What is the ratio of A’s to B’s? ABBBAAA

Andy got a 32% discount on all his purchases at his favorite shop. If he saved $64 on the total bill, the bill amount without th

Julli [10]

$200 without the discount and he payed $136.

7 0

1 year ago

the garza family drives to a state park. the remaining distance, in miles, to reach the state park is 285−60x , where x represen

alukav5142 [94]

285 - 60x ; where x represents the number of driving hours.

285 ⇒ The total distance to the state park.
60x ⇒ The number of miles driven after x hours.
60 ⇒ The number of miles driven after 1 hour.

y = 285 - 6x
y ⇒ The number of miles left to drive each day.

5 0

2 years ago

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Two cross sections of a right hexagonal pyramid are obtained by cutting the pyramid with planes parallel to the hexagonal base.

Tanya [424]

Answer:

The larger cross section is 24 meters away from the apex.

Step-by-step explanation:

The cross section of a right hexagonal pyramid is a hexagon; therefore, let us first get some things clear about a hexagon.

The length of the side of the hexagon is equal to the radius of the circle that inscribes it.

The area is

$A=\frac{3\sqrt{3} }{2} r^2$

Where $r$ is the radius of the inscribing circle (or the length of side of the hexagon).

Now we are given the areas of the two cross sections of the right hexagonal pyramid: $A_1=216\:ft^2\: \:\:\:A_2=486\:ft^2$

From these areas we find the radius of the hexagons:

$r_1=\sqrt{\frac{2A_1}{3\sqrt{3} } } =\sqrt{\frac{2*216}{3\sqrt{3} } }=\boxed{9.12ft}$

$r_2=\sqrt{\frac{2A_2}{3\sqrt{3} } } =\sqrt{\frac{2*486}{3\sqrt{3} } }=\boxed{13.68ft}$

Now when we look at the right hexagonal pyramid from the sides ( as shown in the figure attached ), we see that $r_1$ $r_2$ form similar triangles with length $H$

Therefore we have:

$\frac{H-8}{r_1} =\frac{H}{r_2}$

We put in the numerical values of $r_1$ , $r_2$ and solve for $H$ :

$\boxed{H=\frac{8r_2}{r_2-r_1} =\frac{8*13.677}{13.68-9.12} =24\:feet.}$

8 0

1 year ago

Use the table to answer the question. House A $124,270 Annual appreciation 4% House B $114,270 Annual appreciation 5% In which o

Varvara68 [4.7K]

For the house A we have:
f (x) = 124270 (1.04) ^ x
Evaluating for 7, 8, 9 and 10 we have:
f (7) = 124270 (1.04) ^ 7 = 163530.8422
f (8) = 124270 (1.04) ^ 8 = 170072.0759
f (9) = 124270 (1.04) ^ 9 = 176874.9589
f (10) = 124270 (1.04) ^ 10 = 183949.9573

For house B we have:
f (x) = 114270 (1.05) ^ x
Evaluating for 7, 8, 9 and 10 we have:
f (7) = 114270 (1.05) ^ 7 = 160789.3653
f (8) = 114270 (1.05) ^ 8 = 168828.8336
f (9) = 114270 (1.05) ^ 9 = 177270.2752
f (10) = 114270 (1.05) ^ 10 = 186133.789

We observe that for years 7 and 8 the value of house A is greater than the value of house B.

Answer:
7 and 8

4 0

1 year ago

For ΔABC, ∠A = 4x - 10, ∠B = 5x + 10, and ∠C = 7x + 20. If ΔABC undergoes a dilation by a scale factor of 1 3 to create ΔA'B'C'

VLD [36.1K]

We know that the angles of a triangle sum to 180°. For ΔABC, this means we have:
(4x-10)+(5x+10)+(7x+20)=180

Combining like terms,
16x+20=180

Subtracting 20 from both sides:
16x=160

Dividing both sides by 16:
x=10
This means ∠A=4*10-10=40-10=30°; ∠B=5*10+10=50+10=60°; and ∠C=7*10+20=70+20=90.

For ΔA'B'C', we have
(2x+10)+(8x-20)+(10x-10)=180

Combining like terms,
20x-20=180

Adding 20 to both sides:
20x=200

Dividing both sides by 20:
x=10

This gives us ∠A'=2*10+10=20+10=30°; ∠B'=8*10-20=80-20=60°; and ∠C'=10*10-10=100-10=90°.

Since the angle are all congruent, ΔABC~ΔA'B'C' by AAA.

5 0

1 year ago

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