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Aleks04 [339]
3 years ago
14

Can you help me to answer this question ?????????

Mathematics
1 answer:
Iteru [2.4K]3 years ago
7 0

Answer:

11/45

Step-by-step explanation:

-\frac{5}{6} *\frac{2}{3} +\frac{14}{5} *\frac{2}{7} \\\\-\frac{5}{3} *\frac{1}{3} +\frac{2}{5} *\frac{2}{1} \\\\-\frac{5}{9} +\frac{4}{5} \\\\-\frac{25}{45} +\frac{36}{45} \\\\\frac{11}{45} \\\\

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5000 at 12% compound quarterly compared interest earned after 5 year's
hichkok12 [17]

Answer:

A = P(1+r÷100)^n

where A is amount after some days

r is the rate

n is the number of years

p is the principle (the amount of money before the interest)

A = 5000 ( 1 + 12÷100)^5

A = 8811.70

A = 8812

8 0
3 years ago
Question 8
sasho [114]

we are given

system of equations

First equation is

3x-2y=31

Second equation is

3x+2y=-1

now, we can find augmented matrix

A=\begin{pmatrix}3&-2&31\\ 3&2&-1\end{pmatrix}

now, we can change it into reduced row echelon form

step-1: multiply the 1st row by 1/3

A=\begin{pmatrix}1&-2/3&31/3\\ 3&2&-1\end{pmatrix}

step-2:add -3 times the 1st row to the 2nd row

A=\begin{pmatrix}1&-2/3&31/3\\ 0&4&-32\end{pmatrix}

step-3:multiply the 2nd row by 1/4

A=\begin{pmatrix}1&-2/3&31/3\\ 0&1&-8\end{pmatrix}

step-4: add 2/3 times the 2nd row to the 1st row

A=\begin{pmatrix}1&0&5\\ 0&1&-8\end{pmatrix}

so, we get

x=5,y=-8

Answer is (5,-8)

6 0
4 years ago
Read 2 more answers
Pls help i only have 2 minutes!!!!!!!!!!!!!!
umka21 [38]

Answer:

40 + ( -63) is less

Step-by-step explanation:

.... .....

4 0
3 years ago
Read 2 more answers
5a - 24b =-15<br> 10a + 45c =21<br> 48b + 35c = 49
andrew-mc [135]

5a + 15 = 24b so,

10a + 30 = 48b,

sub to third equation,

10a + 30 + 35c = 49

10a + 35c = 19

perform elimination with second equation

10a + 45c = 21

10a + 35c = 19

________________-

10c = 2

c = 1/5

sub to second equation,

10a + (45)(1/5) = 21

10a + 9 = 21

10a = 12

a = 6/5

sub to first equation

(5)(6/5) - 24b = -15

6 + 15 = 24b

21 = 24b

b = 7/8

redo the calculations, correct me if I'm wrong

6 0
3 years ago
Suppose y varies directly with x. If y = 6 when x = 2, find x when y = 12. 4 3
kobusy [5.1K]

\bf \qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad \stackrel{\textit{constant of variation}}{y=\stackrel{\downarrow }{k}x~\hfill } \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \textit{we also know that } \begin{cases} y=6\\ x=2 \end{cases}\implies 6=k(2)\implies \cfrac{6}{2}=k\implies 3=k \\\\\\ therefore\qquad \boxed{y=3x} \\\\\\ \textit{when y = 12, what is \underline{x}?}\qquad 12=3x\implies \cfrac{12}{3}=x\implies 4=x

6 0
3 years ago
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