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3 years ago

-7 = x -3 find the value of x

Mathematics

1 answer:

Mekhanik [1.2K]3 years ago

3 0

Answer

-4

Step-by-step explanation:

to cancle out 3, you have to add three to it.

−7=−3

−7 + 3 = −3 + 3

-4 =

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What is the simplified form of the quantity 1 over b minus 1 over a all over the quantity 1 over b plus 1 over a?

jolli1 [7]

Option B: (b-a)/(b+a) is the correct answer

Step-by-step explanation:

The given expression is;

$\frac{\frac{1}{b}-\frac{1}{a}}{\frac{1}{b}+\frac{1}{a}}$

First of all we have to take LCM in both numerator and denominator

So,

$=\frac{\frac{b-a}{ab}}{\frac{b+a}{ab}}$

The reciprocal of the denominator will be multiplied with the numerator for simplification

So,

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Option B: (b-a)/(b+a) is the correct answer

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3 years ago

Vector question. Let v=<-2,1> u=<3,-5> , find:

lyudmila [28]

Answer:

1. ) $\vec {v}.\vec{u}= -11$

2.) The angle between v and u is 147.52°.

Step-by-step explanation:

Given:

$\vec {v}= ( -2 , 1 )\\\vec {u}= ( 3 , -5 )$

To Find:

1. $\vec {v}.\vec{u}= ?$

2. $\theta = ?$

Solution:

$\vec {v}.\vec{u}$ is scalar product given as,

$\vec {v}.\vec{u}=|\vec {v}||\vec {u}|\cos \theta$

$|\vec {v}|=|(-2, 1)|=\sqrt{(-2)^{2} +1^{2}}=\sqrt{5}\\|\vec {u}|=|(3, 5)|=\sqrt{(3)^{2} +(-5)^{2}}=\sqrt{34}$

$\vec {v}.\vec{u}=(-2i +j).(3i-5j)\\$

Here only i.i = j.j =1 and i.j = j.i = 0

∴ $\vec {v}.\vec{u}=(-2\times 3 +1\times -5)\\\vec {v}.\vec{u}=-6-5=-11 \\$

Now, Substituting the above values we get

$-11=\sqrt{5}\times \sqrt{34}\cos \theta\\ \cos \theta=\frac{-11}{\sqrt{170}} \\ \cos \theta =-0.84366\\\therefore \theta =cos^{-1}(-0.84366)\\\therefore \theta =147.52\°$

As it is negative mean $\theta$ is in Second Quadrant Because Cosine is negative in Second Quadrant.

1. ) $\vec {v}.\vec{u}= -11$

2.) The angle between v and u is 147.52°.

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