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2 years ago

Find the difference of the two functions

Mathematics

2 answers:

xeze [42]2 years ago

5 0

Answer:

Fourth option is the correct answer

$(f - g)(x) = 3 {x}^{2} + 6x - 3$

Step-by-step explanation:

$f(x) = 3{x}^{2} + 7x \\ g(x) = x + 3 \\ \\ (f - g)(x) = f(x) - g(x) \\ = 3{x}^{2} + 7x - (x + 3) \\ = 3{x}^{2} + 7x - x - 3 \\ \purple{ \bold{ (f - g)(x) = 3 {x}^{2} + 6x - 3 }}$

jarptica [38.1K]2 years ago

3 0

Answer:

I thought fourth option because there is only one square

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A drought effects 9 out of 12

masya89 [10]

Answer:

You just have to divide 9 by 12 to get your answer of 75%

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3 years ago

Read 2 more answers

Sandy is driving cross country. She has covered 200 miles from the east coast towards the west coast. She has driven this distan

lilavasa [31]

Okay so here it is

the answer is M=100D + 200

Why?

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3 years ago

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Mamont248 [21]

Answer:

I love this app welcome to the club

8 0

3 years ago

Verify the identity<br><br> cos quantity x plus pi divided by two = -sin x

In-s [12.5K]

Answer:

Identity is verified.

Step-by-step explanation:

We have to verify the identity $cos(x+\frac{\pi }{2}) = (- sinx)$

To prove any identity we always prove one side(either left hand side or right hand side) of the equation equal to the other side.

In this identity we take the left hand side first

$cos(x+\frac{\pi}{2})$

$=cosx\times cos(\pi/2)-sinx\times sin(\pi/2))$ (as we know cos(a+b) = cosa×cosb-sina×sinb)

$= cosx\times0-sinx\times1$

$= 0-sinx$

= - sinx ( Right hand side)

Hence identity is proved.

6 0

3 years ago

Rationalize the denominator of $\frac{5}{2+\sqrt{6}}$. The answer can be written as $\frac{A\sqrt{B}+C}{D}$, where $A$, $B$, $C$

horrorfan [7]

Answer:

$A +B+C+D = 3$ is the correct answer.

Step-by-step explanation:

Given:

$\frac{5}{2+\sqrt{6}}$

To find:

$A+B+C+D = ?$ if given term is written as following:

$\frac{A\sqrt{B}+C}{D}$

<u>Solution:</u>

We can see that the resulting expression does not contain anything under $\sqrt$ (square root) so we need to rationalize the denominator to remove the square root from denominator.

The rule to rationalize is:

Any term having square root term in the denominator, multiply and divide with the expression by changing the sign of square root term of the denominator.

Applying this rule to rationalize the given expression:

$\dfrac{5}{2+\sqrt{6}} \times \dfrac{2-\sqrt6}{2-\sqrt6}\\\Rightarrow \dfrac{5 \times (2-\sqrt6)}{(2+\sqrt{6}) \times (2-\sqrt6)} \\\Rightarrow \dfrac{10-5\sqrt6}{2^2-(\sqrt6)^2}\ \ \ \ \ (\because \bold{(a+b)(a-b)=a^2-b^2})\\\Rightarrow \dfrac{10-5\sqrt6}{4-6}\\\Rightarrow \dfrac{10-5\sqrt6}{-2}\\\Rightarrow \dfrac{-5\sqrt6+10}{-2}\\\Rightarrow \dfrac{5\sqrt6-10}{2}$

Comparing the above expression with:

$\frac{A\sqrt{B}+C}{D}$

A = 5, B = 6 (Not divisible by square of any prime)

C = -10

D = 2 (positive)

GCD of A, C and D is 1.

So, $A +B+C+D = 5+6-10+2 = \bold3$

5 0

3 years ago