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2 years ago

What is the solution to this inequality 9-3x>=2(x-3)

Mathematics

2 answers:

NeTakaya2 years ago

3 0

$9-3x \geq 2(x-3)$

$9-3x \geq 2x-6$

$9 \geq 5x-6$

$15 \geq 5x$

$3 \geq x$

$\boxed{x \leq 3}$

Arturiano [62]2 years ago

3 0

$9-3x\geq2(x-3)\\
9-3x\geq2x-6\\
-5x\geq-15\\
x\leq3$

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Suppose a and b are the solutions to the quadratic equation 2x^2-3x-6=0. Find the value of (a+2)(b+2).

KonstantinChe [14]

Answer:

(a+2)(b+2) = 4

Step-by-step explanation:

We are given the following quadratic equation:

$2x^2-3x-6=0$

Let a a and b be the solution of the given quadratic equation.

Solving the equation:

$2x^2-3x-6=0\\\text{Using the quadratic formula}\\\\x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\\text{Comparing the equation to }ax^2 + bx + c = 0\\\text{We have}\\a = 2\\b = -3\\c = -6\\x = \dfrac{3\pm \sqrt{9-4(2)(-6)}}{4} = \dfrac{3\pm \sqrt{57}}{4}\\\\a = \dfrac{3+\sqrt{57}}{4}, b = \dfrac{3-\sqrt{57}}{4}$

We have to find the value of (a+2)(b+2).

Putting the values:

$(a+2)(b+2)\\\\=\bigg(\dfrac{3+\sqrt{57}}{4}+2\bigg)\bigg(\dfrac{3-\sqrt{57}}{4}+2\bigg)\\\\=\bigg(\dfrac{11+\sqrt{57}}{4}\bigg)\bigg(\dfrac{11-\sqrt{57}}{4}\bigg)\\\\=\dfrac{121-57}{156} = \dfrac{64}{16} = 4$

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2 years ago

PLZZ HELP find the area of the triangle QRS?

defon

Answer:

180 is the answer

Step-by-step explanation:

Step-by-step explanation:

To find out areas of rectangles, you have to count the left side of the rectangle which is (12) and then count the bottom of the rectangle which is (15)

Then multiply 12 times 15 to find out the area

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2 years ago

Please help with this square problem, the picture is shown.

Galina-37 [17]

Area of the upper rectangle = x(16 - 6) = 10x yd²

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Total area = 10x + 6x + 36 = 16x + 36 yd²

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3 years ago