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3 years ago

Ten to the first power is written as

Mathematics

1 answer:

Triss [41]3 years ago

4 0

$10^1$

I hope this helps :)

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6y - 5x = 5 and x= 2y + 7

jolli1 [7]

Y = -10 and x = -13 (substitute 2y + 7 for x)

5 0

4 years ago

Choose 3 values that would make thi inequalitie true 28+x>42

Y_Kistochka [10]

9514 1404 393

Answer:

14.01, 493, 87

Step-by-step explanation:

Subtracting 28 from both sides tells you the range of values you need to be looking at.

28 + x > 42

x > 14

Any values more than 14 will make the inequality true. Three of them are ...

14.01, 493, 87

4 0

3 years ago

Find thhe remainder when 7^203 is divided by 4

Aleksandr [31]

Using the square-and-multiply approach, we have

$7^{203}=7\times(7^{101})^2$
$7^{101}=7\times(7^{50})^2$
$7^{50}=(7^{25})^2$
$7^{25}=7\times(7^{12})^2$
$7^{12}=(7^6)^2$
$7^6=(7^3)^2$
$7^3=7\times7^2$

and so, using the property that, if $a_1\equiv b_1\mod n$ and $a_2\equiv b_2\mod n$ , then $a_1a_2\equiv b_1b_2\mod n$ , we get

$7\equiv3\mod4$
$7^2\equiv9\equiv1\mod4$
$7^3\equiv7\times1\equiv7\equiv3\mod4$
$7^6\equiv9\equiv1\mod4$
$7^{12}\equiv1\mod4$
$7^{25}\equiv7\times1\equiv7\equiv3\mod4$
$7^{50}\equiv9\equiv1\mod4$
$7^{101}\equiv7\times1\equiv7\equiv3\mod4$
$7^{203}\equiv7\times9\equiv3\times1\equiv3\mod4$

4 0

3 years ago

Heyy I really need help. I don't understand any of this.

mart [117]

Answer:

The parenthesis means to multiply the numbers. I don't understand any of these either but I hope this helps.

Step-by-step explanation:

4 0

3 years ago

8+a=21 plssssssssssssszzzz

attashe74 [19]

Answer:

Step-by-step explanation:

because a is the missing number. Well, 21 minus or - 8 is equal to 13.

8 0

3 years ago