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2 years ago

If x = 2, y = 8, find (i) x³+y³ (ii) ∛y

Mathematics

2 answers:

zlopas [31]2 years ago

7 0

Answer:

( i ) 520

( ii ) 2

Step-by-step explanation:

We can find this solution by plugging in known values -

If x = 2, y = 8

x³+y³ = ( 2 )³ + ( 8 )³ = 8 + 512

= 520

Know let us move on to the second half -

We only need one part of this information now, y = 8. If so,

∛y = ∛8

2 x 2 x 2 = 8 - and thus 2 should be our solution for this portion.

KiRa [710]2 years ago

5 0

Answer:

(i) 520

(ii) 2

Step-by-step explanation:

(i) x³ + y³

Plug x as 2, and y as 8.

(2)³ + (8)³

Solve for exponents.

8 + 512

Add.

= 520

(ii) ∛y

Plug y as 8.

∛(8)

Solve for cube root.

= 2

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3 years ago

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3 years ago

Read 2 more answers

This monitor has a diagonal length of 19 inches and an aspect ratio of 5:4. What are the width and height of the monitor? Round

nasty-shy [4]

<em>Rounded to the nearest tenth of an inch the Witdth of the monitor is 14.8 inches while the Height of the monitor is 11.9 inches.</em>

<h2>Explanation:</h2><h2 />

The aspect ratio is defined as:

<em>"The proportional relationship between the width and height of an image"</em>

It is usually expressed in a mathematical language as:

$x:y \\ \\ x: \ Width \\ \\ y: \ Height$

So, it is true that:

$x:y=5:4 \\ \\ or: \\ \\ \frac{x}{y}=\frac{5}{4} \\ \\ \therefore x=\frac{5y}{4}$

In this case, we know that the monitor has a diagonal length (D):

$D=19in$

So, by Pythagorean theorem:

$D^2=361 \\ \\ 361=x^2+y^2 \\ \\ Since \ x=\frac{5y}{4}$

$D^2=361 \\ \\ 361=x^2+y^2 \\ \\ Since \ x=\frac{5y}{4} \\ \\ 361=\left(\frac{5y}{4}\right)^2+y^2 \\ \\ 361=\frac{25y^2}{16}+y^2 \\ \\ \frac{41}{16}y^2=361 \\ \\ y^2=\frac{261\times 16}{41} \\ \\ y=\sqrt{140.87} \\ \\ y=11.869in$

Finding x:

$x=\frac{5(11.869)}{4} \\ \\ x=14.836in$

<em>Finally, rounded to the nearest tenth of an inch the Witdth of the monitor is 14.8 inches while the Height of the monitor is 11.9 inches.</em>

<h2>Learn more:</h2>

Pythagorean Theorem: brainly.com/question/10182890

#LearnWithBrainly

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