All categories

3 years ago

One of the roots of the quadratic equation x^2−5mx+6m^2=0 is 36. Find the greatest possible value of the second root. 100PTS!!!

Mathematics

2 answers:

rosijanka [135]3 years ago

7 0

Answer:

The greatest possible value is 54

Step-by-step explanation:

Solve the quadratic equation

Given

x² - 5mx + 6m² = 0

We can rewrite this as

x² - 3mx - 2mx + 6m² = 0

(x² - 3mx) - (2mx - 6m²) = 0

x(x - 3m) - 2m(x - 3m) = 0

(x - 2m)(x - 3m) = 0

x - 2m = 0 or x - 3m = 0

So,

x = 2m or x = 3m .

2m and 3m are the roots of the equation.

Since one of the roots is 36

Assume

2m = 36

m = 36/2 = 18

3m is

3(18) = 54

If 3m = 36

m = 12

And

2m = 2(12) = 24.

The greatest possible value is 54

Maru [420]3 years ago

6 0

Answer:

Step-by-step explanation:

You might be interested in

4y-5x=-8

Dovator [93]

Answer:

4y−5x=−8

Step 1: Add -4y to both sides.

−5x+4y+−4y=−8+−4y

−5x=−4y−8

Step 2: Divide both sides by -5.

−5x

−5

−4y−8

−5

Answer:

Step-by-step explanation:

3 0

2 years ago

Find the limit (enter 'DNE' if the limit does not exist)

vaieri [72.5K]

Answer:

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\sqrt{-2x^2-6y^2+1}+1\right)=2$

Step-by-step explanation:

We need to first simplify the expression using rationalization(i.e. if a square root term exists in the denominator, then multiply and divide the whole expression by the denominator(but the change the sign of its middle term))

here, we need to find:

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\dfrac{-2x^2-6y^2}{\sqrt{-2x^2-6y^2+1}-1}\right)$

first we'll rationalize our expression:

$\dfrac{-2x^2-6y^2}{\sqrt{-2x^2-6y^2+1}-1}\left(\dfrac{\sqrt{-2x^2-6y^2+1}+1}{\sqrt{-2x^2-6y^2+1}+1}\right)$

$\dfrac{-(2x^2+6y^2)(\sqrt{-2x^2-6y^2+1}+1)}{(\sqrt{-2x^2-6y^2+1}+1)^2-(1)^2}$

$\dfrac{-(2x^2+6y^2)(\sqrt{-2x^2-6y^2+1}+1)}{-2x^2-6y^2+1-1}$

$\dfrac{-(2x^2+6y^2)(\sqrt{-2x^2-6y^2+1}+1)}{-(2x^2+6y^2)}$

$\sqrt{-2x^2-6y^2+1}+1$

this is our simplified expression, now we can apply our limits:

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\sqrt{-2x^2-6y^2+1}+1\right)$

$\sqrt{-2(0)^2-6(0)^2+1}+1$

$1+1$

$2$

the limit does exists and it is 2.

5 0

2 years ago

Write and solve a proportion to complete the statement. Round to the nearest hundredth if necessary.

wariber [46]

Answer:

Step-by-step explanation:

<h3>Convert kilometer to miles:</h3>

1 km = 0.62 mi

So, to convert 6 km to mi, multiply 6 by 0.62.

6km = 6* 0.62

= 3.72 mi

<h3>Convert liter to gallon:</h3>

1 L = 0.26 gal

To convert 2.5 L to gallon, multiply 2.5 by 0.26.

2.5 L = 2.5 * 0.26

= 0.65 gal

<h3>Convert pound to kilogram:</h3>

1 lb = 0.45 Kg

To convert 90 lb to Kg, multiply 90 by 0.45.

90 lb = 90 *0.45

= 40.5 Kg

3 0

1 year ago

Help pleaseeeeeeeeeeee

Lady bird [3.3K]

Answer:

Step-by-step explanation:

Charlene has 8 more dollars so you would have to add it all together to get 86.

8 0

3 years ago

A cylinder with 3 radius and and height is 2 inches what is the volume?

VashaNatasha [74]

Answer:

$\huge\boxed{\sf V = 56.5 \ in.^3}$

Step-by-step explanation:

Radius = r = 3 in.

Height = h = 2 in.

<u>Volume:</u>

$\sf Volume \ of \ the \ cylinder = \pi r^2 h\\\\V = (3.14)(3)^2(2)\\\\V =(3.14)(9)(2)\\\\V = (3.14)(18)\\\\V = 56.5 \ in.^3\\\\\rule[225]{225}{2}$

3 0

1 year ago