Answer:
VW=36
VT=27
Step-by-step explanation:
VW is the midsegment of the triangle
so
VW=72/2
=36
VXT is a right angle triangle
so
a²+b²=c²
a=√c²-b²
a=√45²-36²
a=√(45-36)(45+36)
a=√9×81
a=√9×√81
a=27
therefore
VT=27
Answer:
y=3,x=1.3,z=6.2 :)
Step-by-step explanation:
y=3,x=1.3,z=6.2 :)
So, we have:
: x+y+z=10 [equation 1]
: 5x+3y=14 [equation 2]
: x−2y+z=2.5 [equation3]
(I'm going to assume that you know how to solve equations with 2 variables)
1. eliminate a variable to get 2 variables
Let's reverse equation 2:
-x+2y-z= -2.5
and take equation 1, to try to eliminate "z"
x+y+z=10
add up the 2 equations to get:
3y=7.5
and solve to get
y=2.5
2. substitute what we already solved into another equation
Luckily, equation 2 only has 2 variables, and one is y, so we can sub it in.
5x+3y=14 --> 3x+3(7.5)=14
solve to get
x=1.3
3. Substitute both of our variables in to get the final variable: z
Let's take the first equation.
x+y+z=10 --> 1.3+2.5+z=10
solve to get
z=6.2
I hope this helps you!
Since opposite angles and sides are equal, then the quadrilateral ABCD is a parallelogram.
Given that, ABCD is a quadrilateral, segment AB is congruent to segment CD, ∠1 is congruent to ∠2.
We need to prove that ABCD is a parallelogram.
<h3>
How to prove a given quadrilateral is a parallelogram?</h3>
If one pair of opposite sides of a quadrilateral are both parallel and congruent, then it’s a parallelogram.
From the given quadrilateral ABCD:
AB=CD (Given)
∠1=∠2 (Given)
AC=AC (Diagonal)
From ASA congruency ΔABC and ΔADC are congruent.
By CPCT, AD=BC and ∠ABC=∠ADC.
Since opposite angles and sides are equal, then the quadrilateral ABCD is a parallelogram.
To learn more about parallelogram visit:
brainly.com/question/1563728.
#SPJ1
Step-by-step explanation:
The equation will form a circle because it in the form of

First, let set the equation equal.

Here the center will be (7,-5), and the radius of 5. The boundary line will be dashed
Since this is in the inequalities, we must find the solution set.
Plug in 0,0 for x and y and see if it's true.



This is a true so we shade the region that includes 0,0
Since 0,0 has a greater distance from the center of the circle, 0,0 is outside of the circle, so our solution set will
be outside of circle.
Here a picture of graph,