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lubasha [3.4K]
3 years ago
15

I will give the correct answer brainliest !!

Mathematics
1 answer:
zubka84 [21]3 years ago
6 0

why do they be giving us this hardwork :(

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Multiply and simplify: (6x + 3y)(6x − 3y)
nasty-shy [4]

Answer:

36x² - 9y²

Step-by-step explanation:

We see that the given expression (6x + 3y)(6x - 3y) looks just like the expression (a + b)(a - b). When multiplied out, this is a difference of squares identity that I highly recommend you memorise:

(a + b)(a - b) = a² - b²

Here, a = 6x and b = 3y, so plug these in:

(6x + 3y)(6x - 3y) = (6x)² - (3y)² = 36x² - 9y²

5 0
3 years ago
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Use the distributive property to simplify each expression<br> (2b - 10) 3.2
Darya [45]

Answer: 6.4b - 32

Step-by-step explanation:

4 0
3 years ago
Is number 4 correct?
Strike441 [17]

Answer:

sure

Step-by-step explanation:

6 0
3 years ago
There is a group of 10 people who are ordering pizza. If each person gets 2 slices and each pizza has 4 slices, how many pizzas
Sonbull [250]

Answer:

5 pizzas

Step-by-step explanation: If you have 10 people and each will get two slices then you will need 20 pieces of pizza because 10x2 is 20. If each pizza has 4 slices then you will need 5 pizzas because 4/20 is 5.

6 0
3 years ago
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Find the number b such that the line y = b divides the region bounded by the curves y = 36x2 and y = 25 into two regions with eq
Gemiola [76]

Answer:

b = 15.75

Step-by-step explanation:

Lets find the interception points of the curves

36 x² = 25

x² = 25/36 = 0.69444

|x| = √(25/36) = 5/6

thus the interception points are 5/6 and -5/6. By evaluating in 0, we can conclude that the curve y=25 is above the other curve and b should be between 0 and 25 (note that 0 is the smallest value of 36 x²).

The area of the bounded region is given by the integral

\int\limits^{5/6}_{-5/6} {(25-36 \, x^2)} \, dx = (25x - 12 \, x^3)\, |_{x=-5/6}^{x=5/6} = 25*5/6 - 12*(5/6)^3 - (25*(-5/6) - 12*(-5/6)^3) = 250/9

The whole region has an area of 250/9. We need b such as the area of the region below the curve y =b and above y=36x^2 is 125/9. The region would be bounded by the points z and -z, for certain z (this is for the symmetry). Also for the symmetry, this region can be splitted into 2 regions with equal area: between -z and 0, and between 0 and z. The area between 0 and z should be 125/18. Note that 36 z² = b, then z = √b/6.

125/18 = \int\limits^{\sqrt{b}/6}_0 {(b - 36 \, x^2)} \, dx = (bx - 12 \, x^3)\, |_{x = 0}^{x=\sqrt{b}/6} = b^{1.5}/6 - b^{1.5}/18 = b^{1.5}/9

125/18 = b^{1.5}/9

b = (62.5²)^{1/3} = 15.75

8 0
3 years ago
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