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3 years ago

Add (4x^2+x-9) and (3x^2+x+4)

Mathematics

1 answer:

HACTEHA [7]3 years ago

7 0

Answer:

7x^2 + 2x - 5

Step-by-step explanation:

(4x^2 + x - 9) + (3x^2 + x + 4)

7x^2 + 2x - 5

Pretty simple, just add 4 and 3, then x (or 1) and x, lastly -9 and 4. Then you get your final equation.

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34 – 16 = ?<br> Tens Ones

Lorico [155]

Answer:

1 ten

8 ones

Step-by-step explanation:

34 - 16 is 18. Therefore, you need 1 ten and 8 ones to make the number 18.

4 0

3 years ago

11th grade Algebra 2, please help.<br> If g(t) = 2/(t + 3) then g( blank) - 2/(4a + 3)

Helen [10]

Answer:

blank = 4a

Step-by-step explanation:

If g(t) = 2/(t + 3)

then g(4a) = 2/(4a+3)

5 0

3 years ago

Change 20 5/9 to an<br> improper fraction.

RideAnS [48]

Answer:

185/9

Step-by-step explanation:

20 5/9

Multiply 20 by the 9 (the denominator), and you''ll get 180.

Add the 5 (the numerator) to 180, and you'll get 185.

Then, place 185 over the denominator which is 9 and you get 185/9!

4 0

3 years ago

Read 2 more answers

Find the solution of the problem (1 3. (2 cos x - y sin x)dx + (cos x + sin y)dy=0.

lakkis [162]

Answer:

$2*sin(x)+y*cos(x)-cos(y)=C_1$

Step-by-step explanation:

Let:

$P(x,y)=2*cos(x)-y*sin(x)$

$Q(x,y)=cos(x)+sin(y)$

This is an exact differential equation because:

$\frac{\partial P(x,y)}{\partial y} =-sin(x)$

$\frac{\partial Q(x,y)}{\partial x}=-sin(x)$

With this in mind let's define f(x,y) such that:

$\frac{\partial f(x,y)}{\partial x}=P(x,y)$

and

$\frac{\partial f(x,y)}{\partial y}=Q(x,y)$

So, the solution will be given by f(x,y)=C1, C1=arbitrary constant

Now, integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to x in order to find f(x,y)

$f(x,y)=\int\ 2*cos(x)-y*sin(x)\, dx =2*sin(x)+y*cos(x)+g(y)$

where g(y) is an arbitrary function of y

Let's differentiate f(x,y) with respect to y in order to find g(y):

$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y} (2*sin(x)+y*cos(x)+g(y))=cos(x)+\frac{dg(y)}{dy}$

Now, let's replace the previous result into $\frac{\partial f(x,y)}{\partial y}=Q(x,y)$ :

$cos(x)+\frac{dg(y)}{dy}=cos(x)+sin(y)$

Solving for $\frac{dg(y)}{dy}$

$\frac{dg(y)}{dy}=sin(y)$

Integrating both sides with respect to y:

$g(y)=\int\ sin(y) \, dy =-cos(y)$

Replacing this result into f(x,y)

$f(x,y)=2*sin(x)+y*cos(x)-cos(y)$

Finally the solution is f(x,y)=C1 :

$2*sin(x)+y*cos(x)-cos(y)=C_1$

7 0

4 years ago

Find h(-1) if h(x) =- 3|3x – 6|– 3?

Dahasolnce [82]

Answer:

Step-by-step explanation:

h(-1) = - 3|3(-1) - 6| - 3

h(-1) = - 3|-3 - 6| - 3

h(-1) = - 3|-9| - 3

h(-1) = - 3(9) - 3

h(-1) = - 27 - 3

h(-1) = - 30

6 0

3 years ago