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Dima020 [189]
3 years ago
13

What is the lowest common multiple of 6 and 9 ?

Mathematics
1 answer:
aliina [53]3 years ago
4 0

Answer:

18

Step-by-step explanation:

Find the least common multiple of 6 and 9. We see that the numbers 18 and 36 are both common multiples of 6 and 9. The least common multiple is the smallest which is 18.

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Viefleur [7K]

Answer:

6. Multiply 2 by 6

7. 71 + 9(8) = 71 + 72 = 143

8. 52 + 3(4) = 12 + 52 = 64

9. 19.25 + 2.75(7) = 19.25 + 19.25 = 38.50

10. 2(0.6)^2 + 4(0.6)(5) = 0.72 + 12 = 12.72

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6 0
2 years ago
Two identical condominiums were built in a new high-rise building. Condo A’s living room is a rectangle 24 ft. long and 18 ft. w
nadezda [96]

Answer:

The perimeter of a rectangular playground is 46 m. ... 46 – 14 = 32 32 / 2 = 16 ... new carpeting for her family room. Her family room is a 12 ft by 21 ft rectangle. ... Charles has a rectangular flower garden that is 5 yd long and 12 yd wide. ... After installing the carpet, Betty decides to also install floor wood molding for her living.

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6 0
3 years ago
(Picture will be inserted) Look at how the equation was solved. Describe another way you could have solved the equation. Tell wh
jenyasd209 [6]

The value of m is 2450

<h3>What is Equation?</h3>

Two or more expressions with an Equal sign is called as Equation.

The given equation is

2(14m/400)=171.5

Distribute 2 with numerator on LHS

28m/400=171.5

14m/200=171.5

7m/100=171.5

Multiply 100 on both sides

7m=17150

m=2450

Hence, the value of m is 2450

To learn more on Equation:

brainly.com/question/10413253

#SPJ1

8 0
1 year ago
I give brainliest!!!<br> Can someone tell me what a radical is
wariber [46]
A radical is also known as a square root. An example of a radical is √25 or ∛36.
4 0
3 years ago
Read 2 more answers
Find an n^th degree polynomial with real coefficients satisfying the given conditions. n = 3; -2 and 2 i are zeros; f(-1) = 15.
Ira Lisetskai [31]
So, n = 3, is a 3rd degree polynomial, roots are -2 and 2i

well 2i is a complex root, or imaginary, and complex root never come all by their lonesome, their sister is always with them, the conjugate, so if 0+2i is there, 0-2i is there too

so, the roots are -2, 2i, -2i

now... \bf \begin{cases}&#10;x=-2\implies x+2=0\implies &(x+2)=0\\&#10;x=2i\implies x-2i=0\implies &(x-2i)=0\\&#10;x=-2i\implies x+2i=0\implies &(x+2i)=0&#10;\end{cases}&#10;\\\\\\&#10;(x+2)\underline{(x-2i)(x+2i)}=0\\\\&#10;-----------------------------\\\\&#10;\textit{difference of squares}&#10;\\ \quad \\&#10;(a-b)(a+b) = a^2-b^2\qquad \qquad &#10;a^2-b^2 = (a-b)(a+b)\\\\&#10;-----------------------------\\\\&#10;(x+2)[x^2-(2i)^2]=0\implies (x+2)[x^2-(2^2i^2)]=0&#10;\\\\\\&#10;(x+2)[x^2-(4\cdot -1)]=0\implies (x+2)(x^2+4)=0&#10;\\\\\\&#10;x^3+2x^2+4x+8=0

now, if we check f(-1), we end up with 5, not 15
hmmm

so, how to turn our 5 to 15? well, 3*5, thus

\bf 3(x^3+2x^2+4x+8)=f(x)\implies 3(5)=f(-1)\implies 15=f(-1)

usually, when we get the roots, or zeros, if any common factor that is a constant is about, they get in a division with 0 and get tossed, and aren't part of the roots, thus, we can simply add one, in this case, the common factor of 3, to make the 5 turn to 15
6 0
3 years ago
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