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3 years ago

Challenge A bag contains pennies, nickels, dimes, and quarters. There are 50 coins in all. Of the

Mathematics

1 answer:

seraphim [82]3 years ago

6 0

Answer:pennies: (0.12)(50) = 6

dimes: (0.34)(50) = 17

nickels: 6 + 2 = 8

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Which of the following expressions is equivalent to 3 m ( m − 2 ) − ( m ^2 + 1 ) ?

leonid [27]

Answer:

Step-by-step explanation:

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3 years ago

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Whats the commisson of 660 in 10%

Sholpan [36]

Answer:

0,10×660=66

660-66= 594

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4 years ago

Fill in the integers -10, -8, -7, -6, -5, -4, -3, -2 and 0 into a magic square so that every row, column, and diagonal has a sum

ioda

-2 -10 -3

-7 0 -8

-6 -5 -4

That's the square, it was a fun puzzle :)

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3 years ago

Can someone help me the 2nd problem I don’t understand it

Galina-37 [17]

Answer:

So this means the bus B covered 390-120=270 miles when bus A has already reached 390 miles.

270 miles

Step-by-step explanation:

So is A is going faster than B so A will reach the destination first.

When will A reach it's destination?

Let's find out.

To solve this problem, the following will come in handy:

Speed=distance/time or time*Speed=distance or time=distance/speed .

time=distance/speed

$T_A=\frac{390}{S_A}$

$T_A=\frac{390}{65}$

$T_A=6$

So it will take bus A 6 hours to cover the distance of 390 miles.

How much time would have it taken bus B to reach that same distance?

$T_B=\frac{390}{45}$

$T_B=8.\overline{6}$

So it would have taken bus B $8.\overline{6}$ hours to cover a distance of 390 miles.

So the time difference is $8.\overline{6}-6=2.\overline{6}$ hours.

It will take $2.\overline{6}$ more hours than bus A for bus B to complete a distance of 390 miles.

So bus B traveled $2.\overline{6} \cdot 45=120$ miles (used the time*speed=distance) after bus A got to it's destination.

So this means the bus B covered 390-120=270 miles when bus A has already reached 390 miles.

4 0

3 years ago

What value of k makes the equation true? (5a^2 b^3)(6a^k b)=30a^6 b^4

Ratling [72]

we are given

$(5a^2 b^3)(6a^k b)=30a^6 b^4$

Firstly, we will simplify left side

and then we can solve for k

Left side is

$(5a^2 b^3)(6a^k b)$

we can arrange like terms

$(5a^2 6a^k)(b^3 b)$

$(5\times 6 a^2a^k)(b^3 b)$

$(30 a^2a^k)(b^3 b^1)$

now, we can use property of exponent

$a^m \times a^n=a^{m+n}$

$(30 a^{2+k})(b^{3+1})$

$30a^{2+k}b^{4}$

now, we can equate it with right side

and we get

$30a^{2+k}b^{4}=30a^6 b^4$

We can see that both sides have 30 , a and b

and they are equal

so, exponent of a must also be equal

$2+k=6$

now, we can solve for k

$2+k-2=6-2$

$k=4$ ................Answer

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3 years ago

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