Answer:
Question 4: -11
Question 5: -7
Step-by-step explanation:
Four
Every triangle has 180 degrees.
So all three angles add to 180
<em><u>Equation</u></em>
60 + 80 + x + 51 = 180
<em><u>Solution</u></em>
Combine the like terms on the left. This is the first time I've seen x be a negative value. Almost all of the time it isn't, which should make you wonder.
191 + x = 180
Subtract 191 from both sides.
191 - 191 + x = 180 - 191
x = - 11
Five
If a triangle is a right triangle and one of the angles is 45, then so is the other one.
<em><u>Proof</u></em>
a + 45 + 90 = 180 Combine like terms on the left
a + 135 = 180 Subtract 135 on both sides.
a + 135-135=180-135 Combine the like terms
a = 45
<em><u>Statement</u></em>
That means 52 + x = 45 and here is another negative answer. Subtract 52 from both sides
52 - 52 + x = 45 - 52 Combine like terms.
x = - 7
First Chart: Perimeter
Square Portion:
Original Side Lengths: P = 4 (1 + 1 + 1 + 1 ) =4
Double Side Lengths: P = 8 (2 x 4 = 8)
Triple Side Lengths: P = 12 (4 x 3 = 12)
Quadruple Side Lengths: P = 16 (4 x 4 = 16)
Rectangle Portion:
Original Side Lengths: P = 6 (1 x 2 + 2 x 2 = 6)
Double Side Lengths: P = 12 (2 x 2 + 4 x 2 = 12)
Triple Side Lengths: P = 24 (4 x 2 + 8 x 2 = 24)
Quadruple Side Lengths: P = 48 (8 x 2 + 16 x 2 = 48)
Second Chart: Area
Square Portion:
Original Side Lengths: A = 1 (1 x 1 = 1)
Double Side Lengths: A = 4 (2 x 2 = 4)
Triple Side Lengths: A = 9 (3 x 3 = 9
Quadruple Side Lengths: A = 16 ( 4 x 4 = 16)
Rectangle Portion:
Original Side Lengths: A = 2 ( 1 x 2 = 2 )
Double Side Lengths: A = 8 ( 2 x 4 = 8)
Triple Side Lengths: A = 18 ( 3 x 6 = 18)
Quadruple Side Lengths: A = 32 (4 x 8 = 32)
Answer:
it changes by 4
Step-by-step explanation:
That's an expression, so the only thing you can do is simplify it, but that already looks simplified.
Exponentail thingies
easy, look at all them them, see that they have 5 in common?
rremember how esay it was to factor
ax^2+bx+c=0
now we have
5^(2x)-6(5^x)+5=0
remember that 5^(2x)=(5^2)^x or (5^x)^2
in other words, we can rewrite it as
1(5^x)^2-6(5^x)+5=0
if yo want, replace 5^x with a and factor
1a^2-6a+5=0
(a-1)(a-5)=0
a=5^x
(5^x-1)(5^x-5)=0
set each to zero
5^x-1=0
5^x=1
take the log₅ of both sides
x=log₅1
5^x-4=0
5^x=4
take the log₅ of both sides
x=log₅4
x=log₅1 and/or log₅4
second quesiton
same thing
1(2^x)-10(2^x)+16=0
factor
(2^x-8)(2^x-2)=0
set each to zero
2^x-8=9
2^x=8
x=3
2^x-2=0
2^x=2
x=1
x=3 or 1
first one
x=log₅1 and/or log₅4
second one
x=1 and/or 3
