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seraphim [82]
2 years ago
15

Create a equation where the solution is x= -5

Mathematics
1 answer:
Illusion [34]2 years ago
5 0

x + 3 = 8

OR

4x - 4 = 16

hope this helps!

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PLZ HELP ASAP DENSITY WORD PROBLEMS
yuradex [85]
We know the mass and the volume, using which we can find the density.

Density = Mass/Volume

So,

Density = 16950/1.5 = 11300 kg/m³

So the coins of made of Lead as the density of coins is equal to the density of Lead

So correct answer is option C
5 0
3 years ago
Salma's Coffee Shop makes a blend that is a mixture of two types of coffee. Type A coffee costs Salma $5.15 per pound, and type
son4ous [18]

.........................

7 0
3 years ago
Read 2 more answers
If f(x) = 8x2 + 4x - 12. Find f(2)<br> f(2)=
posledela

Answer:

f(2) = 28

Step-by-step explanation:

\rm f(x) = 8 {x}^{2}  + 4x - 12

\rm Evaluate \:  8 x^2 + 4 x - 12 \:  where  \: x = 2: \\  \\  \rm \implies f(2) = 8 \times {2}^{2}  + 4 \times 2 - 12 \\  \\   \rm \implies f(2) = 8 \times 4 + 8 - 12 \\  \\   \rm \implies f(2) = 32 - 4 \\  \\   \rm \implies f(2) = 28

3 0
3 years ago
Can somebody solve these 2 probelms for me with work?
mel-nik [20]

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Sorry I am in class latet

Step-by-step explanation:

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3 0
3 years ago
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
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