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2 years ago

11

The total cost of 5 doughnuts and 6 cookies at a bakery is $11.95.The cost of each cookie is $0.95.Select the equation and its s

olution that can be used to determine the cost,x, of 1 doughnut.

1 answer:

lisabon 2012 [21]2 years ago

6 0

The equation for the situation is $5x+6(0.95)=11.95$

The solution to the equation is $x=1.25$

First, you must set up the equation.

$5x+6(0.95)=11.95$

Then, you must simplify the terms.

$5x+5.7=11.95$

Next, subtract 5.7 on both sides.

$5x=6.25$

Finally, divide 5 on both sides

$x=1.25$

You might be interested in

Pierce currently has $10,000. What was the value of his money five years ago if he has earned 5 percent interest each year?

tia_tia [17]

The first answer is correct, if you go back by 5% each year you will see that.

4 0

2 years ago

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When baking a cake you have a choice of the following pans: a round cake that is 2 inches deep and has a 7-inch diameter, a 6 in

irga5000 [103]

The answer is
round cake - 82.42 in²
rectangular cake - 114 in²

Round cake:
d = 7 in
r = d/2 = 7 in / 2 = 3.5 in
h = 2 in

The surface are of a cylinder is:
A = 2πr² + 2πrh

The surface are of the round cake (which is actually a cylindrical cake) excluding the bottom is:
A = 2πr² + 2πrh - πr²
A = πr² + 2πrh
A = 3.14 * 3.5² + 2 * 3.14 * 3.5 * 2
= 38.46 + 43.96
= 82.42 in²

Rectangular cake:
w = 6 in
l = 9 in
h = 2 in

The surface are of a rectangle is:
A = 2wl + 2wh + 2lh

The surface are of the rectangular cake excluding the bottom is:
A = 2wl + 2wh + 2lh - wl
A = wl + 2wh + 2lh
A = 6 * 9 + 2 * 6 * 2 + 2 * 9 * 2
= 54 + 24 + 36
= 114 in²

5 0

2 years ago

1. You are saving to buy a new house in 7 years. If you invest $4,500 now at 5.5% interest compounded

motikmotik

Answer:

Part 1) $\$6,595.94$

Part 2) $\$3,449.23$

Part 3) $\$17,040.06$

Part 4) $\$20,773.90$

Part 5) The Option A is the best way to invest the money by $4,223.94 than Option B

Step-by-step explanation:

Part 1)

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$t=7\ years\\ P=\$4,500\\ r=5.5\%=5.5/100=0.055\\n=4$

substitute in the formula above

$A=4,500(1+\frac{0.055}{4})^{4*7}$

$A=4,500(1.01375)^{28}$

$A=\$6,595.94$

Part 2)

we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$t=2\ years\\ P=\$3,200\\ r=3.75\%=3.75/100=0.0375$

substitute in the formula above

$A=3,200(e)^{0.0375*2}$

$A=\$3,449.23$

Part 3)

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$t=18\ years\\ A=\$40,000\\ r=4.75\%=4.75/100=0.0475\\n=12$

substitute in the formula above

$40,000=P(1+\frac{0.0475}{12})^{12*18}$

$40,000=P(\frac{12.0475}{12})^{216}$

$P=40,000/[(\frac{12.0475}{12})^{216}]$

$P=\$17,040.06$

Part 4)

we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$t=7\ years\\ A=\$30,000\\ r=5.25\%=5.25/100=0.0525$

substitute in the formula above

$30,000=P(e)^{0.0525*7}$

$30,000=P(e)^{0.3675}$

$P=30,000/(e)^{0.3675}$

$P=\$20,773.90$

Part 5)

<u><em>Option A</em></u>

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$t=8\ years\\ P=\$11,500\\ r=5.6\%=5.6/100=0.056\\n=2$

substitute in the formula above

$A=11,500(1+\frac{0.056}{2})^{2*8}$

$A=11,500(1.028)^{16}$

$A=\$17,889.07$

<u><em>Option B</em></u>

we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$t=5\ years\\ P=\$11,500\\ r=3.45\%=3.45/100=0.0345$

substitute in the formula above

$A=11,500(e)^{0.0345*5}$

$A=11,500(e)^{0.1725}$

$A=\$13,665.13$

Compare the options

Option A ------> $\$17,889.07$

Option B -----> $\$13,665.13$

so

Option A > Option B

Find out the difference

$\$17,889.07-$13,665.13=$4,223.94$

therefore

The Option A is the best way to invest the money by $4,223.94 than Option B

3 0

2 years ago

What are the solutions of the equation x4 + 3x2 + 2 = 0? Use u substitution to solve.

Julli [10]

ANSWER

$x = \: x = \pm \: \sqrt{2} i \: or \: x = \pm \: i$

EXPLANATION

${x}^{4} + 3 {x}^{2} + 2 = 0$

${ ({x}^{2}) }^{2} + 3( {x}^{2}) + 2 = 0$

Let

$u = {x}^{2}$

Then the equation becomes:

${u}^{2} + 3u + 2 = 0$

${u}^{2} + 3u + 2 = 0$

${u}^{2} + 2u +u + 2 = 0$

Factor:

${u}(u + 2)+ 1(u + 2) = 0$

$(u + 1)(u + 2) = 0$

$u = - 1$

or

$u = - 2$

This implies that

${x}^{2} = - 1 \implies \: x = \pm \: i$

or

${x}^{2} = - 2 \implies \: x = \pm \: \sqrt{2} i$

3 0

2 years ago

If f(x) = 5x + 40, what is f(x) when x = –5?<br><br> a.–9<br> b.–8<br> c.7<br> d.15

defon

Just plug the number

_5(5)+40=-25+40

=15

3 0

2 years ago

Read 2 more answers