7
The appropriate choice is ...
B. 3/7
For the first one c-9=(2)(4)+6
You are gonna multiply 2 and 4 first, 2*4=8. Now you have c-9=8+6. 8+6=14. So now the equation is c-9=14. Now we are going to add 9 to both sides. So c=23
For the second equation 10+y=90. We are going to subtract 10 from both sides. So, y=80.
Third, add 4 to both sides. a=10.
Fourth, subtract 8 from both sides. d=-9
Fifth, 5+2(3+4)=-x.
First we are going to add 3 and 4. 3+4=7. 5+2(7)=-x. Next we are going to multiply 2 and 7. 2*7=14. 5+14=-x. 5+14=19. So our equation is 19=-x. Next we are going to multiply both sides by -1. Making our final equation x=-19.
c=23
y=80
a=10
d=-9
x=-19
I hope that helps, if you need any further explanation just ask.<span />
