B2+10b+25/b29-9 . B+3/B+5

Answers

Answer 1

Answer: b4 + b3 - 45b2 - 27b + 25—————————————————————————b2

Answer 2

Answer: the answer is (b+5)(b+5)/b(29-5)

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Answers

Answer:

We know that range is the set of outputs that shown on the y-axis of the graph, so we must find the endpoints of the graph on the y-axis. When we look, we get the endpoints -6 and 10, so we know that this is the range, and so we represent this in an expression:

-6 < y < 10

Step-by-step explanation:

The perimeter of a pool table is 30 feet. The table is twice as long as it is wide. What is the length of the pool table?

Answers

So, 30 is the perimeter. We are told the table is twice as long as it is wide.

So, we have think of 2 numbers, one being twice as big than the other. So, it is a rectangle The two number represents the length and the width. To find the perimeter we add all the lengths and the widths. Since the pool table is a rectangle we have two lengths and two widths.

So,

the tow numbers be 10 and 5. 5 is half of 10 and, when multiplied by 2 it is 10. So just to make sure the numbers 5 and 10 work out lets do a calculation.

So,

10+5+10+5 = 30feet. This proves that the sides are these two numbers.

Find the GCF of the following monomials. -50m 4 n 7 and 40m 2 n 10

Answers

Answer:

$10m^2n^7$

Step-by-step explanation:

$-50m^4n^7 \ and \ 40m^2n^(10)$

WE need to find GCf for both monomials. GCF is the greatest common factor

$-50 ->-1 \cdot 5 \cdot 5 \cdot 2$

$40 ->2 \cdot 2 \cdot 5 \cdot 2$

GCF is 2 times 5 =10

GCF of exponent is the lowest exponent

GCF of m^4 and m^2 is m^2

GCF of n^7 and n^10 is n^7

$-50m^4n^7 \ and \ 40m^2n^(10)$

GCF is $10m^2n^7$

2 , 1, , 2 if it helps <3

(0)Radio direction finders are placed at points A and B, which are 4.32 mi apart on an east-west line, with A west of B. The transmitter has bearings 10.1 from A and 310.1 from B. Find the distance from A.

Answers

2.95 miles

The question involves radio direction finders placed at two points, A and B, which are 4.32 miles apart on an east-west line. The transmitter has bearings 10.1 degrees from A and 310.1 degrees from B. The task is to determine the distance from A.In order to determine the distance from A, the first step is to construct a diagram of the scenario to visualize the placement of the three points, A, B, and the transmitter. To do so, a coordinate system is used, with A being located at the origin (0,0).The bearing of the transmitter from A is 10.1 degrees, which can be plotted on the diagram as a straight line from the origin to an angle of 10.1 degrees to the east. Similarly, the bearing of the transmitter from B is 310.1 degrees, which can be plotted on the diagram as a straight line from point B to an angle of 49.9 degrees to the west.To determine the distance from A, the Law of Cosines can be applied, which states that c^2 = a^2 + b^2 − 2ab cos(C), where c is the unknown side, a and b are the known sides, and C is the angle opposite the unknown side. In this case, c is the distance from A, a is the distance from B, and b is the distance between A and B. The angle C is equal to the sum of the two bearings (10.1 + 49.9 = 60 degrees).Therefore, c^2 = a^2 + b^2 − 2ab cos(C) can be rewritten as:dA^2 = d^2 + 4.32^2 - 2d(4.32)cos(60)dA^2 = d^2 + 4.32^2 - 2d(4.32)(1/2)dA^2 = d^2 + 4.32^2 - 2.16dTo solve for dA, the equation can be rearranged and solved for d:0 = d^2 - 2.16d + dA^2 - 4.32^2d = 1.08 ± sqrt(1.08^2 - dA^2 + 4.32^2)The positive root of this equation can be used to determine dA:dA = 1.08 + sqrt(1.08^2 - d^2 + 4.32^2)dA = 1.08 + sqrt(1.08^2 - 4.32^2 cos^2(10.1))dA ≈ 2.95 milesTherefore, the distance from A is approximately 2.95 miles.

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