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2 years ago

What's (X+3 (X+5) it's really confusing me and I need some help please

1 answer:

5 0

Well when you see parentheses set up like these together that automatically means to multiply. The solution is:
Apply the FOIL method: (a+b)(c+d)=ac+ad+bc+bd
x2+5x+3x+15
Simplify
x2+8x+15

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Name three other common multiples of 8 and 12

Sladkaya [172]

The first few multiples of 8 are: 8, 16, 24, 32, 40, and 48. The first multiples of 12 are: 12, 24, 36, 48, 60, and 72.

Common multiple of 8 and 12 are 24 and every subsequent multiple of 24.

As 8×3=24

12×2=24

So 24 is a common multiple of 8 and 12 this means the other common multiples of 8 and 12 are the multiples of 24 like 48, 72, 96 etc.

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10 months ago

How many modes does the following data set have? 5, 5, 6, 7, 8, 12, 12, 12, 150, 150, 150

Feliz [49]

Answer:

Step-by-step explanation:

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2 years ago

Read 2 more answers

HELP PLEASE must show work I have the answer just need to show work

Kazeer [188]

Answer:

Step-by-step explanation:

To solve these equations involving variables and exponents we need to follow these steps.

1) We need to find out the factor that is common in the equation.

2) After taking common, solve the equation. We can add or subtract only those values that have same bases.

1) $8+6x^4$

here we can see, both numbers are divisible by 2, so taking 2 common

$=2(8/2 + 6x^4/2)\\= 2(4 + 3x^4)$

It cannot be further simplified because both number donot have same bases.

3. $4n^9 + 12 n$

We can take 4n common

$=4n(4n^9/4n + 12 n/4n)\\=4n(n^8 + 3)$

5. -12a -3

Here -3 cam be taken common

= -3(-12a/-3 -3/-3)

= -3(4a +1)

7. $12n^5 + 16n^3$

here the smallest power of n is n^3 so, we can take n^3 common and both coefficients are divisible by 4 so taking 4n^3 common

$4n^3( 3n^2 + 4)$

9. $5k^2 - 40k+10$

Here we cannot take k common, as k is not a multiple of 10. For taking common it should be divisible by each value in the equation. But each value s divisible by 5 so, taking 5 common

$=5(k^2 - 8k + 2)$

11. $-60 + 60n^2 +50n^3$

Here we cannot take n common, as n is not a multiple of -60. For taking common it should be divisible by each value in the equation. But each value s divisible by 10 so, taking 10 common

$=10(-6 + 6n^2 +5n^3)$

13. $-36n^3 -12n-28$

Here we cannot take n common, as n is not a multiple of 28. For taking common it should be divisible by each value in the equation. But each value s divisible by -4 so, taking -4 common

$=-4(9n^3 + 3n +7)$

15. $63n^3+81n+18$

Here we cannot take n common, as n is not a multiple of 18. For taking common it should be divisible by each value in the equation. But each value s divisible by 9 so, taking 9 common

$=9(7n^3 + 9n + 2)$