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ioda
3 years ago
9

To which sets does negative five eighths belong.

Mathematics
1 answer:
mezya [45]3 years ago
3 0
It belongs to rational numbers and intergers
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four times a number minus twice another number is -16 the sum of the two numbers is -1 find the numbers
lianna [129]

Answer: The numbers are -3 and 2

Step-by-step explanation:  Let x be the first number and y be the second number

then according to given statements the linear equations will be:

"Four times a number minus twice another number is -16"

"The sum of the two numbers is -1"

From equation 2:

Putting y = -x-1 in equation 1

Subtracting 2 from both sides

Dividing both sides by 6

Putting x = -3 in equation 2:

Hence,

The numbers are -3 and 2

6 0
2 years ago
The following tile floor pattern is made of 36 square tiles.
Ierofanga [76]

Answer:

C

Step-by-step explanation:

C

4 0
2 years ago
Find cosine of angle a ​
Hunter-Best [27]

Answer:

ଆପଣ ଅଧ୍ୟୟନ କରିପାରିବେ ଏବଂ ଇଣ୍ଟରନେଟ୍ ବ୍ୟବହାର କରି ଆପଣଙ୍କର ସମୟ ନଷ୍ଟ କରିପାରିବେ ନାହିଁ :)

Step-by-step explanation:

6 0
3 years ago
What is the answer to this
____ [38]

Answer:

19

Step-by-step explanation:

Add up all the numbers:

3,2,2,2,1,1,1,4,2,1

Then you get 19.

Or you can say it as

3^2 + 2^4 + 1^4 + 4

7 0
3 years ago
(ASAP PICTURE ADDED) What is the simplified form of the following expression?
bonufazy [111]

Answer:

option c is correct.

Step-by-step explanation:

7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{16x}\right)-3\left(\sqrt[3]{8x}\right)

WE need to simplify this equation.

Solve the parenthesis of each term.

=7\left\sqrt[3]{2x}\right-3\left\sqrt[3]{16x}\right-3\left\sqrt[3]{8x}\right

Now, We will find factors of the terms inside the square root

factors of 2: 2

factors of 16 : 2x2x2x2

factors of 8: 2x2x2

Putting these values in our equation:=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2X2X2X2 x}\right)-3\left(\sqrt[3]{2X2X2 x}\right)\\=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2X2X2} \sqrt[3] {2 x}\right)-3\left(\sqrt[3]{2X2X2} \sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2^3} \sqrt[3] {2 x}\right)-3\left(\sqrt[3]{2^3} \sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2x}\right)-3*2\left(\sqrt[3] {2 x}\right)-3*2\left(\sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2}\sqrt[3]{x}\right)-6\left(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)

Adding like terms we get:

=7\left(\sqrt[3]{2}\sqrt[3]{x}\right)-6\left(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right\\=(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)\\

(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)\\can\,\,be \,\, written\,\, as\,\,\\(\sqrt[3] {2x})-6\left(\sqrt[3]{x}\right)

So, option c is correct

5 0
3 years ago
Read 2 more answers
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