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malfutka [58]
3 years ago
14

(Picture) MULTIPLYING MONOMIALS AND BINOMIALS

Mathematics
1 answer:
motikmotik3 years ago
7 0

Answer:

Second option: 81y^4 - 16x^2, the difference of squares

Step-by-step explanation:

(9y^2-4x)(9y^2+4x) is a special product named difference of squares, then we can apply this formula:

(a-b)(a+b)=a^2-b^2, with a=9y^2 and b=4x, then:

(9y^2-4x)(9y^2+4x)=(9y^2)^2 - (4x)^2

(9y^2-4x)(9y^2+4x)=(9)^2 (y^2)^2 - (4)^2 (x)^2

(9y^2-4x)(9y^2+4x)=81y^(2*2) - 16x^2

(9y^2-4x)(9y^2+4x)=81y^4 - 16x^2


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What are the solutions to the equation?<br> 7x^3=28x
11Alexandr11 [23.1K]

7x³ = 28x is our equation. We want its solutions.

When you have x and different powers, set the whole thing equal to zero.

7x³ = 28x

7x³ - 28x = 0

Now notice there's a common x in both terms. Let's factor it out.

x (7x² - 28) = 0

As 7 is a factor of 7 and 28, it too can be factored out.

x (7) (x² - 4) = 0

We can further factor x² - 4. We want a pair of numbers that multiply to 4 and whose sum is zero. The pairs are 1 and 4, 2 and 2. If we add 2 and -2 we get zero.

x (7) (x - 2) (x + 2) = 0

Now we use the Zero Product Property - if some product multiplies to zero, so do its pieces.

x = 0        -----> so x = 0

7 = 0       -----> no solution

x - 2 = 0   ----> so x = 2   after adding 2 to both sides

x + 2 = 0  ---> so = x - 2  after subtracting 2 to both sides


Thus the solutions are x = 0, x = 2, x = -2.

6 0
3 years ago
Please solve all i will mark brainliest and 15 points.​
OleMash [197]

Answer:

Hope that will help you get

3 0
3 years ago
Can someone please help me answer this my grade depends on it!
ch4aika [34]
Can someone please help me answer this my grade depends on it!

a. 3 gallons 1 cup
b. 15 cups
3 0
4 years ago
Convert 5820 grams to kilograms.
mr_godi [17]

5.82 Kg

because 1Kg= 1000g

5 0
3 years ago
How many integers between 10000 and 99999, inclusive, are divisible by 3 or<br> 5 or 7?
Yuki888 [10]

Answer: Hence, there are approximately 48884 integers are divisible by 3 or 5 or 7.

Step-by-step explanation:

Since we have given that

Integers between 10000 and 99999 = 99999-10000+1=90000

n( divisible by 3) = \dfrac{90000}{3}=30000

n( divisible by 5) = \dfrac{90000}{5}=18000

n( divisible by 7) = \dfrac{90000}{7}=12857.14

n( divisible by 3 and 5) = n(3∩5)=\dfrac{90000}{15}=6000

n( divisible by 5 and 7) = n(5∩7) = \dfrac{90000}{35}=2571.42

n( divisible by 3 and 7) = n(3∩7) = \dfrac{90000}{21}=4285.71

n( divisible by 3,5 and 7) = n(3∩5∩7) = \dfrac{90000}{105}=857.14

As we know the formula,

n(3∪5∪7)=n(3)+n(5)+n(7)-n(3∩5)-n(5∩7)-n(3∩7)+n(3∩5∩7)

=30000+18000+12857.14-6000-2571.42-4258.71+857.14\\\\=48884.15

Hence, there are approximately 48884 integers are divisible by 3 or 5 or 7.

5 0
4 years ago
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