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Firdavs [7]
4 years ago
8

Find the GCF of the terms of the polynomial. 18x4 – 27x3 – 36x2

Mathematics
1 answer:
d1i1m1o1n [39]4 years ago
3 0

Answer:

9 x^ 2

Step-by-step explanation:

18x4 – 27x3 – 36x2

<u>GCF :</u>

<em>Find the prime factors of each term in order to find the greatest common factor (GCF)</em>

= 9 x^ 2

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Convert: 150 oz = _____ lb _____ oz ​
yulyashka [42]

16 oz = 1 lb

Step-by-step explanation:

9.375 lb = 150 oz

          Answer:

    9.375 lb

    150 oz

its d or e

8 0
3 years ago
Read 2 more answers
Hiro and his co-workers had lunch at a restaurant near their work. An 18% tip on their bill was $10.08. How much was the bill, n
Ghella [55]

Answer:

$8.27

Step-by-step explanation:

First we need to find the tip.  We know it was 18% of the total. So...

x (the variable we need to find)         18%

__                                                   =     ___

10.08 (the total bill including tip)        100%

cross multiply

100x = 181.44

--------------------   (divide by 100 to get x by itself)

100

x = 1.81 (The 18% tip that was given)

So now we know how much the tip was, we subtract it from the bill

10.08 - 1.81 = $8.27

5 0
3 years ago
Please help me I really such at geometry
garik1379 [7]

Answer:

DF, DE, EF

Step-by-step explanation:

Nah dude, you got it. :)

6 0
3 years ago
The area of a triangle is 80x^5y^3. The height of the triangle is x^4y. What is the length of the base of the triangle?
svetoff [14.1K]
Area of a triangle is given by 1/2bh where b is the base and h is the perpendicular height of the triangle.
The area is 80x∧5y³ and the height is x∧4y
Thus; 80x∧5y³ = 1/2(x∧4y) b
         160x∧5y³ = (x∧4y)b
                  b = (160x∧5y³)/ x∧4y)
                  b = 160xy²
Therefore, the base of the triangle is 160xy²
7 0
3 years ago
Read 2 more answers
The operating costs for each machine for one day have an unknown distribution with mean 1610 and standard deviation 136 dollars.
BabaBlast [244]

Answer:

20.27

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

Using the Central Limit Theorem for Means, what is the standard deviation for the sample mean distribution?

This is s when \sigma = 136, n = 45. So

s = \frac{136}{\sqrt{45}} = 20.27

So the correct answer is:

20.27

3 0
3 years ago
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