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2 years ago

Why can it help to line up the decimal points before comparing and ordering numbers with decimals

Mathematics

2 answers:

aliina [53]2 years ago

5 0

So you don't loose your place

Reptile [31]2 years ago

3 0

Answer:

so you can't lose your place

Step-by-step explanation:

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Instructions: Find the surface area of the figure below. Round your answers to the nearest tenth, if necessary.

Y_Kistochka [10]

The surface area of the figure given is of: 61.3 yd².

<h3>What is the combined surface area of a figure?</h3>

The combined surface area of a figure is the area of each figure.

This figure is composed by:

A circle of radius 3 yd.

A triangle with base 6 yd and one leg 11.4 yd.

The area of the circle is given as follows:

$A = \pi r^2 = \pi \times 3^2 = 9\pi$

For the triangle, we need to find it's height, which is the leg of a right triangle, in which one of the sides is 3 and the hypotenuse is 11.4, hence:

h² + 3² = 11.4²

$h = \sqrt{11.4^2 - 3^2}$

h = 11 yd.

Hence the area of the triangle is:

A = 0.5 x 6 x 11 = 33 yd²

The surface area of the figure is:

9pi + 33 yd² = 61.3 yd².

More can be learned about the surface area of at brainly.com/question/21208177

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6 0

2 years ago

If logb2=x and logb3=y, evaluate the following in terms of x and y:

Alja [10]

$log_b{162} = x + 4y\\\\log_b324 = 2x+4y\\\\log_b\frac{8}{9} = 3x-2y\\\\\frac{log_b27}{log_b16} = 3y-4x$

Solution:

Given that,

$log_b2 = x\\\\log_b3 = y --------(i)$

Use the following log rules

Rule 1: $log_b(ac) = log_ba + log_bc$

Rule 2: $log_b\frac{a}{c} = log_ba - log_bc$

Rule 3: $log_ba^c = clog_ba$

$a) log_b{162}$

Break 162 down to primes:

$162 = 2^1 \times 3^4$

$log_b{162} =log_b 2^1. 3^4\\\\By\ rule\ 1\\\\ log_b{162} = log_b 2^1 +log_b 3^4\\\\By\ rule\ 3\\\\1log_b2 + 4log_b3\\\\1x+4y\\\\x+4y$

Thus we get,

$log_b162 = x + 4y$

$b) log_b 324$

Break 324 down to primes:

$324 = 2^2 \times 3^4$

$log_b324 = log_b 2^2.3^4\\\\By\ rule\ 1\\\\log_b324 = log_b2^2 + log_b3^4\\\\By\ rule\ 3\\\\log_b324 = 2log_b2 + 4log_b3\\\\From\ (i)\\\\log_b324 = 2x + 4y$

$c) log_b\frac{8}{9}$

By rule 2

$log_b\frac{8}{9} = log_b8 - log_b9\\\\log_b\frac{8}{9} = log_b 2^3 - log_b3^2\\\\By\ rule\ 3\\\\log_b\frac{8}{9} = 3 log_b2 - 2log_b3\\\\From\ (i)\\\\log_b\frac{8}{9} = 3x - 2y$

$d) \frac{log_b27}{log_b16}$

By rule 2

$\frac{log_b27}{log_b16} = log_b27 - log_b16\\\\ \frac{log_b27}{log_b16} = log_b3^3 - log_b2^4\\\\By\ rule\ 2\\\\ \frac{log_b27}{log_b16} = 3log_b3 - 4log_b2 \\\\From\ (i)\\\\\frac{log_b27}{log_b16} = 3y - 4x$