Answer:
The new shape will be cone with a radius of 17 units, tilt height 24 and units of height k.
Step-by-step explanation:
The image related to the exercise is necessary, to be able to solve therefore the attached one.
We have the following information:
Perimeter of the triangle = 58 units
Hypotenuse = 24 units
We have that the other two sides are equal and have x units, therefore we have the following:
x + x + 24 = 58
2 * x = 58-24
2 * x = 34
x = 34/2
x = 17
Now the triangle rotates around line k
, and then it will result in a cone, which is a three-dimensional shape.
Therefore, the new shape will be cone with a radius of 17 units, tilt height 24 and units of height k.
Answer:
![\frac{2A}{h} = b](https://tex.z-dn.net/?f=%5Cfrac%7B2A%7D%7Bh%7D%20%3D%20b)
Step-by-step explanation:
- this can be done by simple manipulation .
- in the given equation, A is the subject.
- so, by the above statement you can understand the meaning of making something as a subject.
- given equation: A = 1/2 bh
- multiply both the sides by 2,
2A = bh
- divide both the sides by h,
![\frac{2A}{h} = b](https://tex.z-dn.net/?f=%5Cfrac%7B2A%7D%7Bh%7D%20%3D%20b)
so, the required form is : ![\frac{2A}{h} = b](https://tex.z-dn.net/?f=%5Cfrac%7B2A%7D%7Bh%7D%20%3D%20b)
Answer:
a)The component form is
![\vec {QP}=\binom{ - 1}{ - 8}](https://tex.z-dn.net/?f=%5Cvec%20%7BQP%7D%3D%5Cbinom%7B%20-%201%7D%7B%20-%208%7D%20)
b)The magnitude is √65
c) <2,14>
Step-by-step explanation:
Recall that:
![\vec {QP}=\vec {OP}-\vec{OQ}](https://tex.z-dn.net/?f=%5Cvec%20%7BQP%7D%3D%5Cvec%20%7BOP%7D-%5Cvec%7BOQ%7D)
We substitute the position vectors to get:
![\vec {QP}=\binom{ - 8}{7} - \binom { - 7}{15}](https://tex.z-dn.net/?f=%5Cvec%20%7BQP%7D%3D%5Cbinom%7B%20-%208%7D%7B7%7D%20-%20%20%5Cbinom%20%7B%20-%207%7D%7B15%7D%20)
We subtract the corresponding components to obtain:
![\vec {QP}=\binom{ - 8 - - 7}{7 - 15}](https://tex.z-dn.net/?f=%5Cvec%20%7BQP%7D%3D%5Cbinom%7B%20-%208%20-%20%20-%207%7D%7B7%20-%2015%7D%20)
This gives:
![\vec {QP}=\binom{ - 8 + 7}{7 - 15}](https://tex.z-dn.net/?f=%5Cvec%20%7BQP%7D%3D%5Cbinom%7B%20-%208%20%20%2B%207%7D%7B7%20-%2015%7D%20)
This simplifies to:
![\vec {QP}=\binom{ - 1}{ - 8}](https://tex.z-dn.net/?f=%5Cvec%20%7BQP%7D%3D%5Cbinom%7B%20-%201%7D%7B%20-%208%7D%20)
The magnitude of a vector in the component form:
![\binom{x}{y}](https://tex.z-dn.net/?f=%20%5Cbinom%7Bx%7D%7By%7D)
is
![\sqrt{ {x}^{2} + {y}^{2} }](https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%7Bx%7D%5E%7B2%7D%20%20%2B%20%20%7By%7D%5E%7B2%7D%20%7D%20)
This means:
![|\vec {QP}|= \sqrt{ {( - 1)}^{2} + {( - 8)}^{2} }](https://tex.z-dn.net/?f=%20%7C%5Cvec%20%7BQP%7D%7C%3D%20%5Csqrt%7B%20%7B%28%20-%201%29%7D%5E%7B2%7D%20%20%2B%20%20%7B%28%20-%208%29%7D%5E%7B2%7D%20%7D%20)
This simplifies to:
![|\vec {QP}| = \sqrt{ 1 + 64 }](https://tex.z-dn.net/?f=%20%7C%5Cvec%20%7BQP%7D%7C%20%3D%20%5Csqrt%7B%201%20%2B%20%2064%20%7D%20)
![|\vec {QP}| = \sqrt{ 65 }](https://tex.z-dn.net/?f=%20%7C%5Cvec%20%7BQP%7D%7C%20%3D%20%5Csqrt%7B%2065%20%7D%20)
c) We have the vectors u = <4, 8>, v = <-2, 6>.
We want to find:
u+v
This implies that:
u+v=<4,8>+<-2,6>
We add the corresponding components to get;
u+v=<4+-2,8+6>
This simplifies to:
u+v=<2,14>
So the waiting time for a bus has density f(t)=λe−λtf(t)=λe−λt, where λλ is the rate. To understand the rate, you know that f(t)dtf(t)dt is a probability, so λλ has units of 1/[t]1/[t]. Thus if your bus arrives rr times per hour, the rate would be λ=rλ=r. Since the expectation of an exponential distribution is 1/λ1/λ, the higher your rate, the quicker you'll see a bus, which makes sense.
So define <span><span>X=min(<span>B1</span>,<span>B2</span>)</span><span>X=min(<span>B1</span>,<span>B2</span>)</span></span>, where <span><span>B1</span><span>B1</span></span> is exponential with rate <span>33</span> and <span><span>B2</span><span>B2</span></span> has rate <span>44</span>. It's easy to show the minimum of two independent exponentials is another exponential with rate <span><span><span>λ1</span>+<span>λ2</span></span><span><span>λ1</span>+<span>λ2</span></span></span>. So you want:
<span><span>P(X>20 minutes)=P(X>1/3)=1−F(1/3),</span><span>P(X>20 minutes)=P(X>1/3)=1−F(1/3),</span></span>
where <span><span>F(t)=1−<span>e<span>−t(<span>λ1</span>+<span>λ2</span>)</span></span></span><span>F(t)=1−<span>e<span>−t(<span>λ1</span>+<span>λ2</span>)</span></span></span></span>.