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3 years ago

2x -3< -1 or -3x < -15

Mathematics

1 answer:

pogonyaev3 years ago

4 0

Answer:

x<1 or x>5

Step-by-step explanation:

2x -3< -1 or -3x < -15

Solve them separately then put them back together

2x-3 < -1

Add 3 to each side

2x-3+3<-1+3

2x<2

Divide by 2

2x/2 <2/2

x<1

-3x < -15

Divide each side by -3, remembering to flip the inequality

-3x/3 > -15/-3

x>5

Put them back together

x<1 or x>5

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Step-by-step explanation:

Let total no.of classmates who liked the jewellery = x

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3 years ago

Which two values of x are roots of the polynomial below? x2-11x + 15

Alex787 [66]

The two values of roots of the polynomial $x^{2}-11 x+15$ are $\frac{11+\sqrt{61}}{2} \text { or } \frac{11-\sqrt{61}}{2}$

Solution:

Given, polynomial expression is $x^{2}-11 x+15$

We have to find the roots of the given expression.

In order to find roots, now let us use quadratic formula.

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Given that $x^{2}-11 x+15$

Here a = 1, b = -11 and c = 15

On substituting the values we get,

$x=\frac{-(-11) \pm \sqrt{(-11)^{2}-4 \times 1 \times 15}}{2 \times 1}$

$\begin{array}{l}{x=\frac{11 \pm \sqrt{121-60}}{2}} \\\\ {x=\frac{11 \pm \sqrt{61}}{2}} \\\\ {x=\frac{11+\sqrt{61}}{2} \text { or } \frac{11-\sqrt{61}}{2}}\end{array}$

Hence, the roots of given polynomial are $\frac{11+\sqrt{61}}{2} \text { or } \frac{11-\sqrt{61}}{2}$

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3 years ago

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All you should have to do is take 180-95 to get 85 for x & then take 85+45= 130 - 180 = 50.

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