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Lemur [1.5K]
4 years ago
11

If the product ab is 0, then either a or b must be

Mathematics
1 answer:
algol [13]4 years ago
8 0
0 because 0 times anything is 0
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What is the arc length of an arc with radius 14 in and central angle 66^o? Show all calculations that lead to the answer and inc
Bas_tet [7]

Answer:

28pi*66/360

Step-by-step explanation

Used the equation Central angle/360*2pi r

5 0
3 years ago
5 million and 300 in standard form
Lera25 [3.4K]

Answer:

Hope it helps.

Step-by-step explanation:

5 million

= 5,000,000

= 5 x 10⁶

300

= 3 x 10²

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4 0
2 years ago
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Anybody good at this?<br> x - ( 9x -10) + 11 = 12x + 3 ( -2 + 1/3)
kenny6666 [7]
X - (9x -10) + 11 = 12x + 3( -2 + 1/3)
x - 9x + 10 + 11 = 12x - 5  ...........................(1/3-2) = -5/3 => x 3 = -5
-8x + 21 = 12x - 5
26 = 20x
x = 13/10
8 0
3 years ago
PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!! PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!! PLSSS HELPPPP I WILLL GIVE YOU B
riadik2000 [5.3K]

Answer:

x=20; y=50 or (20, 50)

Step-by-step explanation:

Substitution means plugging in one variable's value that consists of the opposite variable. Because y=3x-10 is already specified, you can plug it into the second equation's y value. After doing that, it looks like this:

-4x+2y=20→-4x+2(3x-10)=20

Then you would distribute the 3 across the parentheses next to it, like this:

-4x+2(3x-10)=20→-4x+6x-20=20

Then, add like terms, like this:

-4x+6x-20=20→2x=40

Then divide both side by 2 to isolate x.

x=20

Now, you can plug 20 (x) into either equation, but the first one seems simpler so you would pick that. It would look like this:

y=3x-10→y=3(20)-10

Solving would look like this:

y=3(20)-10→y=60-10→y=50

So the answer is x=20; y=50 or (20, 50).

6 0
3 years ago
Read 2 more answers
What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?
Natasha2012 [34]

Answer:

x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}  are zeroes of given quadratic equation.

Step-by-step explanation:

We have been a quadratic equation:

2x^2-10x-3

We need to find the zeroes of quadratic equation

We have a formula to find zeroes of a quadratic equation:

x=\frac{b^2\pm\sqrt{D}}{2a}\text{where}D=\sqrt{b^2-4ac}

General form of quadratic equation is ax^2+bx+c

On comparing general equation with b given equation we get

a=2,b=-10,c=-3

On substituting the values in formula we get

D=\sqrt{(-10)^2-4(2)(-3)}

\Rightarrow D=\sqrt{100+24}=\sqrt{124}

Now substituting D in  x=\frac{b^2\pm\sqrt{D}}{2a} we get

x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}

x=\frac{100\pm\sqrt{124}}{4}

x=\frac{100\pm2\sqrt{31}}{4}

x=\frac{50\pm\sqrt{31}}{2}

Therefore, x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}



5 0
3 years ago
Read 2 more answers
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