Answer:
neither
Step-by-step explanation:
4x - 6y = -24 y = 2/3x + 4
-6y = -4x - 24
-6/-6y = -4/-6x -24/-6 y =2/3x + 4
y = 2/3x +4
Same line
30% + 10% = 40% off
150 * 0.40 = $60 off
150 - 60 = $90
Solution: you pay $90
Answer: First option.
Step-by-step explanation:
To solve for "h" from the given the equation
, you need to:
Apply the Subtraction property of equality and subtract
to both sides of the equation.
Then you need to apply the Division property of equality and divide both sides of the equation by
.
Then:

Answer:
a)$1628.90
b)$1647.00
c)$1648.72
Step-by-step explanation:
The question is on compound interest.
The formula to apply here is;

where
- P=principal /beginning amount
- r=interest rate as a decimal
- n=number of compoundings a year
- t=total number of years
a) If compounded annually, n=1
p=$1000, r=5%=0.05 t=10
Amount will be;

Amount=$1628.90
b) If compounded monthly, n=12
p=$1000, r=5%=0.05, t=10, n=12

Amount=$1647.00
c)If interest compounded continuously, it means the principal is earning interest constantly and the interest keeps earning on the interest earned.Here the formula to apply is;
A=Pe^rt where e is the mathematical constant e=2.71828182846
Hence the amount will be;

Amount=$1648.72
1.
Draw a circle with center C And radius CA, as shown in the attached picture.
Let the lengths of radii AO, OB, OC be R. Triangle ABC is inscribed in the circle with center O and one of its sides is a diameter, this means that the angle ACB is a right angle.
|AO|=|OC|=R, by the Pythagorean theorem |AC|=

.
these are all shown in the picture.
2.
Area of triangle ABC is 1/2 * 2R * R= R^2
3.
Let the area between arc BXA and chord AB be Y. (the yellow region).
and let G be the shaded region between arcs AB and AXB.
G=1/2(Area circle with center O)-Y
=

To find Y:
Notice that the area of the sector ACB is 1/4 of the area of circle with center C, since m(ACB) is 1/4 of 360 degrees.
So Area of sector ACB =

Y =area of sector ABC-Area(triangle ABC)=

4.
Finally,

This proves that the 2 shaded regions have equal area.