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kykrilka [37]
3 years ago
9

Look the the question below

Mathematics
1 answer:
RideAnS [48]3 years ago
6 0

Answer:

Step-by-step explanation:

Since 3 is odd he will lose

-10(3)=-30

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If Alicia can read 3 pages in 1 minute, how long would it take her to read 33 pages?
Bond [772]

divide total pages by pages per minute

33 / 3 = 11 minutes

7 0
3 years ago
2. Andrew made an error in determining the polynomial equation of smallest degree whose roots are 3, 2+2i and 2-2i.
inna [77]

The polynomial equation for the given roots is x³-7x²+20x-24=0.

Given that, the polynomial equation of the smallest degree whose roots are 3, 2+2i and 2-2i.

Now, we need to write the polynomial equation.

<h3>What is a polynomial equation? </h3>

A polynomial equation is an equation where a polynomial is set equal to zero. i.e., it is an equation formed with variables, non-negative integer exponents, and coefficients together with operations and an equal sign. It has different exponents. The highest one gives the degree of the equation. For an equation to be a polynomial equation, the variable in it should have only non-negative integer exponents. i.e., the exponents of variables should be only non-negative and they should neither be negative nor be fractions.

Now, x=3, x=2+2i and x=2-2i

x-3, x-2-2i and x-2+2i

So, the polynomial equation is (x-3)(x-2-2i)(x-2+2i)=0

⇒(x-3)(x(x-2-2i)-2(x-2-2i)+2i(x-2-2i))=0

⇒(x-3)(x²-2x-2xi-2x+4+4i+2xi-4i-4i²)=0

⇒(x-3)(x²-2x-2x+4-4i²)=0

⇒(x-3)(x²-4x+4+4)=0 (∵i²=-1)

⇒(x-3)(x²-4x+8)=0

⇒x(x²-4x+8)-3(x²-4x+8)=0

⇒x³-4x²+8x-3x²+12x-24=0

⇒x³-7x²+20x-24=0

Therefore, the polynomial equation for the given roots is x³-7x²+20x-24=0.

To learn more about the polynomial equation visit:

brainly.com/question/20630027.

#SPJ1

3 0
2 years ago
Determine x in the following equation 2x - 4 = 10
kari74 [83]

Answer:

7

Step-by-step explanation:

10+4 = 14

14/2  = 7

x = 7

7 0
3 years ago
Nuance said she drew a square pyramid and that all of the faces are triangles.Is this possible?
sergeinik [125]
No because a square pyramid must hav a square for its base and a square is not a triangle

answer is not possible
3 0
3 years ago
Read 2 more answers
I am stuck at this question​
Marrrta [24]

Answer:

Step-by-step explanation:

B(2,10); D(6,2)

Midpoint(x1+x2/2, y1+y2/2) = M ( 2+6/2, 10+2/2) = M(8/2, 12/2) = M(4,6)

Rhombus all sides are equal.

AB = BC = CD =AD

distance = √(x2-x1)² + (y2- y1)²

As A lies on x-axis, it y-co ordinate = 0; Let its x-co ordinate be x

A(X,0)

AB = AD

√(2-x)² + (10-0)² = √(6-x)² + (2-0)²

√(2-x)² + (10)² =  √(6-x)² + (2)²

√x² -4x +4 + 100 =  √x²-12x+36 + 4

√x² -4x + 104 =  √x²-12x+40

square both sides,

x² -4x + 104 =  x²-12x+40

x² -4x - x²+ 12x = 40 - 104

             8x = -64

               x = -64/8

               x = -8

A(-8,0)

Let C(a,b)

M is AC midpoint

(-8+a/2, 0 + b/2)  = M(4,6)

     (-8+a/2, b/2)  = M(4,6)

Comparing;  

-8+a/2 = 4          ; b/2 = 6

  -8+a = 4*2       ; b = 6*2

  -8+a = 8          ; b = 12

        a = 8 +8

       a = 16

Hence, C(16,12)

6 0
3 years ago
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