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3 years ago

How many different ways can you make 75cents using nickles quaters and dimes?

1 answer:

4 0

3 dimes, 1 nickel
2 dimes, 3 nickels
1 dime, 5 nickels
7 nickels
(don't forget to add in the initial quarter, dime, and nickel)

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Evaluate ƒ(x ) = x 2 + 1 for ƒ(10). 2√2 101 21

Vladimir [108]

F(10) = (10)^2 + 1
= 100 + 1
= 101

101 is the answer

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3 years ago

A transformation of ASTV results in AUTV.

Fantom [35]

Answer: Reflection

Step-by-step explanation:

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3 years ago

2. Simplify 3x (x-5) 2x - 3 - where x = 0 and x* 5. 3x

Diano4ka-milaya [45]

Answer:

(x-5)

2x - 3= 3x,x-5,2x-3

Step-by-step explanation:

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3 years ago

Perform the indicated operations. Write the answer in standard form, a+bi. 5-3i / -2-9i

Vsevolod [243]

$\huge \boxed{\mathfrak{Answer} \downarrow}$

$\large \bf\frac { 5 - 3 i } { - 2 - 9 i } \\$

Multiply both numerator and denominator of $\sf \frac{5-3i}{-2-9i} \\$ by the complex conjugate of the denominator, -2+9i.

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2-9i\right)\left(-2+9i\right)}) \\$

Multiplication can be transformed into difference of squares using the rule: $\sf\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$ .

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2\right)^{2}-9^{2}i^{2}}) \\$

By definition, i² is -1. Calculate the denominator.

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{85}) \\$

Multiply complex numbers 5-3i and -2+9i in the same way as you multiply binomials.

$\large \bf \: Re(\frac{5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9i^{2}}{85}) \\$

Do the multiplications in $\sf5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9\left(-1\right)$ .

$\large \bf \: Re(\frac{-10+45i+6i+27}{85}) \\$

Combine the real and imaginary parts in -10+45i+6i+27.

$\large \bf \: Re(\frac{-10+27+\left(45+6\right)i}{85}) \\$

Do the additions in $\sf-10+27+\left(45+6\right)i$ .

$\large \bf Re(\frac{17+51i}{85}) \\$

Divide 17+51i by 85 to get $\sf\frac{1}{5}+\frac{3}{5}i \\$ .

$\large \bf \: Re(\frac{1}{5}+\frac{3}{5}i) \\$

The real part of $\sf \frac{1}{5}+\frac{3}{5}i \\$ is $\sf \frac{1}{5} \\$ .

$\large \boxed{\bf\frac{1}{5} = 0.2} \\$

3 0

3 years ago

For which equation is y = 7 a solution?

Natasha2012 [34]

Answer:B

Step-by-step explanation:

y - 26 = -19

y + 26 - 26 = -19 + 26

y=7

4 0

3 years ago

Read 2 more answers