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4 years ago

Write a situation that can be represented by the expression 3a+6 where a=age in years

1 answer:

7 0

This equation can be used when comparing ages.

An example to illustrate this:
Assume that adding 6 to 3 times the age of Jack will give us the age of his grandfather.
When translating this into equations, assuming that the age of jack is "a" and the age of his grandfather is "b", we will find that:
b = 6 + 3a

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What is the length of Xy when w=1.3units Explain why F is correct

Aleksandr-060686 [28]

Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation

8 0

3 years ago

A 4.5 kg cat requires medication at a dose rate of 5 mg/kg once daily for the next 7 days. The tablets are available in 10 mg st

Marta_Voda [28]

Answer:

Approximately 16 tablets of medicine

Step-by-step explanation:

Based on the information above , we can use a Rule of Three to first figure out how many mg the cat would need to ingest per day.

$\frac{5mg}{1kg} = \frac{x}{4.5kg}$

$\frac{5mg*4.5kg}{1kg} = 22.5mg$

Now we see that the cat needs to take a daily dose of 22.5mg. Since the Tablets come at a strength of 10 mg per tablet then,

$22.5 / 10 = 2.25$

The cat would need to take 2.25 tablets per day. Since she needs to take the dose for the next 7 days then,

$2.25 * 7 = 15.75$

She would need to take approximately 16 tablets of medicine during the week .

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

5 0

4 years ago

Inqualities question pls help!

Margaret [11]

Answer:

Step-by-step explanation:

The closed circle at - $\frac{3}{2}$ indicates that x can equal this value

The open circle at $\frac{7}{2}$ indicates that x cannot equal this value.

All values of x between - $\frac{3}{2}$ and $\frac{7}{2}$ are valid, thus

- $\frac{3}{2}$ ≤ x < $\frac{7}{2}$ → B

5 0

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

8 0

4 years ago

Mark and Jen are both

Sedbober [7]

Can you tell me what book this is from so I can try to help!:)

6 0

3 years ago