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ella [17]
2 years ago
7

• I need a well explained ans ...• spams no needed ​

Mathematics
2 answers:
WITCHER [35]2 years ago
4 0

Answer:

Sry if it is wrong..........

Alex73 [517]2 years ago
3 0

Answer:

6. 0.22222222222

7. -0.0037037037

Step-by-step explanation:

Im nice at such questions so I gave the answer

hope it helps you... please mark me as BRAINLIEST

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-12 degrees. Each hour, the temperature dropped by 3 degrees so it'd look like: -3 -6 -9 -12
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How many black t shirts would Hannah and Dario need to sell to make up the loss in profit?
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How much are they selling the shirts for ? how much debt do they have ?
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Lamar is writing a coordinate proof to show that a segment from the midpoint of the hypotenuse of a right triangle to the opposi
Zanzabum

1. N is a midpoint of the segment KL, then N has coordinates

\left(\dfrac{x_K+x_L}{2},\dfrac{y_K+y_L}{2} \right) =\left(\dfrac{0+2a}{2},\dfrac{2b+0}{2} \right) =(a,b).

2. To find the area of △KNM, the length of the base MK is 2b, and the length of the height is a. So an expression for the area of △KNM is

A_{KMN}=\dfrac{1}{2}\cdot \text{base}\cdot \text{height}=\dfrac{1}{2}\cdot 2b\cdot a=ab.

3. To find the area of △MNL, the length of the base ML is 2a and the length of the height is b. So an expression for the area of △MNL is

A_{MNL}=\dfrac{1}{2}\cdot \text{base}\cdot \text{height}=\dfrac{1}{2}\cdot 2a\cdot b=ab.

4. Comparing the expressions for the areas you have that the area A_{KMN} is equal to the area A_{MNL}. This means that the segment from the midpoint of the hypotenuse of a right triangle to the opposite vertex forms two triangles with equal areas.

6 0
3 years ago
Read 2 more answers
A particle moves with velocity vector
asambeis [7]

Answer:

\vec{s(t)}=\frac{1}{2}t^2\hat{i}-(2t+1)\hat{j}+(\frac{1}{3}t^3+1)\hat{k}

Step-by-step explanation:

We are given that velocity vector of a particle

\vec{v(t)}=t\hat{i}-2\hat{j}+t^2\hat{k}

When t=0 then the particle is at the point (0,-1,1).

We have to find the position of particle  at time t.

We know that

Velocity =\frac{Displacement }{time}=\frac{ds}{dt}

Therefore,\vec{v}=\frac{\vec{ds}}{dt}

\int{ds}=\int (t\hat{i}-2\hat{j}+t^2\hat{k})dt

Integrate on both sides then we get

\vec{s(t)}=\frac{1}{2}t^2\hat{i}-2t\hat{j}+\frac{1}{3}t^3\hat{k}+C

\int x^n dx=\frac{x^{n+1}}{n+1}+C

Substitute the value of point at time t=0 then we get

C=-\hat{j}+\hat{k}

Substitute the value of C then we get

\vec{s(t)}=\frac{1}{2}t^2\hat{i}-2t\hat{j}+\frac{1}{3}t^3\hat{k}-\hat{j}+\hat{k}

Therefore, the position of particle at time t

\vec{s(t)}=\frac{1}{2}t^2\hat{i}-(2t+1)\hat{j}+(\frac{1}{3}t^3+1)\hat{k}

8 0
3 years ago
Worth 60points!!!!!
EastWind [94]

Answer:

X=4

Step-by-step explanation:

2÷\frac{1}{2}=4

6 0
3 years ago
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