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masya89 [10]
2 years ago
6

Tristan has some dimes and some quarters. He has at least 20 coins worth at most $4.25 combined. If Tristan has 10 dimes, determ

ine the maximum number of quarters that he could have. If there are no possible solutions, submit an empty answer.
Mathematics
1 answer:
Dovator [93]2 years ago
7 0

Answer:

The maximum number of quarters that he could have is 13

Step-by-step explanation:

Let x represent the number of quarters he could have. If he has 10 dimes and has at least 20 coins, then

→ x + 10 >= 20

→ x >= 10    (1)

His coins worth at most $4.25. Also, it is known that 1 dollar is equal to 100 cents, 1 dime is equal to 10 cents and 1 quarter is equal to 25 cents.

→ (x * 25) + (10 * 10)  <= (4.25 * 100)

→ 25x  + 100 <= 425

→ 25x <= 325

→ x <= 13  (2)

If we combine the equation 1 and 2:

10 <= x <= 13

The maximum value of x is 13

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Thepotemich [5.8K]

Answer:

Ruler: £0.90

Pencil: £0.60

Step-by-step explanation:

Ruler = x

Pencil = y

2x + y = 2.4

5x + y = 5.1

3x = 2.7

x = 0.9

y =  0.6

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2 years ago
1+5+5VE+Y5-5U<br> Help me please
professor190 [17]

Answer:

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Step-by-step explanation:

In this riddle we have to do addition and subtraction to give answer .

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8 0
2 years ago
Read 2 more answers
Katie invested $33,750 at 11.17% compounded continuously.
ser-zykov [4K]

Answer:

Katie's account balance in 10 years will be $3,769,875

Step-by-step explanation:

you multiply the amount of money you invest by the percentage, and the amount of years.

EX: $33,750 × 11.17% × 10= $3,769,875

5 0
2 years ago
PLEASE HELP ME!!!!!!!!!!!
Ahat [919]

Answer:

I'm not sure

Step-by-step explanation:

3 0
1 year ago
Please answer all of them need this
VikaD [51]

First Question

For a better understanding of the solution provided here please find the first attached file which has the diagram of the the isosceles trapezoid.

We dropped perpendiculars from C and D to intersect AB at Q and P respectively.

As can be seen in \Delta BCQ, we can easily find the values of CQ and BQ.

Since, Sin(75^0)=\frac{CQ}{8}

\therefore CQ=8\times Sin(75^0)\approx 7.73 ft

In a similar manner we can find BQ as:

Cos(75^0)=\frac{BQ}{8}

BQ\approx2.07 ft

All these values can be found in the diagram attached.

Thus, because of the inherent symmetry of the isosceles trapezoid, PQ can be found as:

PQ=22-(AP+QB)=22-(2.07+2.07)=17.86

Let us now consider\Delta AQC

We can apply the Pythagorean Theorem here to find the length of the diagonal AC which is the hypotenuse of \Delta AQC.

AC=\sqrt{(AQ)^2+(QC)^2}=\sqrt{(AP+PQ)^2+(QC)^2}=\sqrt{(2.07+17.86)^2+(7.73)^2}\approx21.38 feet.

Thus, out of the given options, Option B is the closest and hence is the answer.

Second Question

For this question we can directly apply the formula for the area of a triangle using sines which is as:

Area=\frac{1}{2}(First Side)(Second Side)(Sine of the angle between the two sides)

Thus, from the given data,

Area=\frac{1}{2}\times 218.5\times 224.5\times sin(58.2^0)\approx20845 m^2

Therefore, Option D is the correct option.

Third Question

For this question we will apply the Sine Rule to the \Delta ABC given to us.

Thus, from the triangle we will have:

\frac{AB}{Sin(\angle C)}=\frac{BC}{Sin(\angle A)}

\frac{c}{Sin(\angle C)}=\frac{a}{Sin(\angle A)}

\frac{17}{Sin(25^0)}=\frac{a}{Sin(45^0)}

This gives a to be:

a\approx28.44

Which is not close to any of the given options.

Fourth Question

Please find the second attachment for a better understanding of the solution provided her.

As can be clearly seen from the attached diagram, we can apply the Cosine Rule here to find the return distance of the plane which is CA.

AC=\sqrt{(AB)^2+(BC)^2-2(AB)(BC)\times Cos(\angle B)}

\therefore AC=\sqrt{(172.20)^2+(111.64)^2-2(172.20)(111.64)\times Cos(177.29^0)}\approx283.8 miles.

Thus, Option D is the answer.





8 0
2 years ago
Read 2 more answers
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