After prime planting season was over, a horticulturist sold lilac bushes for $15.00. If the original price was $39.00, what is the percentage mark down?

Answers

Answer 1

Answer: Find what percent of $39.00 the sale price was.
=> 15/39*100=38.4%
now subtract from 100% to get the mark down percentage.
100-38.4=61.6
Final answer:
Marked down 61.6%
Hope I helped :)

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Margaret is planning a rectangular garden. Its length is 4 feet less than twice its width. Its area is 170ft^2. what are the dimensions of the garden?.

Answers

If you would like to know the dimensions of the garden, you can calculate this using the following steps:

length ... (2 * x - 4) feet
width ... x feet
area ... 170 ft^2

area = length * width
170 = (2 * x - 4) * x
170 = 2 * x * x - 4 * x
170 = 2 * x^2 - 4 * x
0 = 2 * x^2 - 4 * x - 170 /2
0 = x^2 - 2 * x - 85
1. x = 1 - sqrt(86) = -8.27
2. x = 1 + sqrt(86) = 10.27

length: 2 * x - 4 = 2 * 10.27 - 4 = 16.54 feet
width: x = 10.27 feet

The dimensions of the garden are 16.54 feet and 10.27 feet.

6. A hot air balloons each descend 100 feet per minute for 5 minutes. Find the total change in altitude for the hot air balloon.

Answers

Brainly is forcing me to answer this question, don’t report me

Answer: 500 feet in 5 minutes

Step-by-step explanation:

Multiply 100 times 5 to find out how much feet the hot air balloon descends per minute.

2+3=8,

3+7=27,

4+5=32,

5+8=60,

6+7=72,

7+8=????????

SOLVE IT IF YOU CAN

Answers

here 2+3=2(3+1)=2(4)=8
3+7=3(7+2)=3*9=27
and so on
7+8=7(8+6)=98

An airplane flew 4 hours with a 25 mph tail wind. The return trip against the same wind took 5 hours. Find the speed of the airplane in still air. This similar to the current problem as you have to consider the 25 mph tailwind and headwind. Plane on outbound trip of 4 hours with 25 mph tailwind and return trip of 5 hours with 25 mph headwind Let r = the rate or speed of the airplane in still air. Let d = the distance a. Write a system of equations for the airplane. One equation will be for the outbound trip with tailwind of 25 mph. The second equation will be for the return trip with headwind of 25 mph.
b. Solve the system of equations for the speed of the airplane in still air.

Answers

Answer:

r = 225 Mil/h speed of the airplane in still air

Step-by-step explanation:

Then:

d is traveled distance and r the speed of the airplane in still air

so the first equation is for a 4 hours trip

as d = v*t

d = 4 * ( r + 25) (1) the speed of tail wind (25 mil/h)

Second equation the trip back in 5 hours

d = 5 * ( r - 25 ) (2)

So we got a system of two equation and two unknown variables d and

We solve it by subtitution

from equation (1) d = 4r + 100

plugging in equation 2

4r + 100 = 5r - 125 ⇒ -r = -225 ⇒ r = 225 Mil/h

And distance is :

d = 4*r + 100 ⇒ d = 4 * ( 225) + 100

d = 900 + 100

d = 1000 miles

Which of the binomial below is a factor of this trinomial 4x^3-20x^2+24xa) x+3
b)x-3
c)2x-3
d)2x+3

Answers

$4x^3-20x^2+24x$

$4x (x^2-5x+6)$

$4x(x-2)(x-3)$

$(b)x-3$

b) x - 3

see attached for proof

When solving for x, the sine of 33º is used. What other angle and trigonometric function could be used to solve for x? a. cos 57º c. tan 57º b. tan 33º d. cos 33ºhurry

Answers

Answer:

The answer is A.Cos 57.

Step-by-step explanation:

Took the test..

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