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3 years ago

Solve -12 = z ÷ 3 for z. z =

Mathematics

2 answers:

iren2701 [21]3 years ago

8 0

Answer:

-36

Step-by-step explanation:

mash [69]3 years ago

6 0

Answer: z = −36

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

-12 = 1/3z

Step 2: Flip the equation.

1/3x = -12

Step 3: Multiply both sides by 3.

3* (1/3z) = (3) * (-12)

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F(x)=2x+3/4x+5<br>find f(-9)<br>

mylen [45]

$\implies {\blue {\boxed {\boxed {\purple {\sf { f(-9)= 0.48}}}}}}$

$\large\mathfrak{{\pmb{\underline{\orange{Step-by-step\:explanation}}{\orange{:}}}}}$

$f(x) = \frac{2x + 3}{4x + 5} \\$

For $f(-9)$ , put " $-9$ " for every value of " $x$ ".

$↬f( - 9) = \frac{2( - 9) + 3}{4( - 9) + 5}\\$

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$\bold{ \green{ \star{ \red{Mystique35}}}}⋆$

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3 years ago

Read 2 more answers

Find thhe remainder when 7^203 is divided by 4

Aleksandr [31]

Using the square-and-multiply approach, we have

$7^{203}=7\times(7^{101})^2$
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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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