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Vedmedyk [2.9K]
4 years ago
6

Can anyone plz help me with this problem im a little stuck on it...

Mathematics
1 answer:
Greeley [361]4 years ago
6 0

Answer:

18

Step-by-step explanation:

3×6 = 18

If you have any questions about the way I solved it, don't hesitate to ask me in the comments below ÷)

You might be interested in
2 times the quotient of a number and 8
mote1985 [20]

Answer:

answer is 2x / 8.

a number = x

quotient = division of 8

times = multiplication

hope this helps, i also hated my algebraic expressions unit haha

3 0
3 years ago
Read 2 more answers
Write the general equation for the circle that passes through the points (1, 1), (1, 3), and (9, 2).
Burka [1]

Step-by-step explanation:

As we know that

  • The circle center is equidistant from all three points, the distance being the circle radius.
  • Any point equidistant from two points must lie on the perpendicular bisector of the line segment which join those two points.
  • Which is, on the line through the midpoint of the line segment, perpendicular to the line segment.

The perpendicular bisector of the line segment joining the points (1, 1) and (1, 3) will be:

                   \:y=\:\frac{1+3}{2}=\frac{4}{2}=2

The perpendicular bisector of the line segment joining the points (1, 3), and (9, 2) will be:

                   x=\:\frac{1+9}{2}=\frac{10}{2}=5

These intersect at the center of the circle (5, 2).

The distance between (1, 1) and (5, 2) will be:

\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}

\mathrm{The\:distance\:between\:}\left(1,\:1\right)\mathrm{\:and\:}\left(5,\:2\right)\mathrm{\:is\:}

=\sqrt{\left(5-1\right)^2+\left(2-1\right)^2}

=\sqrt{4^2+1}

=\sqrt{16+1}

=\sqrt{17}

So the equation of the circle can be written as:

\left(x-5\right)^2+\left(y-2\right)^2=17

x^2-10x+y^2+29-4y=17

x^2-10x+y^2-4y+12\:=0

x^2+y^2-10x-4y+12=0

8 0
3 years ago
11. Mr Lee bought a second hand car for
Marrrta [24]

Answer :

Depends on the rate compound interest is accrued. See answers.

Step-by-step explanation:

We need the compound interest formula which is:

A = P (1+\frac{r}{n} )^{nt}

A =  final amount

P =  initial principal balance

r =  interest rate

n =  number of times interest applied per time period

t =  number of time periods elapsed

It doesn't say how frequently the interest is compounded, so we will do monthly, quarterly, and yearly.

MONTHLY

So we know the car costs $25,480 and he made a down payment of $10,000. By subtracting the down payment from the purchase price we find the loan amount.

$25,480 - $10,000 = $15,480

P = 15,480

r = 4.75% = .0475

n = 12  (12 months in a year)

t = 2  

Plug everything into the compound interest formula.

A = P (1+\frac{r}{n} )^{nt}

A = 15480(1+\frac{.0475}{12})^{12*2}

A = 15480(1+\frac{.0475}{12})^{24}

A = 15480(1+.00396)^{24}

A = 15480(1.00396)^{24}

A = 15480*1.0992

A = $17016.14

Mr. Lee paid $17,016.14 when the bill came due at 2 years with interest compounded monthly.

QUARTERLY

P = 15,480

r = 4.75% = .0475

n = 4  (4 quarters in a year)

t = 2  

A = P (1+\frac{r}{n} )^{nt}

A = 15480(1+\frac{.0475}{4})^{4*2}

A = 15480(1+\frac{.0475}{4})^{8}

A = 15480(1+.0119})^{8}

A = 15480(1.0119)^{8}

A = 15480*1.099

A = 15480(1.099)

A = 17013.20

Mr. Lee paid $17,013.20 when the bill came due at 2 years with interest compounded quarterly.

YEARLY

P = 15,480

r = 4.75% = .0475

n = 1  

t = 2  

A = P (1+\frac{r}{n} )^{nt}

A = 15480(1+\frac{.0475}{1})^{1*2}

A = 15480(1+\frac{.0475}{12})^{2}

A = 15480(1+.0475})^{2}

A = 15480(1.0475)^{2}

A = 15480*1.0973

A = 16985.53

Mr. Lee paid $16,985.53  when the bill came due at 2 years with interest compounded yearly.

4 0
4 years ago
Find the next three terms in the numbers sequence.. 4,8,14,22,32​
S_A_V [24]
The next three terms are 44,58,74
6 0
3 years ago
Problem asked——Find a quick and easy method to compute the sum of the first 100 counting numbers. No calculators allowed.
Darya [45]

Answer

Step-by-step explanation:

split the numbers into two groups (1 to 50 and 51 to 100), he could add them together vertically to get a sum of 101.

1     + 2   + 3   + 4   + 5   + … + 48 + 49 + 50

100 + 99 + 98 + 97 + 96 + … + 53 + 52 + 51

1 + 100 = 101

2 + 99 = 101

3 + 98 = 101

.

.

.

48 + 53 = 101

49 + 52 = 101

50 + 51 = 101

Final total would be 50(101) = 5050.

50 addition problems all adding up to 101 thats how i got 50x101

4 0
4 years ago
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