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DiKsa [7]
4 years ago
6

Rules for multiplying and dividing by 1 and 0

Mathematics
2 answers:
liubo4ka [24]4 years ago
7 0
Mutiplying: the product is always 0
natita [175]4 years ago
3 0
Multiplying anything by 1 is always the number your multiplying it by so for example 1*2=2
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HELP PLS :(
Kipish [7]

Answer:

I believe it's linear

Step-by-step explanation:

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3 years ago
Write an algebraic expression for each person’s share, if P people share 16 slices of pizza equally.
svp [43]

Answer:

P /16

Step-by-step explanation:

Take the total amount, P and divide by the number of people sharing, 16

P /16

3 0
3 years ago
8 1/2+6 1/2=<br><br><br><br><br><br> ..........NO LINK
LuckyWell [14K]

Answer:

15

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Polynomial Equation question<br>(Screenshot included)
Illusion [34]

y = {x}^{3}  - 4 {x}^{2}  - 45x \\ y = x( {x}^{2}  - 4x - 45) \\ y = x(x + 5)(x - 9) \\ x = 0, - 5,9
So, to find our x-intercepts, y must be zero
0 = x \\  \\ 0 = (x + 5) \\  - 5 = x \\  \\ 0 = (x - 9) \\ 9 = x
ANSWER: (0,0) (-5.0) (9,0) are your x-intercepts
6 0
4 years ago
What is the result of substituting for y in the bottom equation<br> y=x+3<br> y=x^2+2x-4
8_murik_8 [283]

The solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

<em><u>Solution:</u></em>

Given that,

y = x + 3 ------- eqn 1\\\\y = x^2 + 2x - 4  ----- eqn 2

<em><u>We have to substitute eqn 1 in eqn 2</u></em>

x + 3 = x^2 + 2x - 4

\mathrm{Switch\:sides}\\\\x^2+2x-4=x+3\\\\\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}\\\\x^2+2x-4-3=x+3-3\\\\\mathrm{Simplify}\\\\x^2+2x-7=x\\\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\\\x^2+2x-7-x=x-x\\\\\mathrm{Simplify}\\\\x^2+x-7=0

\mathrm{Solve\:with\:the\:quadratic\:formula}\\\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=1,\:c=-7\\\\x =\frac{-1\pm \sqrt{1^2-4\cdot \:1\left(-7\right)}}{2\cdot \:1}

x = \frac{-1 \pm \sqrt{ 1 + 28}}{2}\\\\x = \frac{ -1 \pm \sqrt{29}}{2}

x = \frac{ -1 \pm 5.385 }{2}\\\\We\ have\ two\ solutions\\\\x = \frac{ -1 + 5.385 }{2}\\\\x = 2.1925

Also\\\\x = \frac{ -1 - 5.385 }{2}\\\\x = -3.1925

Substitute x = 2.1925 in eqn 1

y = 2.1925 + 3

y = 5.1925

Substitute x = -3.1925 in eqn 1

y = -3.1925 + 3

y = -0.1925

Thus the solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

6 0
4 years ago
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