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2 years ago

Jace has nn nickels and dd dimes. He has at most 20 coins altogether. Write this situation as an inequality.

Mathematics

1 answer:

Marina CMI [18]2 years ago

5 0

Answer:

nn + dd ≤ 20

≤ = at most or

less than or = to

Step-by-step explanation:

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I NEED PART B AND C DONE I DID PART A, PLEASE HELP AND TELL ME WHAT PART YOU ARE SOLVING ALSO LOTS OF POINTS!!!

RideAnS [48]

Answer:

(8,10) is not a solution

Step-by-step explanation:

Cuz it’s not on both inequality shaded part but just one of them and also when we substitute one is true and the other one is false

2x + y _< 8

2(8) + 10 _< 8

18 + 10 _< 8

28 _< 8 false

X + y _> 4

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18 _> 4 true

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3 years ago

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Please help me! And can you please show how you found the answer?

Pani-rosa [81]

Answer:

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Step-by-step explanation:

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3 years ago

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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

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