1. 4^2 +x^2 = 6^2
16 + x^2 = 36
x^2 = 20, x = sqrt(20) = 4.47 miles
2. using the same equation as above: 20 inches
3. needs to be a right triangle so A
4. using distance formula answer is 7.07 units
5.volume cylinder: 62.82, volume sphere: 33.51, difference:=62.82 -33.51 = 29.31 cubic inches
Answer:
Step-by-step explanation:
Hello, your question is not mathematically solvable.
Let's solve for x.
−x+9y=−5
Step 1: Add -9y to both sides.
−x+9y+−9y=−5+−9y
−x=−9y−5
Step 2: Divide both sides by -1.
−x ÷ −1 = −9y−5 ÷ −1
x=9y+5
THE ANSWER FOR THE OTHER EQUATION: (x-5y=1)
Let's solve for x.
x−5y=1
Step 1: Add 5y to both sides.
x−5y+5y=1+5y
x=5y+1
Answer:
x=5y+1
I hope this helped I was a little confused on what your problem meant...so if this is not what you asked for just lmk so I can fix it for you :)
Answer:
m<N = 76°
Step-by-step explanation:
Given:
∆JKL and ∆MNL are isosceles ∆ (isosceles ∆ has 2 equal sides).
m<J = 64° (given)
Required:
m<N
SOLUTION:
m<K = m<J (base angles of an isosceles ∆ are equal)
m<K = 64° (Substitution)
m<K + m<J + m<JLK = 180° (sum of ∆)
64° + 64° + m<JLK = 180° (substitution)
128° + m<JLK = 180°
subtract 128 from each side
m<JLK = 180° - 128°
m<JLK = 52°
In isosceles ∆MNL, m<MLN and <M are base angles of the ∆. Therefore, they are of equal measure.
Thus:
m<MLN = m<JKL (vertical angles are congruent)
m<MLN = 52°
m<M = m<MLN (base angles of isosceles ∆MNL)
m<M = 52° (substitution)
m<N + m<M° + m<MLN = 180° (Sum of ∆)
m<N + 52° + 52° = 180° (Substitution)
m<N + 104° = 180°
subtract 104 from each side
m<N = 180° - 104°
m<N = 76°
What the question? Tell me then I can maybe help
