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2 years ago

How many quarters,dimes,and nickels are in $4.60

Mathematics

2 answers:

Burka [1]2 years ago

6 0

Answer:

18 quarters 1 dime not that hard

lutik1710 [3]2 years ago

5 0

Answer:

16 quarters= $4

extra 2 quarters= $0.50 cents

1 dime= $0.10

TOTAL AMOUNT OF MONEY = 4.60

Step-by-step explanation:

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Plz help me plzzzzz asap

Varvara68 [4.7K]

Answer:

2 $\frac{2}{9}$

Step-by-step explanation:

First, we will turn 5 2/5 into an improper fraction:

5 * 5 = 25 + 2 = 27/5

So they worked for 27/5 days.

We want to divide the days worked by the distance to get the unit rate of distance per day:

$\frac{12}{\frac{27}{5} } = \frac{12}{1} * \frac{5}{27}=\frac{60}{27}=\frac{20}{9}$

So our answer is 2 $\frac{2}{9}$

Hope this helps, have a great day! :D

6 0

2 years ago

cake batter poured into a pan creates a circular batter pattern whose radius increases at a constant rate of 1 in/sec. at what r

docker41 [41]

Answer:

8 π .

Step-by-step explanation:

radius increases at a constant rate of 1 in/sec.

dr / dt = 1

where r is radius .

radius after 4 sec

r = 1 x 4 = 4 in

area A = π r²

differentiating both sides

dA / dt = 2 π r dr / dt

Putting dr / dt = 1 , r = 4 in

dA / dt = 2 π x 4 x 1

= 8 π sq in per sec.

Area increases at the rate of 8 π per sec.

6 0

3 years ago

Systems with three variables. please help

kirza4 [7]

X= 6
Y= 0
Z= -2

CHECK
3x + y -2z =22
3(6) + 0 -2(-2) =22
18 + 0 +4= 22
22=22

x+5y+z =4
6 +5(0)+ -2 =4
6 + 0 +-2 = 4
4=4

x+3z=0
6 + 3(-2) =0
6 -6 = 0
0=0

3 0

3 years ago

A teacher places n seats to form the back row of a classroom layout. Each successive row contains two fewer seats than the prece

Alex_Xolod [135]

Answer:

The number of seat when n is odd $S_n=\frac{n^2+2n+1}{4}$

The number of seat when n is even $S_n=\frac{n^2+2n}{4}$

Step-by-step explanation:

Given that, each successive row contains two fewer seats than the preceding row.

Formula:

The sum n terms of an A.P series is

$S_n=\frac{n}{2}[2a+(n-1)d]$

$=\frac{n}{2}[a+l]$

a = first term of the series.

d= common difference.

n= number of term

l= last term

$n^{th}$ term of a A.P series is

$T_n=a+(n-1)d$

n is odd:

n,n-2,n-4,........,5,3,1

Or we can write 1,3,5,.....,n-4,n-2,n

Here a= 1 and d = second term- first term = 3-1=2

Let $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

Here $T_n=n$ , n=t, a=1 and d=2

$n=1+(t-1)2$

⇒(t-1)2=n-1

$\Rightarrow t-1=\frac{n-1}{2}$

$\Rightarrow t = \frac{n-1}{2}+1$

$\Rightarrow t = \frac{n-1+2}{2}$

$\Rightarrow t = \frac{n+1}{2}$

Last term l= n,, the number of term $=\frac{ n+1}2$ , First term = 1

Total number of seat

$S_n=\frac{\frac{n+1}{2}}{2}[1+n}]$

$=\frac{{n+1}}{4}[1+n}]$

$=\frac{(1+n)^2}{4}$

$=\frac{n^2+2n+1}{4}$

n is even:

n,n-2,n-4,.......,4,2

Or we can write

2,4,.......,n-4,n-2,n

Here a= 2 and d = second term- first term = 4-2=2

Let $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

Here $T_n=n$ , n=t, a=2 and d=2

$n=2+(t-1)2$

⇒(t-1)2=n-2

$\Rightarrow t-1=\frac{n-2}{2}$

$\Rightarrow t = \frac{n-2}{2}+1$

$\Rightarrow t = \frac{n-2+2}{2}$

$\Rightarrow t = \frac{n}{2}$

Last term l= n, the number of term $=\frac n2$ , First term = 2

Total number of seat

$S_n=\frac{\frac{n}{2}}{2}[2+n}]$

$=\frac{{n}}{4}[2+n}]$

$=\frac{n(2+n)}{4}$

$=\frac{n^2+2n}{4}$

4 0

3 years ago

What is the log 5x=log (2x+9)

Ray Of Light [21]

5x = 2x + 9, where 5x > 0 and 2x + 9 > 0 ;
Then, 3x = 9 ;
Then, x = 3 ;
Verify : 5 × 3 = 2 × 3 + 9 ;correct !
5 × 3 > 0 ; correct !
2 × 3 + 9 > 0; correct !

8 0

3 years ago