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ivann1987 [24]
3 years ago
6

Plz plz Plz help!

Mathematics
1 answer:
likoan [24]3 years ago
4 0

Answer:

$16,666.67

Step-by-step explanation:

PMT= PV*i Where PMT is the withdrawals ,PV is present value and i is the dicounting rate

PMT = $1,000.00

PV= ?

i = 6%

hence  $1,000 = PV*6%

PV=1,000/6%

PV = 16,666.67

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Whole-wheat bread costs $2.06/lb. How much would 3.8 lb of whole-wheat bread cost?
Xelga [282]
Whole-wheat bread = $2.06/lb

We need the cost of 3.8 lbs of whole-wheat bread.

So, we simply need to multiply the cost with the amount needed.

(cost per pound) × (amount of bread)

($2.06) × (3.8)

Simplify.

$7.828

Round to the nearest tenths.

$7.83

So, 3.8 pounds of whole wheat bread costs $7.83

~Hope I helped!~


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3 years ago
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PLEASE HELP!!! I ONLY HAVE A FEW MINUTES!!
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Answer:

y=0.25x-6

Step-by-step explanation:

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Keely scored 128 points, which was 4 times as many points as Grayson scored. Which equation can be used to find the number of po
Ludmilka [50]

Answer:

32

Step-by-step explanation:

take 128 divide by 4

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3 years ago
Theo earns £20 one weekend.
Colt1911 [192]

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He spend on bus fares £7.50

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A 15 kilogram object is suspended from the end of a vertically hanging spring stretches the spring 1/3 meters. At time t = 0, th
Yuri [45]

Answer:

15\frac{d^{2}y(t)}{dt^{2} }  - 441.45y(t) = ± 170 cos(5t)

y(0)=0, y'(0)=0

Step-by-step explanation:

See the attached image

This problem involves Newton's 2nd Law which is: ∑F = ma, we have that the acting forces on the mass-spring system are: F_{r} (t) that correspond to the force of resistance on the mass by the action of the spring and F(t) that is an external force with unknown direction (that does not specify in the enounce).

For determinate F_{r} (t) we can use Hooke's Law given by the formula F_{r} (t) = k y(t) where k correspond to the elastic constant of the spring and y(t) correspond to  the relative displacement of the mass-spring system with respect of his rest state.

We know from the problem that an 15 Kg mass stretches the spring 1/3 m so we apply Hooke's law and obtain that...

k = \frac{F_{r}}{y} = \frac{mg}{y} = \frac{15 Kg (9.81 \frac{m}{s^{2} } )}{\frac{1}{3} m}  = 441.45 \frac{N}{m}

Now we apply Newton's 2nd Law and obtaint that...

F_{r} (t) ± F(t) = ma(t)

F_{r} (t) = ky(t) = 441.45y(t)

F(t) = 170 cos(5t)

m = 15 kg

a(t) = \frac{d^{2}y(t)}{dt^{2} }

Finally... 15\frac{d^{2}y(t)}{dt^{2} }  - 441.45y(t) = ± 170 cos(5t)

We know from the problem that there's not initial displacement and initial velocity, so... y(0)=0 and y'(0)=0

Finally the Initial Value Problem that models the situation describe by the problem is

\left \{ 15\frac{d^{2}y(t)}{dt^{2} }  - 441.45y(t) = \frac{+}{} 170 cos(5t) \atop {y(0)=0, y'(0)=0\right.

6 0
3 years ago
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