<span>x + (x + 1)
.............
</span>
Since the two triangles are proportional, there is a scale for going from one triangle to the other
Answer:
Step-by-step explanation:
Given
Required
Find all product of real values that satisfy the equation
Cross multiply:
Subtract 7 from both sides
Reorder
Multiply through by -1
The above represents a quadratic equation and as such could take either of the following conditions.
(1) No real roots:
This possibility does not apply in this case as such, would not be considered.
(2) One real root
This is true if
For a quadratic equation
By comparison with
Substitute these values in
Add 56 to both sides
Divide through by 4
Take square roots
Hence, the possible values of r are:
or 
and the product is:
5y - 2x + 1 = 0.
5y = 2x-1.
y = 2/5 x - 2/5
this tells us that tan(theta) = 2/5
sin(theta)/sqrt(1-sin^2(theta)) = 2/5
x/sqrt(1-x^2)=2/5
x^2/(1-x^2)=4/25
25x^2=4-4x^2
29x^2=4
x^2=4/29
x = -2/sqrt(29) because quadrant III
cos(x) = sqrt(1-4/29) = -5/sqrt(29)
tan(theta)=2/5
cos(theta)=-5/sqrt(29)
sin(theta)=-2/sqrt(29)
cot(theta)=5/2
sec(theta)=-sqrt(29)/5
csc(theta)=-sqrt(29)/2
Answer:
PQRS and TUVW are similar.
Step-by-step explanation:
We are give two polygons.
Sides of first polygon are :
PQ = 8 ft
QR= 13ft
RS = 8ft
PS = 13ft.
Sides of second polygon:
TW = 2.4 ft
TU = 5.2 ft
UV = 2.4 ft
VW = 5.2 ft.
We can see that PQRS and TUVW are parallelograms.
Note: <em>If an angle of a parallelogram is equal to an angle of another parallelogram then parallelograms would be similar.</em>
We can see that <Q ≅ <T = 120°.
<h3>Therefore, PQRS and TUVW are similar.</h3>
