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3 years ago

–5(–2) = please help :) :D

Mathematics

2 answers:

ioda3 years ago

3 0

Answer:

Step-by-step explanation:

-5 * -2 = 10

cuz - * - = +

and 5 * 2 = 10

hence 10

solniwko [45]3 years ago

3 0

Answer:

-7

Step-by-step explanation:

-5(-2)

-5-2

-7

..........

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What is the amplitude, period, and phase shift of f(x) = -4 sin(2x + π) - 5?

son4ous [18]

F(x) = -4sin(2x + π) - 5

Amplitude
A = -π

Period
2π = 2π = π
B 2

Phase Shift
-C = -π = ≈ 1.57
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3 years ago

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7f + g = 5, for f literal equations

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Answer:

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Step-by-step explanation:

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3 years ago

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Answer:

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3 years ago

Find thhe remainder when 7^203 is divided by 4

Aleksandr [31]

Using the square-and-multiply approach, we have

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

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

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

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3 years ago