All categories

3 years ago

Multiply the polynomials (a2 + 3a – 7) and (2a2 – a + 4). Simplify the answer.

Mathematics

1 answer:

Alex_Xolod [135]3 years ago

8 0

Answer correct or incorrect

You might be interested in

How do I solve -4-(2m+3)=-m-(5+3m)

drek231 [11]

-4-(2m+3)=-m-(5+3)
-4-2m-3=-m-5-3m
-2m-7=-4m-5
-2m+4m=-5+7
2m=2
m=2/2
m=1

3 0

2 years ago

Image.......................

babunello [35]

Answer:

10.75

Step-by-step explanation:

The cost of an adult ticket is £6 more than that of a child ticket, so will be shown by c+6.

Now, we are told that the cost of four child tickets and two adult tickets is £40.50, so we can put this in an equation and solve for c:

(c+6)+(c+6)+c+c+c+c=40.50

6c+12=40.50

6c=28.50

c=4.75

so, a childs ticket is 4.75

now to find the cost of an adult ticket you add 6,

4.75 + 6

= 10.75

3 0

2 years ago

What is 10 × 3 1/6 in it simplest form

Ray Of Light [21]

Answer:

Step-by-step explanation:10x3 1/6=30 10/6=31 2/6=31 1/3

3 0

2 years ago

Hey guys i need this fast its due in 15 minutes and please show step by step process

dolphi86 [110]

You can use a calculator for this if your teacher lets you to make it easier

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Ctiny%5Cint_%7Be%7D%5E%7B%7Be%7D%5E%7B2%7D%7D%20%5Cleft%28%20%5Csqrt%7Bx%20%2B%20%5Cfrac%

dlinn [17]

We have the identity

$\left(\sqrt x + \dfrac1{2^n}\right)^2 = x + \dfrac1{2^{n-1}} \sqrt x + \dfrac1{2^{2n}}$

Take the square root of both sides and rearrange terms on the right to get

$\sqrt x + \dfrac1{2^n} = \sqrt{x + \dfrac1{2^{n-1}} \left(\sqrt{x} + \dfrac1{2^{n+1}}\right)}$

Decrementing n gives

$\sqrt x + \dfrac1{2^{n-1}} = \sqrt{x + \dfrac1{2^{n-2}} \left(\sqrt{x} + \dfrac1{2^{n}}\right)}$

and substituting the previous expression into this, we have

$\sqrt x + \dfrac1{2^{n-1}} = \sqrt{x + \dfrac1{2^{n-2}} \sqrt{x + \dfrac1{2^{n-1}} \left(\sqrt x + \dfrac1{2^{n+1}}\right) } }$

Continuing in this fashion, after k steps we would have

$\sqrt x + \dfrac1{2^{n-k}} = \sqrt{x + \dfrac1{2^{n-(k+1)}} \sqrt{x + \dfrac1{2^{n-k}} \sqrt{x + \dfrac1{2^{n-(k-1)}} \sqrt{\cdots \dfrac1{2^{n-1}} \left(\sqrt x + \dfrac1{2^{n+1}}\right)}}}}$

After a total of n - 2 steps, we arrive at

$\sqrt x + \dfrac14 = \sqrt{x + \dfrac12 \sqrt{x + \dfrac1{2^2} \sqrt{x + \dfrac1{2^3} \sqrt{\cdots \dfrac1{2^{n-1}} \left(\sqrt x + \dfrac1{2^{n+1}}\right)}}}}$

Then as n goes to infinity, the first nested radical converges to √x + 1/4. Similar reasoning can be used to show the other nested radical converges to √x - 1/4. Then the integral reduces to

$\displaystyle \int_e^{e^2} \left(\sqrt x - \frac14\right) + \left(\sqrt x + \frac14\right) \, dx = 2 \int_e^{e^2} \sqrt x \, dx = \boxed{\frac43 \left(e^3 - e^{\frac32}\right)}$

5 0

2 years ago