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3 years ago

Name the smallest angle of ABC. The diagram is not to scale. с 11 10 B

Mathematics

1 answer:

Mrac [35]3 years ago

6 0

Answer:

uhhhh Where's a picture?

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Determine the next term in the sequence: 4, 8, 12,<br> 16<br> 18<br> 24<br> 48

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Hoped it help :)

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3 years ago

Does any body know the answer??

Mashcka [7]

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3 years ago

3+-1/1/3 the first half is 3+-1 the second half is 1/3

kolbaska11 [484]

$\bf \cfrac{3\pm1}{\frac{1}{3}}\implies 
\begin{cases}
\cfrac{3+1}{\frac{1}{3}}\\\\
\cfrac{3-1}{\frac{1}{3}}
\end{cases}\\\\
-------------------------------\\\\$

$\bf \cfrac{3+1}{\frac{1}{3}}\implies \cfrac{4}{\frac{1}{3}}\implies \cfrac{\frac{4}{1}}{\frac{1}{3}}\implies \cfrac{4}{1}\cdot \cfrac{3}{1}\implies \cfrac{4\cdot 3}{1\cdot 1}\implies \cfrac{12}{1}\implies \boxed{12}
\\\\\\
\cfrac{3-1}{\frac{1}{3}}\implies \cfrac{2}{\frac{1}{3}}\implies \cfrac{\frac{2}{1}}{\frac{1}{3}}\implies \cfrac{2}{1}\cdot \cfrac{3}{1}\implies \cfrac{2\cdot 3}{1\cdot 1}\implies \cfrac{6}{1}\implies \boxed{6}$

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3 years ago

A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.

Usimov [2.4K]

Answer:

a) $\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

b) $v(t=3) = 3(3)^2 -18(3) +15=-12$

c) $t =1s, t=5s$

d) $[0,1) \cup (5,\infty)$

e) $D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

f) $a(t) = \frac{dv}{dt}= 6t -18$

g) The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

Step-by-step explanation:

For this case we have the following function given:

$f(t) = s = t^3 -9t^2 +15 t$

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

$\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

$v(t=3) = 3(3)^2 -18(3) +15=-12$

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

$3t^2 -18 t +15 =0$

We can divide both sides of the equation by 3 and we got:

$t^2 -6t +5=0$

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

$(t-5)*(t-1) =0$

And for this case we got $t =1s, t=5s$

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

$t^2 -6t +5 >0$

$(t-5) *(t-1)>0$

So then the intervals positive are $[0,1) \cup (5,\infty)$

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

$D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6$

And if we replace we got:

$D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

$a(t) = \frac{dv}{dt}= 6t -18$

Part g and h

The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

5 0

3 years ago

What is the result of the mathematical expression below:

uysha [10]

5 is the correct answer.

6 0

3 years ago

Read 2 more answers