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Gelneren [198K]
3 years ago
7

Suppose H(x) =(6x +6) 3.

Mathematics
1 answer:
Lapatulllka [165]3 years ago
5 0

Answer:

I try to search the answer but I couldn't find it

You might be interested in
Expand using pascal triangle (3x+2)^4
Liula [17]

Answer:

The answer is

<h3>81x⁴ + 216x³ + 216x² + 94x + 16</h3>

Step-by-step explanation:

(3x + 2)⁴

<u>Using the Pascal's triangle</u>

For an exponent of 4 we have

( a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b³

Using a = 3x and b = 2 we have

( 3x +2)⁴ = ( 3x)⁴ + 4 [ (3x)³(2) ] + 6[ (3x)²(2)²] + 4 [ 3x(2)³ ] + 2⁴

Expand and simplify

That's

➢ ( 3x +2)⁴ = 81x⁴ + 4[ (27x³)(2) ] + 6 [ (9x²)(4) ] + 4 [ 3x(8) ] + 16

➢ 81x⁴ 4( 54x³) + 6(36x²) + 4( 24x) + 16

We have the final answer as

<h3>81x⁴ + 216x³ + 216x² + 94x + 16</h3>

Hope this helps you

6 0
4 years ago
What is the surface area of this cube
Sedaia [141]

Answer:

surface area of cube is 6a² where a is side of the cube

6 0
3 years ago
What is the value of x and y if 21x+7y=42 and -5x+5y=10 are equal
nikdorinn [45]

Answer:

-5x/5 +5y/5=10/5=2

y-x=2

y=x+2

21x+7*(x+2)=42

21x+7x+14=42

28x=42-14

28x=28

x=1

y=1+2

y=3

Step-by-step explanation:

7 0
3 years ago
Which reason justifies statement 2?
astraxan [27]

Answer:

B.consecutive interior angles

3 0
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
3 years ago
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