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3 years ago

PLSSSSS HELP PLS QUICKLY PLS

Mathematics

2 answers:

masya89 [10]3 years ago

8 0

Answer:

-64

Step-by-step explanation:

we should put -2 for j and -1 for k,

NeTakaya3 years ago

3 0

Answer:

the answer is 64

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22-[(87-32)/5]x2
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3 years ago

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1 year ago

I need help on number 6

taurus [48]

Answer:

x = 0

Step-by-step explanation:

First we set up our equation. From the picture we can see that:

11 + x + 1 = 2x + 12

Now we can gather our like terms:

x + 12 = 2x + 12

Lets subtract x from both sides to isolate the x on the left:

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Check the answer

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3 years ago

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likoan [24]

Answer:

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Step-by-step explanation:

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3 years ago

Perform the indicated operations. Write the answer in standard form, a+bi.<br> 5-3i / -2-9i

Vsevolod [243]

$\huge \boxed{\mathfrak{Answer} \downarrow}$

$\large \bf\frac { 5 - 3 i } { - 2 - 9 i } \\$

Multiply both numerator and denominator of $\sf \frac{5-3i}{-2-9i} \\$ by the complex conjugate of the denominator, -2+9i.

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2-9i\right)\left(-2+9i\right)}) \\$

Multiplication can be transformed into difference of squares using the rule: $\sf\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$ .

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2\right)^{2}-9^{2}i^{2}}) \\$

By definition, i² is -1. Calculate the denominator.

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{85}) \\$

Multiply complex numbers 5-3i and -2+9i in the same way as you multiply binomials.

$\large \bf \: Re(\frac{5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9i^{2}}{85}) \\$

Do the multiplications in $\sf5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9\left(-1\right)$ .

$\large \bf \: Re(\frac{-10+45i+6i+27}{85}) \\$

Combine the real and imaginary parts in -10+45i+6i+27.

$\large \bf \: Re(\frac{-10+27+\left(45+6\right)i}{85}) \\$

Do the additions in $\sf-10+27+\left(45+6\right)i$ .

$\large \bf Re(\frac{17+51i}{85}) \\$

Divide 17+51i by 85 to get $\sf\frac{1}{5}+\frac{3}{5}i \\$ .

$\large \bf \: Re(\frac{1}{5}+\frac{3}{5}i) \\$

The real part of $\sf \frac{1}{5}+\frac{3}{5}i \\$ is $\sf \frac{1}{5} \\$ .

$\large \boxed{\bf\frac{1}{5} = 0.2} \\$

3 0

2 years ago