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3 years ago

Help me understand Lateral Area, and surface are pleassee

1 answer:

4 0

The lateral surface of an object is the area of all the sides of the object, excluding the area of its base and top. For a cube, the lateral surface area would be the area of four sides.For a pyramid, the lateral surface area is the sum of the areas of all of the triangular faces but excluding the area<span> of the base.</span>

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SOLVE EACH PROBLEM: 1. Edwin pays tax at the rate of 25% of his income. What fraction of Edwin’s income is this? 2. Vida correct

Elena L [17]

Step-by-step explanation:

1/4

90%

1.5

0.005

6/10

25%

2/10

49.5

3/4

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1 year ago

A certain amount of money is divided among Rio, Kim and Leo in the ratio 5:7:3. If Leo gets Php 24,000.00, how much is the total

cricket20 [7]

Answer:

120,000

Step-by-step explanation:

Let's say Rio gets 5x, Kim gets 7x, and Leo gets 3x. In the ratio 5:7:3:1, x = 1

5x + 7x + 3x = total = 15x

3x = 24000

divide both sides by 3 to get x

x = 8000

15x = 8000*15 = 120000

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2 years ago

WILL GIVE BRAINLIEST!

MAXImum [283]

Answer:

Texting speed

Step-by-step explanation:

It's the dependent variable

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3 years ago

Read 2 more answers

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .