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3 years ago

Hunter has a quarters and y dimes, having at least 18 coins worth at most

Mathematics

1 answer:

garik1379 [7]3 years ago

5 0

Answer:The number of nickles is 6

The number of dims is 12 .

Step-by-step explanation:

Given as :

Sum of total number of dims and nickels coins = 18

The value of total combination = $1.50

Let The total number of dims = d

Let The total number of nickels = n

1 nickles = $0.05

1 dims = $0.1

According to question

Total number of dims and nickels coins = number of dims + number of nickels

Or, d + n = 18 .........1

And

$0.1 ×d + $0.05 ×n = $1.50

Or, 0.1 d + 0.05 n = 1.50 .........2

Now, Solving eq 1 an eq 2

0.1 × (d + n) - (0.1 d + 0.05 n) = 0.1 × 18 - 1.50

Or, (0.1 d - 0.1 d) + (0.1 n - 0.05 n) = 1.8 - 1.50

Or, 0 + 0.05 n = 0.3

Or, 0.05 n = 0.3

∴ n =

i.e n = 6

So, The number of nickles = n = 6

Put the value of n int eq 1

∵ d + n = 18

Or, d = 18 - n

Or, d = 18 - 6

i.e d = 12

So, The number of dims = d = 12

Hence, The number of nickles is 6 and the number of dims is 12 .

Step-by-step explanation:

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3 years ago

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Please help me solve this exercise.!!

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<h3>Answer: -4/3</h3>

=============================================================

Explanation:

Let's square both sides and do a bit of algebra to get the following.

$\sin(x) + \cos(x) = 1/5\\\\\left(\sin(x) + \cos(x)\right)^2 = \left(1/5\right)^2\\\\\sin^2(x) + 2\sin(x)\cos(x) + \cos^2(x) = 1/25\\\\\sin^2(x) + \cos^2(x) + 2\sin(x)\cos(x) = 1/25\\\\1 + 2\sin(x)\cos(x) = 1/25\\\\\sin(2x) = 1/25 - 1\\\\\sin(2x) = 1/25 - 25/25\\\\\sin(2x) = -24/25\\\\$

Now apply the pythagorean trig identity to determine cos(2x) based on this. You should find that cos(2x) = -7/25

This then means tan(2x) = sin(2x)/cos(2x) = 24/7.

From here, you'll use this trig identity

$\tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}\\\\$

which is the same as solving

$\tan(2x) = \frac{2w}{1-w^2}\\\\$

where w = tan(x)

Plug in tan(2x) = 24/7 and solve for w to get w = -4/3 or w = 3/4

So either tan(x) = -4/3 or tan(x) = 3/4.

If we were to numerically solve the original equation for x, then we'd get roughly x = 2.21; then notice how tan(2.21) = -1.345 approximately when your calculator is in radian mode.

Since tan(x) < 0 in this case, we go for tan(x) = -4/3

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3 years ago

2. Harry just deposited $1500 into a savings account giving 6% interest compounded quarterly.a) How much will be in the account

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The formula for compound interest is:

$\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ \text{where,} \\ A=\text{ Final amount} \\ r=\text{ Interest rate} \\ n=\text{ Number of times interest applied per period} \\ t=\text{ Number of time period elapsed} \\ P=\text{ Intial principal balance} \end{gathered}$

Given data:

$\begin{gathered} P=\text{ \$1500} \\ r=6\text{ \%}=0.06 \\ n=4\text{ times (compounded quarterly)} \end{gathered}$

a. After ten years, that is t = 10 years, the amount in the account will be

$\begin{gathered} A=1500(1+\frac{0.06}{4})^{4\times10} \\ A=\text{ }1500(1+0.015)^{40} \\ A=\text{ }1500(1.015)^{40} \\ A=\text{ \$2721.03} \end{gathered}$

b. After twenty years, that is t = 20 years, the amount in the account will be:

$\begin{gathered} A=1500(1+\frac{0.06}{4})^{4\times20} \\ A=1500(1.015)^{4\times20} \\ A=1500(1.015)^{80} \\ A=\text{ \$}4935.99 \end{gathered}$

c. The time it takes for Harry's initial account value to double will be:

$\begin{gathered} A=2\text{ x initial value = 2 }\times\text{ \$1500 = \$3000} \\ 3000=1500(1.015)^{4t} \\ (1.015)^{4t}=\frac{3000}{1500} \\ (1.015)^{4t}=2 \\ \text{ Find the logarithm of both sides} \\ \ln (1.015)^{4t}=\ln 2 \\ 4t=\frac{\ln 2}{\ln 1.015} \\ 4t=46.56 \\ t=\frac{46.56}{4}=11.64 \end{gathered}$

Therefore, the time it takes Harry's initial account to double is approximately 11 years

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