Ask question

Ask question

All categories

3 years ago

13

The least common multiple of 20, 24, and 45 is _____. A. 30 B. 180 C. 360 D. 21,600

1 answer:

Alexus [3.1K]3 years ago

8 0

Answer:c because the least common multiple is 360

Step-by-step explanation:

You might be interested in

Help pls, will mark brainliest

BaLLatris [955]

Here , we are provided with a table which shows 5 consecutive terms of an arithmetic sequence . But before solving further , let's recall that ;

The n'th term of a Arithmetic Sequence let's say it be ${\sf T_n}$ is given by ;

${\boxed{\bf T_{n}=T_{1}+(n-1)d}}$

Where , <u>d</u> is the common difference

Now , here we are given with ;

${\quad \qquad \sf \blacktriangleright \blacktriangleright \blacktriangleright T_{1}=8 \: and \: T_{5}=-4}$

We have to find the 2nd , 3rd and 4th term respectively ,

Now , by using the above formula , 5th term can be written as ;

${: \implies \quad \sf T_{1}+(5-1)d=T_{5}}$

Putting the values and transposing 1st term to RHS , we have ;

${: \implies \quad \sf 4d = -4-8}$

${: \implies \quad \sf d=-\dfrac{12}{4}}$

${: \implies \quad \sf d=-3}$

Now , as we got the common difference , so we can find out the missing terms now ;

${: \implies \quad \sf T_{2}= T_{1}+(2-1)d}$

${: \implies \quad \sf T_{2}= 8 +d}$

${: \implies \quad \sf T_{2}= 8-3}$

${: \implies \quad \bf \therefore \: T_{2}= 5}$

Now

${: \implies \quad \sf T_{3}= T_{1}+(3-1)d}$

${: \implies \quad \sf T_{3}= 8 +2d}$

${: \implies \quad \sf T_{3}= 8-6}$

${: \implies \quad \bf \therefore \: T_{3}= 2}$

Also ,

${: \implies \quad \sf T_{4}= T_{1}+(4-1)d}$

${: \implies \quad \sf T_{4}= 8 +3d}$

${: \implies \quad \sf T_{4}= 8-9}$

${: \implies \quad \bf \therefore \: T_{4}= -1}$

Now , The given table can be written as ;

${\begin{array}{|c|c|c|c|c|c|}\cline{1-6} \bf n & \sf 1 & 2 & 3 & 4 & 5 \\ \cline{1-6} \bf T_{n} & \sf 8 & 5 & 2 & -1 & -4 \end{array}}$

Note :- Kindly view the answer from web , if you're not able to see the full answer from here ;

brainly.com/question/26750175

8 0

3 years ago

Y=-1/3x^2-4x-5 in vertex form

grigory [225]

Answer:

$Y=-\frac{1}{3}(x+6)^2+7$ [Vertex form]

Step-by-step explanation:

Given function:

$Y=-\frac{1}{3}x^2-4x-5$

We need to find the vertex form which is.,

$y=a(x-h)^2+k$

where $(h,k)$ represents the co-ordinates of vertex.

We apply completing square method to do so.

We have

$Y=-\frac{1}{3}x^2-4x-5$

First of all we make sure that the leading co-efficient is =1.

In order to make the leading co-efficient is =1, we multiply each term with -3.

$-3\times Y=-3\times\frac{1}{3}x^2-(-3)\times4x-(-3)\times 5$

$-3Y=x^2+12x+15$

Isolating $x^2$ and $x$ terms on one side.

Subtracting both sides by 15.

$-3Y-15=x^2+12x-15-15$

$-3Y-15=x^2+12x$

In order to make the right side a perfect square trinomial, we will take half of the co-efficient of $x$ term, square it and add it both sides side.

square of half of the co-efficient of $x$ term = $(\frac{1}{2}\times 12)^2=(6)^2=36$

Adding 36 to both sides.

$-3Y-15+36=x^2+12x+36$

$-3Y+21=x^2+12x+36$

Since $x^2+12x+36$ is a perfect square of $(x+6)$ , so, we can write as:

$-3Y+21=(x+6)^2$

Subtracting 21 to both sides:

$-3Y+21-21=(x+6)^2-21$

$-3Y=(x+6)^2-21$

Dividing both sides by -3.

$\frac{-3Y}{-3}=\frac{(x+6)^2}{-3}-\frac{21}{-3}$

$Y=-\frac{1}{3}(x+6)^2+7$ [Vertex form]

8 0

3 years ago

Hey yall so im taking a test and imm very confused on hwo to find the letter n for the equation

kotegsom [21]

Answer:

n = -7,6

Step-by-step explanation:

n(n+1)+3=45

distributive property: a(b+c) = ab+ac

n²+n+3 = 45

n²+n -42 = 0

factor now

(n+7)(n-6) = 0

n = -7,6

8 0

4 years ago

Necesito ayuda con está preguntas

prisoha [69]

Answer:

para el primer dibujo seria 1/12

para el segundo dibujo seria 1/16

para el tercer dibujo seria 1/10, pero tengo dudas con esta porque no se ve toda la figura.

Step-by-step explanation:

4 0

3 years ago

PLS PLS I NEED HELP I HAVE NO TIME I'LL MARK YOU BRAINLIST !!!

Dominik [7]

Answer:

1) y = 5/2x - 3/2

2) y = 2/5x - 4/5

Step-by-step explanation:

1)

(1, 1) and (3, 6)

m = (y2 - y1)/(x2 - x1)

m= (6 - 1)/(3-1)

m= 5/2

m= 5/2

y - y1 = m(x - x1)

y - 1 = 5/2(x - 1)

y - 1 = 5/2x - 5/2

y = 5/2x - 3/2

2)

(2, 0) and (7, 2)

m = (y2 - y1)/(x2 - x1)

m= (2 - 0)/(7 - 2)

m = 2/5

y - y1 = m(x - x1)

y - 0 = 2/5(x - 2)

y = 2/5x - 4/5

5 0

2 years ago

Read 2 more answers