There are two numbers. Their GCF is 5. Their LCM is 150. The first number is 25. What is the second number?

Mathematics

2 answers:

Galina-37 [17]3 years ago

7 0

Answer:

Step-by-step explanation:

svp [43]3 years ago

5 0

Answer:

if 150 is a multiple, then all of the factors of the second number must also be factors of 150 which are: 5 (twice), 2 and 3.

Since 150 is the LCM, all of those factors must be either in our number or 25. Since 25 has just 5 (twice) as a factor, both 2 and 3 must be factors.

The remaining question is how many 5s we have. Since the GCF with 25 is 5, our number must exactly one 5.

So the factors of our number are: 2, 3, and 5. There is one number with exactly these factors = 2*3*5 = 30.

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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:

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