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3 years ago

-cz+6z= tz+83 solve for z

Mathematics

1 answer:

jenyasd209 [6]3 years ago

7 0

Answer:

z = 83 /( -c+6-t)

Step-by-step explanation:

-cz+6z= tz+83

Subtract tz from each side

-cz+6z -tz= tz-tz+83

-cz+6z- tz=83

Factor out a z

z( -c+6-t) = 83

Divide by ( -c+6-t)

z( -c+6-t)/( -c+6-t) = 83 /( -c+6-t)

z = 83 /( -c+6-t)

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M is a degree 3 polynomial with m ( 0 ) = 53.12 and zeros − 4 and 4 i . Find an equation for m with only real coefficients (i.E.

Nitella [24]

Answer:

Therefore the required polynomial is

M(x)=0.83(x³+4x²+16x+64)

Step-by-step explanation:

Given that M is a polynomial of degree 3.

So, it has three zeros.

Let the polynomial be

M(x) =a(x-p)(x-q)(x-r)

The two zeros of the polynomial are -4 and 4i.

Since 4i is a complex number. Then the conjugate of 4i is also a zero of the polynomial i.e -4i.

Then,

M(x)= a{x-(-4)}(x-4i){x-(-4i)}

=a(x+4)(x-4i)(x+4i)

=a(x+4){x²-(4i)²} [ applying the formula (a+b)(a-b)=a²-b²]

=a(x+4)(x²-16i²)

=a(x+4)(x²+16) [∵i² = -1]

=a(x³+4x²+16x+64)

Again given that M(0)= 53.12 . Putting x=0 in the polynomial

53.12 =a(0+4.0+16.0+64)

$\Rightarrow a = \frac{53.12}{64}$

=0.83

Therefore the required polynomial is

M(x)=0.83(x³+4x²+16x+64)

5 0

3 years ago

Solve the system of linear equations.

sweet-ann [11.9K]

Answer:

dependent system
x = 2 -a
y = 1 +a
z = a

Step-by-step explanation:

Let's solve this by eliminating z, then we'll go from there.

Add 6 times the second equation to the first.

(3x -3y +6z) +6(x +2y -z) = (3) +6(4)

9x +9y = 27 . . . simplify

x + y = 3 . . . . . . divide by 9 [eq4]

Add 13 times the second equation to the third.

(5x -8y +13z) +13(x +2y -z) = (2) +13(4)

18x +18y = 54

x + y = 3 . . . . . . divide by 18 [eq5]

Equations [eq4] and [eq5] are identical. This tells us the system is dependent, and has an infinite number of solutions. We can find them in terms of z:

y = 3 -x . . . . solve eq5 for y

x +2(3 -x) -z = 4 . . . . substitute into the second equation

-x +6 -z = 4

x = 2 - z . . . . . . add x-4

y = 3 -(2 -z)

y = z +1

So far, we have written the solutions in terms of z. If we use the parameter "a", we can write the solutions as ...

x = 2 -a

y = 1 +a

z = a

_____

<em>Check</em>

First equation:

3(2-a) -3(a+1) +6a = 3

6 -3a -3a -3 +6a = 3 . . . true

Second equation:

(2-a) +2(a+1) -a = 4

2 -a +2a +2 -a = 4 . . . true

Third equation:

5(2-a) -8(a+1) +13a = 2

10 -5a -8a -8 +13a = 2 . . . true

Our solution checks algebraically.

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2 years ago

Help me? idk the answer :P

Alexus [3.1K]

$\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}\\\\
-------------------------------\\\\
\left( 2^8\cdot 3^{-5}\cdot 6^0 \right)^{-2}\left( \cfrac{3^{-2}}{2^3} \right)^4\cdot 2^{28}\impliedby \textit{let's do the first group}
\\\\
-------------------------------\\\\
$

$\bf \left( 2^8\cdot \cfrac{1}{3^5}\cdot 1 \right)^{-2}\implies \left( \cfrac{2^8}{3^5} \right)^{-2}\implies \left( \cfrac{3^5}{2^8} \right)^{2}\implies \cfrac{3^{2\cdot 5}}{2^{2\cdot 8}}\implies \boxed{\cfrac{3^{10}}{2^{16}}}\\\\
-------------------------------\\\\
\textit{now the second group}\qquad \left( \cfrac{3^{-2}}{2^3} \right)^4\implies \left( \cfrac{\frac{1}{3^2}}{2^3} \right)^4\implies \left( \cfrac{1}{2^3\cdot 3^2} \right)^4$

$\bf \cfrac{1^4}{2^{4\cdot 3}\cdot 3^{4\cdot 2}}\implies \boxed{\cfrac{1}{2^{12}\cdot 3^8}}\\\\
-------------------------------\\\\
\textit{so we end up with\qquad }\cfrac{3^{10}}{2^{16}}\cdot \cfrac{1}{2^{12}\cdot 3^8}\cdot 2^{28}\implies \cfrac{3^{10}\cdot 2^{28}}{2^{16}\cdot 2^{12}\cdot 3^8}
\\\\\\
\cfrac{3^{10}\cdot 2^{28}}{2^{16+12}\cdot 3^8}\implies \cfrac{3^{10}\cdot 2^{28}}{2^{28}\cdot 3^8}\implies 3^{10}\cdot 2^{28}\cdot 2^{-28}\cdot 3^{-8}$

$\bf 3^{10-8}\cdot 2^{28-28}\implies 3^2\cdot 1\implies 3^2\implies 9$

7 0

2 years ago

41°F equals how many degrees Celsius?

GrogVix [38]

<span>So we wan't to know hot to convert degrees Fahrenheit to degrees Celsius. Let's transform first from Celsius to Fahrenheit: 0 degrees Celsius is 32 degrees Fahrenhiet and 1.8 degrees Celsius is 1 degree Fahrenheit so: T(F)=1.8T(C) + 32. To transform from Celsius to Fahrenheit we need to invert the former equation: 1.8T(C)= T(F)-32. Now we divide the whole equation with 1.8 and we get: T(C)=(T(F) - 32)/1.8 and input our T(F)=41 F and we get: T(C) =(41-32)/1.8 and that is equal to: T(C)= 5 C. So the correct answer is B. 5 degrees C</span>

3 0

3 years ago

Read 2 more answers

5. Graph the given system of linear inequalities on the coordinate plane below. (2 points)sy <-3x - 4< 21-15NA11 2 3 4 5 6

xxTIMURxx [149]

ANSWERS

(5) Graph:

(6) Solutions: (-1, -5) and (-2, -7)

Not solutions: (1, 1) and (-4, -3)

EXPLANATION

(5) To graph this system, first, we have to graph each of the inequalities. Both are linear, so we have to graph each line with a dashed line - this is because they are "less than" the line, so the line is not included, and shade the area below each of them.

The first inequality is y < -3x-4,

The second inequality is y < 2x - 1,

The graph of the system is the area shared by both graphs,

(6) Any point that is inside the shaded area is a solution of the system. For example, points (-1, -5) and (-2,-7) are solutions to the system.

On the other hand, any point outside that area, or on the dashed lines is not a solution. For example, points (1, 1) and (-4, -3) are not solutions

8 0

10 months ago