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alukav5142 [94]
3 years ago
13

Solve the equation for a. K = 4a+9ab

Mathematics
2 answers:
Elanso [62]3 years ago
8 0
So try to isolate a by division and such
K=4a+9ab
we use reverse dstributive which is ab+ac=a(b+c) so
undistribute a
4a+9ab=a(4+9b)
k=a(4+9b)
divide both sides by (4+9b)
\frac{k}{4+9b}=a
a=\frac{k}{4+9b}
boyakko [2]3 years ago
4 0
K=9ab+4a
A=k/9b+4
B=-4/9 + k/9a
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Bethany and her family are attending the Funville Town Fall Festival, and the first thing they do is get in line for the Maize (
erik [133]

Answer:

Part 1: 31.4 feet

Part 2: 61 feet

Part 3: 99,000 stalks of corn

Step-by-step explanation:

<u>Part 1:</u>

Diagram attached.

You can follow the problem closely and draw a diagram like this. The RED LINE is the distance we are looking to find.

The red line is the hypotenuse of a right triangle we can create (see the dotted line). We can now use the pythagorean theorem to solve for the distance.

Pythagorean Theorem is  hypotenuse^2 = leg^2 + leg^2

Let red line [what we want to find] be h and the other 2 sides are 10-5=5 ft & 25 + 6 = 31ft. So we have:

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Hence Bethany has to walk 31.4 feet

<u>Part 2:</u>

See 2nd diagram attached.

The blue line is the path of Bethany and the green line is path of Ben. The RED line is what we are looking to find, the shortest distance between them.

We use Pythagorean Theorem twice on the two triangles and add up the results.

<u>Left Triangle:</u>

If we make a triangle, we have base (one leg) = 30 and the other leg is 20-10=10. So we have:

h^2=30^2 + 10^2\\h=\sqrt{30^2 + 10^2} \\h=31.62

<u>RightSide triangle:</u>

two legs are 25 and 15 respectively, so we use the pythagorean theorem again:

h^2=25^2 + 15^2\\h=\sqrt{25^2 + 15^2} \\h=29.15

Thus, the shortest distance = 31.62 + 29.15 = 60.77 = 61 feet (rounded to nearest whole number)

<u>Part 3:</u>

The area of the Maize MINUS the area of the path is the Area of the Corn [what we are looking for].

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Now, each sq. yd. holds 18 stalks, so 5,500 sq. yd. will hold:

18 * 5,500 = 99,000 stalks of corn

<u />

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