All categories

3 years ago

13.05 Completing the Square:

1 answer:

6 0

The way these equations are arranged helps.
First only deal with the left handside.

X^2+20x= (x+a)^2 -b where we are looking for a,b
You need to know the general formula for (x+a)^2=x^2+2ax+a^2
20x = 2*10x so a will be 10
(x+10)^2= x^2+20x+100

Now lets look at the original question:
X^2+20x=10
we know now that (x+10)^2= x^2+20x+100
Thus X^2+20x=(x+10)^2 -100
So X^2+20x=10 becomes

(x+10)^2 -100 =10
(x+10)^2=110
For the other question you need to follow the same logic.

You might be interested in

Evaluate the expression for x = 2 and y = 4. Enter your answer in the box. 16x0 + 2x2 • y –1

swat32

Given that,

The given expression is : $16x^0+2x^2{\cdot}y^{-1}$

To find,

The value of the above expression when x = 2 and y = 4

Solution,

We have,

$16x^0+2x^2{\cdot}y^{-1}$

Put x = 2 and y = 4 in the above expression.

$16x^0+2x^2{\cdot}y^{-1}\\\\=16\times (2)^0+2(2)^2\times (4)^{-1}\\\\=16\times 1+2(2)^2\times\dfrac{1}{4}\\\\=16+2\\\\=18$

So, the value of the above expression is 18.

6 0

2 years ago

Matt gave 5 apples to his son and sold 8 times as many apples. If Matt had 1/2 of the apples left, how many did he have in the b

Gala2k [10]

Answer:

Step-by-step explanation:

8*5 = 40

40 + 5 = 45

1/2 x = 45

x = 90

4 0

2 years ago

Read 2 more answers

Is the number a rational number? Help me out please!

krek1111 [17]

Answer:

first answer

Step-by-step explanation:

Hi, this is correct because repeating decimals are rational numbers.

Rational numbers can be written by a fraction of integers.

If there is a pattern, it means that this is the division of two integers.

hope this helps

8 0

2 years ago

Read 2 more answers

I am so stuck plz help me simplify these

pishuonlain [190]

Answer:

2b + 3c

3d + 2e

14b

10d + 21e

7m + 14n

12p + 18q

13a + 11b + 10c

4 0

2 years ago

Read 2 more answers

If a = √3-√11 and b = 1 /a, then find a² - b²

Serga [27]

If $b=\frac1a$ , then by rationalizing the denominator we can rewrite

$b = \dfrac1{\sqrt3-\sqrt{11}} \times \dfrac{\sqrt3+\sqrt{11}}{\sqrt3+\sqrt{11}} = \dfrac{\sqrt3+\sqrt{11}}{\left(\sqrt3\right)^2-\left(\sqrt{11}\right)^2} = -\dfrac{\sqrt3+\sqrt{11}}8$

Now,

$a^2 - b^2 = (a-b) (a+b)$

and

$a - b = \sqrt3 - \sqrt{11} + \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{9\sqrt3 - 7\sqrt{11}}8$

$a + b = \sqrt3 - \sqrt{11} - \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{7\sqrt3 - 9\sqrt{11}}8$

$\implies a^2 - b^2 = \dfrac{\left(9\sqrt3 - 7\sqrt{11}\right) \left(7\sqrt3 - 9\sqrt{11}\right)}{64} = \boxed{\dfrac{441 - 65\sqrt{33}}{32}}$

5 0

1 year ago