HELP it's asking for he GREATEST COMMON FACTOR PLEASE RESPOND ASAP!

Mathematics

2 answers:

Dominik [7]3 years ago

8 0

In (12)x^5 y^2 (13) 6x^2. (14) x^2 y. (15)15x (16) 3x^4 y^2 (17) xy (18) 2x^2 y^3 (19) 3x^2 (20) 3x (21) 6x^4
I hope it helps

crimeas [40]3 years ago

6 0

Number 6 is wrong. The answer is 2x^3. Number 7 is wrong. The answer is 2x. Number 8 is wrong. The answer is 10y^2. Number 10 is wrong. The answer is 12y^2. Number 11 is wrong. The answer is 7mn. Number 12: x^5*y^2. Number 13: 6x^2. Number 14: x^2*y. Number 15: 15x. Number 16: 3x^4*y^2. Number 17: xy. Number 18: 2x^2*y^3. Number 19: 3x^2. Number 20: 3x. Number 21: 6x^4.

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Determine the surface area of a rectangular prism with the following dimensions:

atroni [7]

Answer:

b. 106 sq.cm

Step-by-step explanation:

A=L×W

A1= 2(5×8) = 2(40) = 80

A2= 2(1×8) = 2(8) = 16

A3= 2(5×1) = 2(5) = 10

Total area = 80 + 16 +10 = 106 sq. cm

6 0

3 years ago

Does anyone know how to solve the equation ln 2 - ln(x - 8) = 4, giving nearly 80 points

RoseWind [281]

Answer:

x= 8.1353

x= 8.135 (rounded to the nearest tenth-thousandths)

x= 8.14 (rounded to the nearest thousandths)

x= 8.1 (rounded to the nearest tenth)

Step-by-step explanation:

Note I am not 100% sure with my answer

2 − ln (x − 8)= 4

−ln (x − 8) + 2= 4

−ln (x − 8) + 2 + −2= 4 + −2

−ln (x − 8)= 2

−ln (x − 8)/−1= 2/−1

ln (x − 8)= −2

ln (x − 8)= −2

e^ln (x − 8)= e^−2

x − 8= e^−2

x − 8= 0.1353

x − 8 + 8= 0.1353 + 8

x= 8.1353

5 0

3 years ago

A pole that is 3m tall casts a shadow that is 1.22m long. At the same time, a nearby building casts a shadow that is 38.5m long.

vichka [17]

Refer to the diagram shown below.

The height of the pole is CD (3 m), and its shadow is AB (1.22 m).
The height of the building is h m, and its shadow is AD (38.5 m).

The two triangles ΔABC and ΔADE are similar because of AAA.
Therefore
$\frac{DE}{BC} = \frac{AD}{AB} \\\\ \frac{h}{3} = \frac{38.5}{1.22}$

Cross multiply.
1.22 h = 3*38.5 = 115.5
h = 115.5/1.22 = 140.91 m

Answer: 141 m (nearest meter)