All categories

3 years ago

I just need help with this question please

Mathematics

2 answers:

gregori [183]3 years ago

8 0

Answer:

$71.29

Step-by-step explanation:

$60 x 1.09 = $65.40 (year one)

$65.40 x 1.09 = $71.286 (year two)

Vsevolod [243]3 years ago

6 0

Answer:

71.286

Step-by-step explanation:

60 x .09 = 65.4

65.4 x .09 = 71.286

You might be interested in

8. Graph the following linear equations. a. 3x+5y=15

kenny6666 [7]

Answer:

slope: -3/5

y-intercept (0,3)

x - y

0 3

5 0

Step-by-step explanation:

I just did it

4 0

3 years ago

A "centroid" is defined as the center of mass of an object, or shape. How might the midpoint formula be used to find the midpoin

lesantik [10]

Answer:

we can use centeroid formula of a triangle

that is (x1+x2+x3)/3

hope that helps : )

6 0

3 years ago

Andrew wants 1/4 of a pizza Brice wants 2/9 and Collins wants 5/12 .How many slices would the pizza need to be cut into?

EastWind [94]

The answer is 36. because 12x3=36, 9x4=36, and 4x9=36, so it is the least common denominator

7 0

3 years ago

Read 2 more answers

What is the least common multiple of 37 and 12

emmasim [6.3K]

To get the Least Common Multiple (LCM) of 37 and 12 we need to factor each value first and then we choose all the factors which appear in any column and multiply them:

37: 3712: 223 LCM: 22337The Least Common Multiple (LCM) is: 2 x 2 x 3 x 37 = 444

4 0

3 years ago

The n term of a geometric sequence is denoted by Tn and the sum of the first n terms is denoted by Sn.Given T6-T4=5/2 and S5-S3=

Leno4ka [110]

1 step: $S_{5}=T_{1}+T_{2}+T_{3}+T_{4}+T_{5}$ , $S_{3}=T_{1}+T_{2}+T_{3}$ , then
$S_{5}-S_{3}=T_{4}+T_{5}=5$ .

2 step: $T_{n}=T_{1}*q^{n-1}$ , then
$T_{6}=T_{1}*q^{5}$
$T_{5}=T_{1}*q^{4}$
$T_{4}=T_{1}*q^{3}$
$T_{3}=T_{1}*q^{2}$
and $\left \{ {{T_{6}-T_{4}= \frac{5}{2} } \atop {T_{5}+T_{4}=5}} \right.$ will have form $\left \{ {{T_1*q^{5}-T_{1}*q^{3}= \frac{5}{2} } \atop {T_{1}*q^{4}+T_{1}*q^{3}=5} \right.$ .

3 step: Solve this system $\left \{ {{T_1*q^{3}*(q^{2}-1)= \frac{5}{2} } \atop {T_{1}*q^{3}*(q+1)=5} \right.$ and dividing first equation on second we obtain $\frac{q^{2}-1}{q+1}= \frac{ \frac{5}{2} }{5}$ . So, $\frac{(q-1)(q+1)}{q+1} = \frac{1}{2}$ and $q-1= \frac{1}{2}$ , $q= \frac{3}{2}$ - the common ratio.

4 step: Insert $q= \frac{3}{2}$ into equation $T_{1}*q^{3}*(q+1)=5$ and obtain $T_{1}* \frac{27}{8}*( \frac{3}{2}+1 ) =5$ , from where $T_{1}= \frac{16}{27}$ .

5 0

3 years ago