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nikdorinn [45]
3 years ago
9

If the area of a rectangular garden is 45x2−12x square meters, and it has a length of 3x meters, what is its width?

Mathematics
2 answers:
nikitadnepr [17]3 years ago
4 0

Answer:

  (9x -4) meters

Step-by-step explanation:

The relationship between area, length, and width is ...

  area = length × width

Dividing by length gives ...

  width = area/length

Filling in the given expressions, we have ...

  w=\dfrac{45x^2-12x}{3x}=\dfrac{45x^2}{3x}-\dfrac{12x}{3x}=9x-4

The width of the garden is 9x-4 meters.

Novay_Z [31]3 years ago
3 0

Answer:

15x-4

Step-by-step explanation:

Like the previous answer, AREA= length*width

We are given the area, which is 45x^2 - 12x. We are also given the length which is 3x. Using long division, we see (45x^2 - 12x)/3x.

The answer, which is the width, is 15x-4.

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Simplify completely quantity 6 x squared minus 54 x plus 84 over quantity 8 x squared minus 40 x plus 48 divided by quantity x s
elixir [45]

Answer:

We can easily simplify the expression by using a computational tool

The expression is

"6 x squared minus 54 x plus 84 over quantity 8 x squared minus 40 x plus 48 divided by quantity x squared plus x minus "

Please, see attached images below, for a full explanation

4 0
3 years ago
There are 55 M&Ms in the holiday pack and they are all green or red. There are 20 green M&Ms and the rest are red. What
Ugo [173]

Answer:

35 : 55

Step-by-step explanation:

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8 0
3 years ago
Find, correct to the nearest degree, the three angles of the triangle with the vertices d(0,1,1), e( 2, 4,3) − , and f(1, 2, 1)
Ksju [112]
Well, here's one way to do it at least... 

<span>For reference, let 'a' be the side opposite A (segment BC), 'b' be the side opposite B (segment AC) and 'c' be the side opposite C (segment AB). </span>

<span>Let P=(4,0) be the projection of B onto the x-axis. </span>
<span>Let Q=(-3,0) be the projection of C onto the x-axis. </span>

<span>Look at the angle QAC. It has tangent = 5/4 (do you see why?), so angle A is atan(5/4). </span>

<span>Likewise, angle PAB has tangent = 6/3 = 2, so angle PAB is atan(2). </span>

<span>Angle A, then, is 180 - atan(5/4) - atan(2) = 65.225. One down, two to go. </span>

<span>||b|| = sqrt(41) (use Pythagorian Theorum on triangle AQC) </span>
<span>||c|| = sqrt(45) (use Pythagorian Theorum on triangle APB) </span>

<span>Using the Law of Cosines... </span>
<span>||a||^2 = ||b||^2 + ||c||^2 - 2(||b||)(||c||)cos(A) </span>
<span>||a||^2 = 41 + 45 - 2(sqrt(41))(sqrt(45))(.4191) </span>
<span>||a||^2 = 86 - 36 </span>
<span>||a||^2 = 50 </span>
<span>||a|| = sqrt(50) </span>

<span>Now apply the Law of Sines to find the other two angles. </span>

<span>||b|| / sin(B) = ||a|| / sin(A) </span>
<span>sqrt(41) / sin(B) = sqrt(50) / .9080 </span>
<span>(.9080)sqrt(41) / sqrt(50) = sin(B) </span>
<span>.8222 = sin(B) </span>
<span>asin(.8222) = B </span>
<span>55.305 = B </span>

<span>Two down, one to go... </span>

<span>||c|| / sin(C) = ||a|| / sin(A) </span>
<span>sqrt(45) / sin(C) = sqrt(50) / .9080 </span>
<span>(.9080)sqrt(45) / sqrt(50) = sin(C) </span>
<span>.8614 = sin(C) </span>
<span>asin(.8614) = C </span>
<span>59.470 = C </span>

<span>So your three angles are: </span>

<span>A = 65.225 </span>
<span>B = 55.305 </span>
<span>C = 59.470 </span>
8 0
3 years ago
Please help me nowwwww
Radda [10]

Answer:

What is your question young

8 0
3 years ago
Read 2 more answers
Please help this isn't my best subject
VLD [36.1K]

Answer:

152 degrees

Step-by-step explanation:

CF is a straight line, which means it equals 180 degrees.  But FXE is an angle of  28 degrees so you subtract 180 from 28 to get 152 degrees.  

8 0
3 years ago
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