r=rate in decimal
t=time in years
n=how many tmes per year it is compounded
they want 110,000=A
rate is 6% or 0.06
time is 12 years
semianually means 2 times per year, so n=2
divide both sides by (1.03)^24 using your graphinc calculator
$54,112.71 should be invested
1) Parallelogram ABCD > Given
2) BT = TD > Diagonals of a parallelogram bisect
3) <1 = <2 > Vertical angles are equal
4) BC ║ AD > Definition of parallelogram
5) <3 = <4 > If lines parallel, then alternate interior angles are equal
6) Triangle BET congruent to Triangle DFT >ASA
7) ET = TF > CPCTE
Hope they help.
-4(2z + 6) - 12 = 4
-4(2z) + -4(6) -12 = 4 distributed -4 into (2z + 6)
-8z - 24 - 12 = 4 simplified the multiplied terms
-8z - 36 = 4 added like terms (-24 and -12)
-8z = 40 added 36 to both sides of the equation
z = -5 divided both sides by -8
z = -5
(x - 2) - 5 = 19
(x - 2) = 24 added 5 to both sides
(x - 2) = 24() multiplied both sides by
x - 2 = 16 simplified the multiplied term on the right
x = 18 added 2 to both sides
x = 18
A B and C
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Hey there! :)
7. -4(2z + 6) - 12 = 4
-8z - 24 - 12 = 4
-8z - 36 = 4
Add 36 to both sides.
-8z - 36 + 36 = 4 + 36
-8z = 40
Divide both sides by -8.
z = -5
8. 3/2(x - 2) - 5 = 19
3/2x - 3 - 5 = 19
3/2x - 8 = 19
Add 8 to both side.
3/2x = 19 + 8
3/2x = 27
Multiply both sides by 2.
3/2x · 2 = 27 · 2
3x = 54
Divide both sides by 3.
x = 18
~hope I helped~
The statements with accurate reasons are show below.
Given: ABCD is a parallelogram, EF contains T.
1. Parallelogram ABCD (Given)
2. BT=TD (Diagonals of a parallelogram bisect each other)
3. ∠1=∠2 (Vertical angles are equal)
4. BC║AD (Definition of parallelogram)
5. ∠3=∠4 (If lines parallel, then alternate interior angles are equal)
Two corresponding angles and their included sides are congruent. So, by ASA both ΔBET and ΔD FT are congruent.
6. ΔBET ≅ ΔD FT (ASA)
7. ET=T F (CPCTE)
This would equal 4 1/6 or as an improper fraction 25/6
The way we do this is by first turning the mixed numbers into improper fractions. We do this by multiplying the whole number by the denominator and then adding it to the numerator.
For the first fraction we multiply 2 by three then add two. The denominator stays the same.
This product and sum becomes your numerator. So now the first fraction is 8/3.
We then change the second fraction. We multiply 2 by 1 then add 1. This will equal 3/2.
Now we have to change the denominator's to make them equal. So we multiply the first fractions denominator and numerator by 2 and the other fractions denominator and numerator by 3.
We multiply the numerator because what ever you do to the bottom you do to the top.
So the equation you should have now is 16/6 + 9/6.
Then you add the numerator's. This would be 33/6. This is your answer. But we have to simplify first. 25 can not be divided by 6 so we convert it to an improper fraction. 6 can go into 25 4 times so the answer would be 4 1/6.
if you have any question's just ask :D
x equals 2y is -4
You would need to measure these angles by hand... Sorry i cant help :/
sorry i dont know my friend
a. we have two
lines: y = 4-x and
y = 8-x^-1
given two simultaneous equations that are both to be
true, then the solution is the points where the lines cross. the intersection is
where the two equations are equal. therefore the solution that works for both
equations is when
4-x = 8-x^-1
this is where the two lines will cross and that is the
common point that satisfies both equations.
b. 4-x = 8-x^-1
x 4-x 8-x^-1
-2 6 8.5
-1 5 9
3 1 7.67
the table shows that none of the x values from -3 to 3 is
the solution because in no case does
4-x = 8-x^-1
to find the solution we need to rearrange the equation to
find for x:
4-x = 8-x^-1
multiply both sides with x:
4x-x^2 = 8x-1
x= -4.236, 0.236
therefore there are two points that satisfies the
y = 4-x = 4 – (-4.236)
y = 8-x^-1 = .236)^-1
y = 4-x = 4 – (0.236) = 3.764
y = 8-x^-1 = 8-(0.236)^-1
thus the two lines cross at 2 points:
(-4.236, 8.236) & (0.236, 3.764)
c. to solve
graphically the equation 4-x = 8-x^-1
we would graph both lines: y = 4-x and y
the point on the graph where the lines cross is the
solution to the system of equations.
just graph the points on part b on a cartesian coordinate
system and extend the two lines. the
solution is, as stated, the point where the two lines cross on the graph.