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aliya0001 [1]
3 years ago
8

At a football stadium, 25% of the fans in attendance were teenagers. If there were 190 teenagers at the football stadium, what w

as the total number of people at the stadium?
Mathematics
2 answers:
Goryan [66]3 years ago
7 0
The answer to problem is 760.
Harman [31]3 years ago
5 0
760.......................
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How many solutions does y=3x-12 and y=x-(12-2x) have??
sp2606 [1]

Answer:

Infinite

Step-by-step explanation:

Plug the negative into the equation.

y=x-12+2x

Add like terms

y=3x-12

Two of the same lines will always have Infinite solutions.

4 0
3 years ago
((Look at the image))
Dmitriy789 [7]

Answer:

  a)  y = 2x⁴ -18x² +8x +24

  b)  4th degree; tends toward +∞ when |x| gets large

Step-by-step explanation:

a) When r is a root of the polynomial, (x -r) is a factor. The given roots mean the factors are ...

  y = a(x -(-1))(x -(-3))(x -2)(x -2) = a(x+1)(x+3)(x-2)²

The product of the constant terms is a(1)(3)(2²) = 12a. This is the y-intercept, which is said to be 24. So ...

  12a = 24   ⇒   a = 2

The polynomial function is then ...

  y = 2(x +1)(x +3)(x -2)²

  y = 2x⁴ -18x² +8x +24

__

b) The leading coefficient of this 4th-degree polynomial is positive, so when x has large positive or negative values, the function value will tend toward positive infinity.

4 0
3 years ago
Write the quadratic equation that has roots (√3+1)/2 and (√3 -1 )/2, if its coefficient with x^2 is equal to:
NISA [10]

Answer:

The equation is: x^2-x+\frac{1}{2}

(A)x^2-x+\frac{1}{2}=0

(B)5x^2-5x+\frac{5}{2}=0

(C) -\frac{1}{2}

-\frac{1}{2}x^2+\frac{1}{2}x-\frac{1}{4}=0

d) √3

(\sqrt{3}) x^2-x\sqrt{3}+\frac{\sqrt{3}}{2}=0

Step-by-step explanation:

If the roots of the quadratic equation are: \frac{\sqrt{3}+1}{2}  and  \frac{\sqrt{3}-1}{2}

This means:

x=\frac{\sqrt{3}+1}{2} OR x=  \frac{\sqrt{3}-1}{2}

x-x-\frac{\sqrt{3}+1}{2}=0 \: or\: x- \frac{\sqrt{3}-1}{2}=0=0 or x-

If a=0 or b=0, then ab=0

Therefore:

(x-\frac{\sqrt{3}+1}{2})(x-\frac{\sqrt{3}-1}{2})=0\\x^2-\frac{x(\sqrt{3}-1)}{2}-\frac{x(\sqrt{3}+1)}{2}+(\frac{\sqrt{3}-1)}{2})(\frac{\sqrt{3}+1}{2})=0\\x^2-x+\frac{1}{2}=0

Therefore the coefficient of x^2  here is 1.

b) To make 5 the coefficient of x^2

Simply multiply x^2-x+\frac{1}{2}=0 by 5

This gives: 5x^2-5x+\frac{5}{2}=0

c) -\frac{1}{2}

To make -\frac{1}{2} the coefficient of x^2

Simply multiply x^2-x+\frac{1}{2}=0 by -\frac{1}{2}

This gives: -\frac{1}{2}x^2+\frac{1}{2}x-\frac{1}{4}=0

d) √3

To make  \sqrt{3} the coefficient of x^2

Simply multiply x^2-x+\frac{1}{2}=0 by \sqrt{3}

This gives: (\sqrt{3}) x^2-x\sqrt{3}+\frac{\sqrt{3}}{2}=0

5 0
4 years ago
If <img src="https://tex.z-dn.net/?f=z%3D-1-%5Csqrt%7B3%7Di" id="TexFormula1" title="z=-1-\sqrt{3}i" alt="z=-1-\sqrt{3}i" align=
Luden [163]

Rewriting z in polar form makes this trivial.

z=|z|e^{i\mathrm{arg}(z)}

We have

|z|=\sqrt{(-1)^2+(-\sqrt3)^2}=2

\mathrm{arg}(z)=\tan^{-1}(-1,-\sqrt3)=-\dfrac{2\pi}3

(not to be confused with the standard inverse tangent function \tan^{-1}x. Here \tan^{-1}(x,y) is the inverse tangent function that takes into account position in the coordinate plane; look up "atan2" for more information)

So we have

z=-1-\sqrt3\,i=2e^{-2\pi/3\,i}

Then

z^6=2^6\left(e^{-2\pi/3\,i}\right)^6=64e^{-4\pi\,i}=64

so that a=64 and b=0.

3 0
3 years ago
How many solutions does the nonlinear system of equations graphed below
yan [13]

Answer:

D. zero

Step-by-step explanation:

Since the graphs do not intersect, there are zero solutions.

6 0
3 years ago
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