How much bigger is TR (200) than TR (199)?

Find the sum and product of the roots. 2x^2 + 6x - 8 = 0

netineya [11]

Suppose you can factorize:

$ax^2+bx+c=a(x-r_1)(x-r_2)$

Then by expanding the right side, you have

$ax^2+bx+c=a(x^2-(r_1+r_2)x+r_1r_2)=ax^2-a(r_1+r_2)x+ar_1r_2$

which means in this case, you have to have

$\begin{cases}6=-2(r_1+r_2)\\-8=2r_1r_2\end{cases}\implies\begin{cases}r_1+r_2=-3\\r_1r_2=-4\end{cases}$

3 0

3 years ago

Suppose that there are two types of tickets to a show: advance and same-day. Advance tickets cost and same-day tickets cost . Fo

AveGali [126]

Answer:

83746+4747+48484=3943848343

Step-by-step explanation:

Easy

8 0

2 years ago

Between which two ordered pairs does the graph of f(x) = 1/2x2 + x – 9 cross the negative x-axis? (–6, 0) and (–5, 0) (–4, 0) an

lbvjy [14]

Answer:

Between (-6,0) and (-5,0).

Step-by-step explanation:

When it crosses the x axis, it's an x-intercept. To find the x-intercept, set y (aka f(x)) to zero. The solutions to the equation are approximately 3.36 and - 5.36. Since you're finding the negative x-intercept, you only need the negative solution - 5.36. This is in between (-6,0) and (-5,0).

3 0

3 years ago

30 tens - 3 tens 3 tenths

lubasha [3.4K]

300-30.3=269.7 because you barrow from 3 in 300 because you can't round into 0 and that equalled 7 and than you do 9-0=9 and there's 009.7 than you do 9-3=6 and that's 069.7 and you do 2-0=2 and there's 269.7

4 0

3 years ago

PLEASE HELP ASAP! I don’t recall how to do this!

MakcuM [25]

Answer:

Step-by-step explanation:

For a. we start by dividing both sides by 200:

$(1.05)^x=1.885$

In order to solve for x, we have to get it out from its position of an exponent. Do that by taking the natural log of both sides:

$ln(1.05)^x=ln(1.885)$

Applying the power rule for logs lets us now bring down the x in front of the ln:

x * ln(1.05) = ln(1.885)

Now we can divide both sides by ln(1.05) to solve for x:

$x=\frac{ln(1.885)}{ln(1.05)}$

Do this on your calculator to find that

x = 12.99294297

For b. we will first apply the rule for "undoing" the addition of logs by multipllying:

$ln(x*x^2)=5$

Simplifying gives you

$ln(x^3)=5$

Applying the power rule allows us to bring down the 3 in front of the ln:

3 * ln(x) = 5

Now we can divide both sides by 3 to get

$ln(x)=\frac{5}{3}$

Take the inverse ln by raising each side to e:

$e^{ln(x)}=e^{\frac{5}{3}}$

The "e" and the ln on the left undo each other, leaving you with just x; and raising e to the power or 5/3 gives you that

x = 5.29449005

For c. begin by dividing both sides by 20 to get:

$\frac{1}{2}=e^{.1x}$

"Undo" that e by taking the ln of both sides:

$ln(.5)=ln(e^{.1x})$

When the ln and the e undo each other on the right you're left with just .1x; on the left we have, from our calculators:

-.6931471806 = .1x

x = -6.931471806

Question d. is a bit more complicated than the others. Begin by turning the base of 4 into a base of 2 so they are "like" in a sense:

$(2^2)^x-6(2)^x=-8$

Now we will bring over the -8 by adding:

$(2^2)^x-6(2)^x+8=0$

We can turn this into a quadratic of sorts and factor it, but we have to use a u substitution. Let's let $u=2^x$

When we do that, we can rewrite the polynomial as

$u^2-6u+8=0$

This factors very nicely into u = 4 and u = 2

But don't forget the substitution that we made earlier to make this easy to factor. Now we have to put it back in:

$2^x=4,2^x=2$

For the first solution, we will change the base of 4 into a 2 again like we did in the beginning:

$2^2=2^x$

Now that the bases are the same, we can say that

x = 2

For the second solution, we will raise the 2 on the right to a power of 1 to get:

$2^x=2^1$

Now that the bases are the same, we can say that

x = 1

5 0

3 years ago