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2 years ago

Janice has 15 pets. Each of her pets is either a dog or a bird. If her pets have a total of 38 legs, how

Mathematics

1 answer:

balu736 [363]2 years ago

3 0

Hi!

These can be set up as systems of equations.

Let b = the number of birds, and let d = the number of dogs.

b + d = 21

2b + 4d = 76 (this is because birds have 2 legs and dogs have 4 legs.)

You can solve this using either elimination or substitution.

Consider marking brainliest.

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3 years ago

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3 years ago

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Diano4ka-milaya [45]

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Step-by-step explanation:

Using the rule of radicals

$\sqrt{a}$ × $\sqrt{b}$ ⇔ $\sqrt{ab}$

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4 years ago

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3 years ago

The 6-lb particle is subjected to the action of its weight and forces F1 = 52i + 6j - 2tk6 lb, F2 = 5t 2 i - 4tj - 1k6 lb, and F

olga2289 [7]

Answer:

r=294.9m

Step-by-step explanation:

The forces on the particle are

$W=mg\hat{j}\\F_{1}=52\hat{i}+6\hat{j}-2t\hat{k}\\F_{2}=5t^{2}\hat{i}-4t\hat{j}-1\hat{k}\\F_{3}=(5-2t)\hat{i}$

Now , we sum all these forces to get the net force

$F_{T}=W+F_{1}+F_{2}+F_{3}\\F_{T}=(52+5t^{2}+5-2t)\hat{i}+((6+6-4t)\hat{j}+(-2t-1)\hat{k}\\F_{T}=(57-2t+5t^{2})\hat{i}+(12-4t)\hat{j}+(-2t-1)\hat{k}\\$

we can use the fact F=m*a and integrate the acceleration

$a(t)=\frac{1}{m}F(t)\\\\v(t)=\int a(t)dt=\frac{1}{m}\int{F_{T}}dt\\\\v(t)=\frac{1}{m}[(57t-t^{2}+\frac{5}{3}t^{3})\hat{i}+(12t-2t^{2})\hat{j}+(-t^{2}-t)\hat{k}]\\\\r(t)=\int v(t)dt=\frac{1}{m}[(\frac{57}{2}t^{2}-\frac{1}{3}t^{3}}+\frac{5}{4}t^{4})\hat{i}+(6t^{2}-\frac{2}{3}t^{3})\hat{j}+(-\frac{1}{3}t^{3}-\frac{1}{2}t^{2})]$

and we evaluate in r(2) an we take the norm to obtain the distance

$r(2)=\frac{1}{m}[\frac{394}{3}\hat{i}+\frac{56}{3}\hat{j}-\frac{14}{3}\hat{k}]\\|r(2)|=\frac{1}{m}\sqrt{[(\frac{394}{3})^{2}+(\frac{56}{3})^{2}+(\frac{14}{3})^{2}]}\\|r(2)|=\frac{132.73}{0.45}=294.9m$

I hope this is useful for you

regards

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4 years ago