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2 years ago

Express x^2+4x-7 in the form (x+a)^2-b where a and b are integers

Mathematics

1 answer:

lukranit [14]2 years ago

4 0

Answer:

(x + 2)^2 - 11.

Step-by-step explanation:

f(x) = x^2 + 4x - 7

Now x^2 + 4x = (x + 2)^2 - 4 so we have:

f(x) = (x + 2)^2 - 4 - 7

f(x) = (x + 2)^2 - 11.

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lyudmila [28]

Answer:

x = 24.

r $\ne$ 0.

Step-by-step explanation:

2. The given equation is:

$\frac{1}{2} (x - 4) = \frac{1}{3} x + 2$

a) To eliminate the fractions multiply the equation throughout by the LCM of the denominators of the fraction. In this case, the LCM of (2, 3). The LCM is 6. So, multiply the entire equation by 6.

b) Half of the difference between an integer and 4 equals the sum of one - third of the integer and 2. Find the integer.

c) We have the equation:

$\frac{1}{2} (x - 4) = \frac{1}{3} x + 2$

Multiplying throughout by 6, we get:

$\frac{6}{2}(x - 4) = \frac{6}{3} x + 6(2)$

$\implies 3(x - 4) = 2x + 12$

$\implies 3x - 12 = 2x + 12$

$\implies x = 24$

Therefore, the solution of the equation is 24.

3. The given equation is: $ry + s = tx - m$

To solve for y:

We can rearrange the equation as:

$ry = tx - m - s$

$\implies y = \frac{tx - m - s}{r}$

or, $y = \frac{tx - (m + s)}{r}$

Note that we have to impose a condition on variable $r$ . It would be that $r$ can never be zero. i.e., $r \ne 0$ . Otherwise, the value of $y$ would be undefined.

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2 years ago

A 2-column table with 7 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2, 3. The sec

DochEvi [55]

The statements that are true about the intervals of the continuous function are Options 2, 4 and 5

f(x) ≤ 0 over the interval [0, 2].
f(x) > 0 over the interval (–2, 0).
f(x) ≥ 0 over the interval [2, ).

<h3>What is the statement about?</h3>

Looking at the values given, the intervals which satisfies the condition are known to be:

f(x)<=0 over the interval [0,2]

f(x)>0 over the interval (-2,0)

f(x)>=0 over the interval [2,∞)

Because:

Since the table with x and f(x) values, we have to examine analyze the table and see the each option that is in line with f(x) or not .

Examine the values of x that is from -3 to 3, the f(x) values are both positive and negative . hence f(x)>0 is false over the interval (-∞,3)

Looking at the the interval from 0 to 2, the f(x) values are 0 and negative. Hence, f(x)<=0 over the interval [0,2]

When you look over the interval (-1,1), the f(x) values are said to be both positive and negative and as such, f(x)<0 is false over the interval (-1,1)

When you look at the interval (-2,0) , the f(x) is positive and as such, f(x)>0 over the interval (-2,0)

Looking at the interval [2,∞), f(x) is positive and as such, f(x)>=0 over the interval [2,∞)

Therefore, Option 2, 4 and 5 are correct.

Learn more about interval from

brainly.com/question/14454639

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