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3 years ago

Express 455 ml as a percentage of 1.5 litres

Mathematics

1 answer:

9966 [12]3 years ago

4 0

Answer:

30.3

Step-by-step explanation:

a liter is 1000 ml

1000*1.5=1500

455/1500*100=30.3333333333333333

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For each of the figures write an absolute value equation to satisfy the given solution set

Lubov Fominskaja [6]

Answer is: I x-11 I = 6

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3 years ago

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The length of the shadow of a pole on level ground increases by 90m when the angle of elevation of the sun changes from 58° to 3

siniylev [52]

<h3>Answer: Approximately 119.76 meters</h3>

===================================================

Work Shown:

x = starting length of the shadow

y = height of the pole

tan(angle) = opposite/adjacent

tan(58) = y/x

1.6003345 = y/x

1.6003345x = y

x = y/1.6003345

x = (1/1.6003345)y

x = 0.62486936y

-------------------------

When the angle changes, the adjacent side gets 90 meters longer

tan(angle) = opposite/adjacent

tan(36) = y/(x+90)

0.72654253 = y/(0.62486936y+90)

0.72654253(0.62486936y+90) = y

0.453994166y + 65.3888277 = y

65.3888277 = y-0.453994166y

65.3888277 = 0.546005834y

0.546005834y = 65.3888277

y = 65.3888277/0.546005834

y = 119.758478075162

y = 119.76

The height of the pole is about 119.76 meters.

8 0

3 years ago

Can anyone explain how to do mathematical induction. The question is below. Brainliest awarded for the right answer. Thanks.

igomit [66]

The sum of the triangular numbers is $\frac{n(n+1)(n+2)}{6}$

According to the question, the series of triangular numbers is given as in the form of the number of dots constituting the equilateral triangles i.e.
1, 3, 6, 10, 15 . . . . .n

<h3>what is triangular numbers?</h3>

triangular number are the sequence and series of the numbers and each number represents and constitute in visualization of series of equilateral triangle.

given series in the figure
1, 3, 6, 10, 15, . . . . . . ., n
each number represents the number of dots containing in triangle
now the series can be given by
$S = \frac{n(n+1)}{2}$ for n = 1, 2, 3, 4,. . . . .n
now , according to the question the sum of the series can be given as
⇒ ∑S
⇒ ∑ $\frac{n^2+n}{2}$
⇒ $\frac{1}{2}[\sum n^2+\sum n]\\\frac{1}{2}[\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}] \\\frac{n(n+1)}{4}[\frac{(2n+1)}{3}+1] \\\frac{n(n+1)}{4}[\frac{(2n+1+3)}{3}] \\\frac{n(n+1)}{4}[\frac{(2n+4)}{3}] \\\frac{n(n+1)}{4}2[\frac{(n+2)}{3}]\\\frac{n(n+1)(n+2)}{6}$