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3 years ago

Need help with this

Mathematics

1 answer:

Vanyuwa [196]3 years ago

4 0

Answer:

Slope-int form: y = 3x+5

Standard form: y - 3x = 5.

Step-by-step explanation:

(reminder: slope-intercept form is expressed as y=mx+b, and standard form is expressed as ax+bx=c.)

Since the slope is 3, the coefficient of x is also 3, which makes the equation y=3x.

But the y coordinate of the equation at x = -2 is -6, so we need to add 5 to the end of the equation, leaving you with:

y=3x+5.

To convert it to standard form, subtract 3x from both sides:

y - 3x = 5.

I hope this helped you.

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Which of the following expressions represents the least common multiple of 15, 18, and 25?

Virty [35]

Answer:

$(2)(3^2)(5^2)$

Step-by-step explanation:

we know that

The <u>Least Common Multiple</u> (LCM) of a group of numbers is the smallest number that is a multiple of all the numbers.

we have

15,18 and 25

Decompose the numbers in prime factors

$15=(3)(5)$

$18=(2)(3^2)$

$25=(5^2)$

Multiply common and uncommon numbers with their greatest exponent

The LCM is equal to

$LCM=(2)(3^2)(5^2)$

$LCM=450$

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3 years ago

Is -3+3-3+3-3+3 a arithmetic?

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Answer:

$- 3 + 3 = 0 - 3 = - 3 \\ - 3 + 3 = 0 - 3 = - 3 \\ - 3 + 3 = 0$

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3 years ago

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The point (0,-4) is located [ANSWER]

natka813 [3]

Answer:

on the x-axis

Step-by-step explanation:

it is located on the x axis because it is not technically in any quandrant. it is however, on an axis.

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3 years ago

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Are the 2 triangles congruent? If so, what postulate did you use?

jok3333 [9.3K]

Answer:

Step-by-step explanation:

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3 years ago

The sequence$$1,2,1,2,2,1,2,2,2,1,2,2,2,2,1,2,2,2,2,2,1,2,\dots$$consists of $1$'s separated by blocks of $2$'s with $n$ $2$'s i

kicyunya [14]

Consider the lengths of consecutive 1-2 blocks.

block 1 - 1, 2 - length 2

block 2 - 1, 2, 2 - length 3

block 3 - 1, 2, 2, 2 - length 4

block 4 - 1, 2, 2, 2, 2 - length 5

and so on.

Recall the formula for the sum of consecutive positive integers,

$\displaystyle \sum_{i=1}^j i = 1 + 2 + 3 + \cdots + j = \frac{j(j+1)}2 \implies \sum_{i=2}^j = \frac{j(j+1) - 2}2$

Now,

$1234 = \dfrac{j(j+1)-2}2 \implies 2470 = j(j+1) \implies j\approx49.2016$

which means that the 1234th term in the sequence occurs somewhere about 1/5 of the way through the 49th 1-2 block.

In the first 48 blocks, the sequence contains 48 copies of 1 and 1 + 2 + 3 + ... + 47 copies of 2, hence they make up a total of

$\displaystyle \sum_{i=1}^48 1 + \sum_{i=1}^{48} i = 48+\frac{48(48+1)}2 = 1224$

numbers, and their sum is

$\displaystyle \sum_{i=1}^{48} 1 + \sum_{i=1}^{48} 2i = 48 + 48(48+1) = 48\times50 = 2400$

This leaves us with the contribution of the first 10 terms in the 49th block, which consist of one 1 and nine 2s with a sum of $1+9\times2=19$ .

So, the sum of the first 1234 terms in the sequence is 2419.

8 0

2 years ago

Need help with this​

Need help with this