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IRINA_888 [86]
3 years ago
15

Solve for z. az + 17 = -4z – b.

Mathematics
1 answer:
Harrizon [31]3 years ago
5 0

Answer:

z = -b - 17 / ( a + 4 )

Step-by-step explanation:

az + 17 = -4z – b

collect like terms

az + 4z = -b -17

z( a + 4 ) = - b - 17

Divide both sides by ( a  + 4 )

z = -b - 17 / ( a + 4 )

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What is the area of the composite figure? (6π 4) cm2 (6π 16) cm2 (12π 4) cm2 (12π 16) cm2
raketka [301]

The area of the considered composite figure is given by: Option: 6π + 16 cm²

<h3>How to calculate the surface area of a composite figure?</h3>

Surface area are derived for some standard shapes like circle, triangle, parallelogram, rectangle, trapezoid, etc.

When some shape comes which isn't standard figure, then we find its area by slicing it (virtually, like by drawing lines) in standard shapes. Then we calculate those composing shapes' area and sum them all.

Thus, we have:

\text{Area of composite figure} = \sum (\text{Area of composing figures})

That ∑ sign shows "sum"

For this case, the missing image is attached below.

As visible, the area of the figure = Sum of area of those 3 semi circles + Area of that square(its square because it has 4 equal sides in which adjacent sides are perpendicular) in between.

The side of the square is of 2+2=4 cm( as half of its side is 2 cm), thus, Area of that square = 4² = 16 cm²

All those 3 semicircles has radius = 2 cm

Since radius is same, so area of those 3 semicircle is same = \dfrac{\pi r^2}{2} = \dfrac{\pi (2)^2}{2} = 2 \pi\: \rm cm^2

Thus, we get:

Area of the composite figure = 2 \pi +2 \pi +2 \pi  + 16 \: \rm cm^2 = 6 \pi+ 16 \: \rm cm^2

Learn more about area of a composite figure here:

brainly.com/question/10254615

6 0
3 years ago
What is the square root of 121?<br> A)<br> 9<br> B)<br> 11<br> C)<br> 18<br> D)<br> 22.
sveta [45]

Answer:

11

Step-by-step explanation: 11 times 11 equals 121

4 0
3 years ago
Solve the equation: 2x^3-5x^2-6x+9=0 given that x=1 is a zero
olga_2 [115]

The given function

2x^3+5x^2-6x+9=0

Since x = 1 is one of its zeroes, then we will use it to find the other zeroes

\begin{gathered} x=1 \\ x-1=1-1 \\ x-1=0 \end{gathered}

The factor is x - 1

Use the long division to find the other factors

\frac{2x^3-5x^2-6x+9}{x-1}

Divide 2x^3 by x, then multiply the answer by (x - 1)

\begin{gathered} \frac{2x^3}{x}=2x^2 \\ 2x^2(x-1)=2x^3-2x^2 \end{gathered}

Subtract the product from the original equation

\begin{gathered} 2x^3-5x^2-6x+9-(2x^3-2x^2)= \\ (2x^3-2x^3)+(-5x^2+2x^2)-6x+9= \\ 0-3x^2-6x+9 \\ \frac{2x^3-5x^2-6x+9}{x-1}=2x^2+\frac{-3x^2-6x+9}{x-1} \end{gathered}

Now, divide -3x^2 by x, then multiply the answer by (x - 1)

\begin{gathered} -\frac{3x^2}{x}=-3x \\ -3x(x-1)=-3x^2+3x \end{gathered}

Subtract it from the denominator of the fraction

\begin{gathered} (-3x^2-6x+9)-(-3x^2+3x)= \\ (-3x^2+3x^2)+(-6x-3x)+9= \\ 0-9x+9 \\ \frac{2x^2-5x^2-6x+9}{x-1}=2x^2-3x+\frac{-9x+9}{x-1} \end{gathered}

Divide -9x by x and multiply the answer by (x - 1)

\begin{gathered} \frac{-9x}{x}=-9 \\ -9(x-1)=-9x+9 \end{gathered}

Subtract it from the numerator

\begin{gathered} -9x+9-(-9x+9)= \\ (-9x+9x)+(9-9)= \\ 0+0 \\ \frac{2x^3-5x^2-6x+9}{x-1}=2x^2-3x-9 \end{gathered}

Then we will factor this trinomial into 2 factors

\begin{gathered} 2x^2=(2x)(x) \\ -9=(-3)(3) \\ (2x)(-3)+(x)(3)=-6x+3x=-3x\rightarrow middle\text{ term} \\ 2x^2-3x^2-9=(2x+3)(x-3) \end{gathered}

Equate each factor by 0 to find the other zeroes of the equation

\begin{gathered} 2x+3=0 \\ 2x+3-3=0-3 \\ 2x=-3 \\ \frac{2x}{2}=\frac{-3}{2} \\ x=-\frac{3}{2} \end{gathered}\begin{gathered} x-3=0 \\ x-3+3=0+3 \\ x=3 \end{gathered}

The zeroes of the equations are 1, 3, -3/2

The solutions of the equations are

\begin{gathered} x=-\frac{3}{2} \\ x=1 \\ x=3 \end{gathered}

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Unit price :)  Hope I Helped!! <3
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