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4vir4ik [10]
3 years ago
8

This reduction was drawn at a scale of 1 inch scaled to 4 feet actual. What is the actual height of the giraffe?

Mathematics
1 answer:
Wittaler [7]3 years ago
5 0

Answer:

5 inches tall i think

Step-by-step explanation:

sorry if its wrong

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Error Analysis: Mr. Hasson is buying a new bike for the summer time. The bike originally costs $150 but he sees it's on sale for
evablogger [386]

Answer:

$45

Step-by-step explanation:

30% ÷100 =0.3. 0.3×150= 45

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A container holds 16 pints of lemonade. How much is this in gallons
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Since one gallon is equal to 8 pints, 16 pints would equal to 2 gallons.
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How do I find the measure of angles​
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6 0
3 years ago
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
Need help with this question click to see
blondinia [14]

Answer:

1) u > 4

2) 7 < u

3) u < 4

4) 7 > u

Step-by-step explanation:

1) subtract 6 - 2

2) add 3 + 4

3) subtract 4 on both sides (8-4)

4) subtract 9 - 2

7 0
3 years ago
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