Ask question

Ask question

All categories

3 years ago

5

Simplify the expression 4x + 9y + 3(x+y)

2 answers:

brilliants [131]3 years ago

7 0

Answer:

Step-by-step explanation:

4x+9y+3x+3y

=7x+12y

vova2212 [387]3 years ago

7 0

Answer:

Simplifying the expression 4x + 9y + 3(x+y) will give us 7x + 12y

Step-by-step explanation:

To simplify this expression;

4x + 9y + 3(x+y)

First, we will start by opening the bracket, so the expression becomes;

4x + 9y + 3x + 3y

Then we will re-arrange the expression

4x + 3x +9y +3y

We will add the the numbers with x- coefficient together and also we will add the number with y-coefficient together. that is;add 4x and 3x together to give us 7x, also add 9y and 3y together to give us 12y

4x + 3x +9y +3y

7x +12y

we can no longer simplify anymore, so our answer is 7x + 12y

You might be interested in

Consider the equation below. f(x) = 2x3 + 3x2 − 336x (a) Find the interval on which f is increasing. (Enter your answer in inter

Vaselesa [24]

We have been given a function $f(x)=2x^3+3x^2-336x$ . We are asked to find the interval on which function is increasing and decreasing.

(a). First of all, we will find the critical points of function by equating derivative with 0.

$f'(x)=2(3)x^{2}+3(2)x^1-336$

$f'(x)=6x^{2}+6x-336$

$6x^{2}+6x-336=0$

$x^{2}+x-56=0$

$x^{2}+8x-7x-56=0$

$(x+8)-7(x+8)=0$

$(x+8)(x-7)=0$

$x=-8,x=7$

So $x=-8,7$ are critical points and these will divide our function in 3 intervals $(-\infty,-8)U(-8,7)U(7,\infty)$ .

Now we will find derivative over each interval as:

$f'(x)=(x+8)(x-7)$

$f'(-9)=(-9+8)(-9-7)=(-1)(-16)=16$

Since $f'(9)>0$ , therefore, function is increasing on interval $(-\infty,-8)$ .

$f'(x)=(x+8)(x-7)$

$f'(1)=(1+8)(1-7)=(9)(-6)=-54$

Since $f'(1)$ , therefore, function is decreasing on interval $(-8,7)$ .

Let us check for the derivative at $x=8$ .

$f'(x)=(x+8)(x-7)$

$f'(8)=(8+8)(8-7)=(16)(1)=16$

Since $f'(8)>0$ , therefore, function is increasing on interval $(7,\infty)$ .

(b) Since $x=-8,7$ are critical points, so these will be either a maximum or minimum.

Let us find values of f(x) on these two points.

$f(-8)=2(-8)^3+3(-8)^2-336(-8)$

$f(-8)=1856$

$f(7)=2(7)^3+3(7)^2-336(7)$

$f(7)=-1519$

Therefore, $(-8,1856)$ is a local maximum and $(7,-1519)$ is a local minimum.

(c) To find inflection points, we need to check where 2nd derivative is equal to 0.

Let us find 2nd derivative.

$f''(x)=6(2)x^{1}+6$

$f''(x)=12x+6$

$12x+6=0$

$12x=-6$

$\frac{12x}{12}=-\frac{6}{12}$

$x=-\frac{1}{2}$

Therefore, $x=-\frac{1}{2}$ is an inflection point of given function.

3 0

3 years ago

Which measurement is the best estimate of the area in square inches of the stove top that does not include the area of the burne

Paraphin [41]

I think D.)576 in.2

5 0

3 years ago

PLZ Give Us The Answer

Anni [7]

The answer is in the snip

7 0

3 years ago

The vertex is given at (2,-1), which tells us that h= and k=

earnstyle [38]

Answer:

H=2 k=-1

Step-by-step explanation:

4 0

3 years ago

Students voted for a new class president and the results are shown in the table.

blsea [12.9K]

Answer:

113:348

Step-by-step explanation:

i got it wrong cause i only put 113 but it showed this is the answer

6 0

2 years ago