Convert the repeating decimal into a fraction: 0.555

Coefficient of x^2 in expansion of (x+2)^5

iVinArrow [24]

The x² term will be C(5,2)x²·2³ = 80x².

The coefficient is 80.

_____
C(n, k) = n!/(k!·(n-k)!)
C(5, 2) = 5·4/(2·1) = 10

8 0

3 years ago

if Dakota earned $15.75 in interest in account A and $28.00 in interest in account B after 21 months. if the simple interest rat

NISA [10]

keeping in mind that 21 months is more than a year, since there are 12 months in a year, then 21 months is really 21/12 years.

$\bf ~~~~~~ \stackrel{\textit{account A}}{\textit{Simple Interest Earned}} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill &15.75\\ P=\textit{original amount deposited}\dotfill \\ r=rate\to 3\%\to \frac{3}{100}\dotfill &0.03\\ t=years\to \frac{21}{12}\dotfill &\frac{7}{4} \end{cases} \\\\\\ 15.75=P(0.03)\left( \frac{7}{4} \right)\implies \cfrac{15.75}{(0.03)\left( \frac{7}{4} \right)}=P\implies \boxed{300=P}$

$\bf ~~~~~~ \stackrel{\textit{account B}}{\textit{Simple Interest Earned}} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill &28\\ P=\textit{original amount deposited}\dotfill \\ r=rate\to 4.9\%\to \frac{4.9}{100}\dotfill &0.049\\ t=years\to \frac{21}{12}\dotfill &\frac{7}{4} \end{cases} \\\\\\ 28=P(0.049)\left( \frac{7}{4} \right)\implies \cfrac{28}{(0.049)\left( \frac{7}{4} \right)}=P\implies \boxed{326.53\approx P}$

so, clearly, you can see who's greater.

3 0

3 years ago

Step 1: –10 + 8x < 6x – 4

Aleks04 [339]

Answer:

3 > x

Step-by-step explanation:

–2(5 – 4x) < 6x – 4

Step 1: –10 + 8x < 6x – 4

Step 2: –10 < –2x – 4

Step 3: –6 < –2x

Step 4: divide each side by by -2, remembering to flip the inequality

-6/-2 > -2/-2

3 > x

6 0

3 years ago

Read 2 more answers

A line passes through the point (2, 3) and has a slope of -2. Which is the equation of the line in point-slope form?

Lelu [443]

The equation of the line would be y=-2x+9

5 0

3 years ago

4.

marta [7]

Answer:

a) $(x-2)(x-6) + (y-5)(y-11) =0\Rightarrow (x-4)^{2}+(y+3)^{2}=68\\C(4,-3) \:r=\sqrt{68}$ b) (a,b) (c, d) As long as (a,b) and (c,d) are the endpoints of of the diameter.

Step-by-step explanation:

a)

1) The reduced formula of the Circumference is given by:

$(x-a)^{2}}+(y-b)^{2}=r^{2}$

2) Let's expand the factored one into one closer to the pattern above:

$x^{2}-8x+12+y^{2}+6y-55=0\Rightarrow x^{2}-8x+y^{2}+6y=43$

3) Completing the square for both trinomials:

$(x-4)^{2}+(y+3)^{2}=43+16+9\Rightarrow (x-4)^{2}+(y+3)^{2}=68$

4) In the Reduced Formula, $(x-a)^{2}}+(y-b)^{2}=r^{2}$ ,

$C(a,b) \Rightarrow C(4,-3) \:the\:radius\:is\:r=\sqrt{68} \Rightarrow r=2\sqrt{17}$

b) Using the previous example to show this:

When we factor this way

$(x-a)(x -c) + (y -b)(y-d) = 0$

We are indeed, naming "a" and "b", the coordinates of (a, b) of the first endpoint and "b" and "d" the second endpoint as well.Id est, D (2, 5) and B (6,-11).

The radius, is $\sqrt{68} \cong 8.25$

So yes, the equation of the circle can be written as

$(x-a)(x -c) + (y -b)(y-d) = 0$

As long as (a,b) and (c,d) are the endpoints of of the diameter.

$d_{AC}=d_{BC}=R$

3 0

3 years ago