Five thousand three hundred nine divided by forty three

How do I solve it and what is the answer plz help by tomorrow morning

Rina8888 [55]

Answer:

$-7, -2, 3, 11$

Step-by-step explanation:

Negative numbers are "smaller" than positive numbers. Just arrange them.

5 0

3 years ago

What is the value of b for the following circle in general form? x2 + y2 + ax +by+c=0

liberstina [14]

Answer:

Step-by-step explanation:

(1). (x - h)² + (y - k)² = r²

where (h, k) are coordinates of a center of the circle with radius r

(2). ( a ± b )² = a² ± 2ab + b²

~~~~~~~~~~~~~~

( 1 , - 2 ) ; r = 3

( x - 1 )² + ( y + 2 )² = 3²

x² - 2x + 1 + y² + 4y + 4 = 9

x² + y² - 2x + 4y - 4 = 0

a = - 2

b = 4

c = - 4

7 0

2 years ago

Which shape has no parallel sides? (2 points)

vazorg [7]

Answer: a quadrilateral with sides of different lengths.

Parallel sides are marked by arrows (single or double) btw

4 0

3 years ago

Read 2 more answers

Tickets to a local movie theater were sold at $6.00 for adults and $4.50 for students. There were 240 tickets sold for a total o

andreev551 [17]

Answer: State that y = 430-x, and substitute it into the second equation 6x +4.5(430-x)= 2175.

Solve for x. Then plug in what you get for x into the x+y=430 equation and solve for y. Your number of adult tickets should equal 160, and your number of child tickets should equal 270.

Step-by-step explanation:

6 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

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

7 0

1 year ago