Answer:
Part A) Yes , the triangles are congruent
Part B) The side-angle-side (SAS) theorem
Part C) The perimeter of ∆PQR is 
Step-by-step explanation:
Step 1
we know that
The side-angle-side (SAS) theorem, states that: If two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, the triangles are congruent.
so in this problem
Traingle PQR and Triangle STU are congruent by the SAS Theorem
because
m<PQR=m<STU -------> included angle
PQ=TS
QR=TU
Step 2
<u>Find the value of y</u>
we know that
If the triangles are congruent
then
The corresponding sides are equal
so
substitute
so
Step 3
Find the perimeter ∆PQR
Remember that
The perimeter of ∆PQR is equal to the perimeter ∆STU
The perimeter is equal to
substitute the values
Answer:
1. a
2. c
3. e
Step-by-step explanation:
1. a because she ran 5 miles on friday and ran x miles before friday. we know she had ran a total of 20 miles so we have x+5=20
2. there is a total of 20 clubs andre's school, 5 times more than is cousin's school. we have 20 =5x
3. 20 cats get 5 cups of food. each got an x amount. we get 5 =20x
Answer: 36 moles oxygen
Step-by-step explanation:
C10H8 + 12O2 ====》 10CO2 + 4H2O
From the equation of the reaction,
1 mole of napth requires 12 moles of O2
3 moles of napth would require x moles O2
X = (3 * 12) / 1
X = 36 moles of oxygen.
Therefore 3 moles of naphtalene (C10H8) would require 36 moles of oxygen
Answer:
Step-by-step explanation:
This question is asking us to find where sin(2x + 30) has a sin of 1. If you look at the unit circle, 90 degrees has a sin of 1. Mathematically, it will be solved like this (begin by taking the inverse sin of both sides):
![sin^{-1}[sin(2x+30)]=sin^{-1}(1)](https://tex.z-dn.net/?f=sin%5E%7B-1%7D%5Bsin%282x%2B30%29%5D%3Dsin%5E%7B-1%7D%281%29)
On the left, the inverse sin "undoes" or cancels the sin, leaving us with
2x + 30 = sin⁻¹(1)
The right side is asking us what angle has a sin of 1, which is 90. Sub that into the right side:
2x + 30 = 90 and
2x = 60 so
x = 30
You're welcome!
