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2 years ago

Max points easy math problem

Mathematics

2 answers:

11Alexandr11 [23.1K]2 years ago

4 0

Solution:

Given:

x = -4
y = 4

<u>Substitute the values into the equation:</u>

x⁶ - x/4y
=> [(-4)⁶] - (-4)/4(4)

<u>Simplify the numerator and the denominator:</u>

=> [(-4) x (-4) x (-4) x (-4) x (-4) x (-4)] + 4/16
=> [(-64) x (-64)] + 4/16
=> [4096] + 4/16
=> 4100/16

<u>Simplify the fraction:</u>

=> 4100/16
=> 1025 x 4/4 x 4
=> 1025/4

Correct option is B.

Nataly [62]2 years ago

4 0

Answer:

$[-\frac{1023}{4} ]$

Step-by-step explanation:

$\frac{x^6-x}{4y}$

x = -4

y = 4

Replace all the variables with the corresponding values given:

$\frac{-4^6-(-4)}{4(4)}$

Now solve,

-4 to the power of 6 is equal to -4096

$\frac{-4096-(-4)}{4(4)}$

we see that we are subtracting a negative number, meaning that it'll cancel out and turn into an addition sign. Therefore:

$-4096 + 4$

$-4092$

$\frac{-4092}{4(4)}$

4(4) is equal to 16

$\frac{-4092}{16}$

When divided, you'll get a mixed fraction of $-255\frac{3}{4}$

To turn into an improper fraction, simply multiply the whole number by the denominator, then add the numerator to the result.

$-255 * 4 + 3$

$[-\frac{1023}{4} ]$

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C. line b and line c

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3 years ago

Identify the domain and range of each function.

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Answer:

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Step-by-step explanation:

The given function is $y=3 \times 5x$ or it can be written as y = 15x.

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3 years ago

Read 2 more answers

What is the sum of 1/8 + 5/16 +3/8 = and can I please have step by step

AlekseyPX

Answer:

13/16

Step-by-step explanation:

(1/8) + (5/16) + (3/8) =

The denominator 8 goes into 16 twice so multiply the top and bottom numbers by 2

(1 x 2) + (5/16) + (3 x 2)

(8 x 2) (8 x 2)

2/16 + 5/16 + 6/16 = 13/16

I hope this helps!

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2 years ago

1. Find four consecutive even integers such that the sum of the first and third is 6 less than the largest

Sophie [7]

These are indeed quite a lot of exercises, but a lot of them are almost identical - only small computations will change. So, I'm glad to help you with one exercise from every cathegory, but I encourage you to solve the others on your own.

<h2>Exercise 1</h2>

You can call four consecutive integers as

$x,\ x+1,\ x+2,\ x+3$

So, the sum of the first and third is $x+(x+2) = 2x+2$

We want this quantity to be six less than the largest, i.e. $(x+3)-6 = x-3$

So, the equality is

$2x+2 = x-3$

Subtract x from both sides:

$x+2 = -3$

Subtract 2 from both sides:

$x = -5$

So, the consecutive integers are

$-5,\ -4,\ -3,\ -2$

In fact, the sum of the first and third is $-5-3 = -8$ , which is indeed six less than the largest: $-8 = -2-6$

<h2>Exercise 2</h2>

If you call the first odd number $x$ , the next consecutive odd numbers will be $x,\ x+2,\ x+4,\ x+6$

In fact, we have to count skipping two's, because we only want odd integers. From here, you go on like exercise 1: you write the largest (which is x+6), and set it to be two more than the sum of the other three (x, x+2 and x+4)

<h2>Exercise 3</h2>

By the same logic of exercise 2, two consecutive even integers are $x,\ x+2$ , assuming that x is even.

So, you set the equation as usual: the smaller (which is x) is 26 less than three times the larger (which means 3(x+2)-26)

<h2>Exercise 4 to 8</h2>

These are all pretty identical to exercise 1: you start by listing three or four consecutive integers:

$x,\ x+1,\ x+2\quad\text{or}\quad x,\ x+1,\ x+2\ x+3$

and then you translate the request of each exercise accordingly. Remember that expressions like "three times the second number" means that you have to multiply: $3(x+1)$ , while expression like "six more than the first" or "thirteen less than the first" imply adding/subtracting: $x+6$ or $x-13$ .