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Zepler [3.9K]
3 years ago
11

Jeez math help please!!!!!

Mathematics
1 answer:
viva [34]3 years ago
4 0
(3y^7)^3      3^3*y^(7*3)              
------------ = ------------------   
    9y^9           9*y^9

(3y^7)^3      3^3*y^(7*3)        27*y^21      (mult. together the exponents 3 and 7)
------------ = ------------------ = --------------
    9y^9           9*y^9                 9*y^9

That 27/9 reduces to 3.     y^(21) / y^9  reduces to y^12.

Thus, we obtain           3*y^12   (answer)

Please do the next one, asking questions if need be, and showing your work.

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Please solve with explanation (high points)
Hunter-Best [27]

Step-by-step explanation:

so, we have a large triangle made of the 2 cables as legs and the ground distance AB as baseline.

the tower is the height to the baseline of that large triangle.

let's call the top of the tower T.

and remember, the sum of all angles in a triangle is always 180°.

we know the angle A = 62°, and angle B = 72°.

assuming that AB is a truly horizontal line that means that the 2 legs (cables) have different lengths, the triangle is not isoceles, and the tower is not in the middle of the baseline.

so, the height (tower) splits the baseline into 2 parts. let's call them p and q.

p + q = 12 m

p = 12 - q

let's simply define that p is the part of the baseline on the A side, and q is the part of the baseline on the B side.

we have now 2 small right-angled triangles the large height (tower) splits the large triangle into.

one has the sides

AT, height (tower), p

angle A = 62°

angle T = 180 - 90 - 62 = 28°

the other has the sides

BT, height (tower), q

angle B = 72°

angle T = 180 - 90 - 72 = 18°

now remember the law of sine :

a/sin(A) = b/sin(B) = c/sin(C)

with the sides and the associated angles being opposite.

p/sin(28) = height/sin(62)

q/sin(18) = height/sin(72)

we know from above that

p = 12 - q

so,

(12 - q)/sin(28) = height/sin(62)

height = (12 - q)×sin(62)/sin(28)

q/sin(18) = height/sin(72)

height = q×sin(72)/sin(18)

and therefore, as height = height we get

(12 - q)×sin(62)/sin(28) = q×sin(72)/sin(18)

(12 - q)×sin(62)×sin(18) = q×sin(72)×sin(28)

12×sin(62)×sin(18) - q×sin(62)×sin(18) =

= q×sin(72)×sin(28)

12×sin(62)×sin(18) = q×sin(72)×sin(28) + q×sin(62)×sin(18) =

= q×(sin(72)×sin(28) + sin(62)×sin(18))

q = 12×sin(62)×sin(18) / (sin(72)×sin(28) + sin(62)×sin(18))

q = 4.551603755... m

p = 12 - q = 7.448396245... m

height = q×sin(72)/sin(18) = 14.00839594... m ≈ 14 m

the cell tower is about 14 m tall.

7 0
2 years ago
What is 34.75 - 0.876
Goryan [66]

Answer:

34.75 - 0.876 = 33.874

5 0
3 years ago
Points G, E, and O are collinear on segment GO, and GE:EO = 3:4. G is located at (-5,-1), O is located at (9,6). What are the co
pishuonlain [190]

Answer:

  E = (1, 2)

Step-by-step explanation:

You want ...

  (E -G) : (O -E) = 3 : 4

  4(E -G) = 3(O -E) . . . . . . . . "cross multiply"

  4E -4G = 3O -3E . . . . . . . . eliminate parentheses

  7E = 4G + 3O . . . . . . . . . . . add 3E+4G

  E = (4G +3O)/7 . . . . . . . . . .divide by 7

  E = (4(-5, -1) +3(9, 6))/7 = (-20+27, -4+18)/7 = (7, 14)/7 . . . . fill in G and O

  E = (1, 2)

4 0
4 years ago
Someone told me to get it done so im asking brainly
Tatiana [17]
The didn’t distribute the -6 properly to the fraction the second part of solving should have been -6(4x)+((-6)-2/13)
6 0
2 years ago
Read 2 more answers
Match the system of equations on the left with the number of solutions on the right
erastova [34]

Answer:

top to bottom, the answers are b, c, a

Step-by-step explanation:

One way to find the solution to a system of equations is to substitute values in. For the first one,

y=2x+3

y=2x+5,

we can substitute 2x+3 =y into the second equation to get

y=2x+5

2x+3 = 2x+5

subtract 2x from both sides

3 = 5

As 3 is not equal to 5, this is never equal and therefore has no solution

For the second one,

y= 2x+7

y = (-2/3)x + 10

We can plug y=2x+7 into the second equation to get

2x + 7 = y = (-2/3)x + 10

2x + 7 = (-2/3)x + 10

add (2/3)x to both sides to make all x values on one side

2x + (2/3)x + 7  = 10

subtract 7 from both sides to make only x values on one side and only constants on the other

2x + (2/3)x = 3

(6/3)x + (2/3)x = 3

(8/3)x = 3

multiply both sides by 3 to remove a denominator

8x = 9

divide both sides by 8 to isolate x

x=9/8

There is only one value for when the equations are equal, so this has one solution

For the third one

y = x-5

2y = 2x - 10

Plug x-5 = y into the second equation

2 * y= 2*(x-5)

2 * (x-5) = 2x - 10

2x-10 = 2x-10

add 10 to both sides

2x=2x

As 2x is always equal to 2x, no matter what x is, there are infinitely many solutions for this system

6 0
3 years ago
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