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3 years ago

Which value of x would make SUV TUW by HL?

2 answers:

4 0

By HL similarity theorem, the hypothesus and one leg of both triangles are congruent.
In the given figure, it is shown that the legs UV and UW are congruent (equivalent).
The hypothenus of triangle SUV is 2x + 9 and the hypothenus of triangle TUW is 4x - 1. For both hypothenuses to be congruent, 2x + 9 = 4x - 1
4x - 2x = 9 + 1
2x = 10
x = 5.

Therefore, the value of x, for which SUV is congruent to TUW by HL is 5.

8090 [49]3 years ago

3 0

Answer: The answer is 5

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Let f be a function of two variables that has continuous partial derivatives and consider the points

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Answer:

The directional derivative of f at A in the direction of $\vec{u}$ AD is 7.

Step-by-step explanation:

Step 1:

Directional of a function f in direction of the unit vector $\vec{u}=(a,b)$ is denoted by $D\vec{u}f(x,y)$ ,

$D\vec{u}f(x,y)=f_{x}\left ( x ,y\right ).a+f_{y}(x,y).b$ .

Now the given points are

$A(8,9),B(10,9),C(8,10) and D(11,13)$ ,

Step 2:

The vectors are given as

AB = (10-8, 9-9),the direction is

$\vec{u}_{AB} = \frac{AB}{\left \| AB \right \|}=(1,0)$

AC=(8-8,10-9), the direction is

$\vec{u}_{AC} = \frac{AC}{\left \| AC \right \|}=(0,1)$

AC=(11-8,13-9), the direction is

$\vec{u}_{AD} = \frac{AD}{\left \| AD \right \|}=\left (\frac{3}{5},\frac{4}{5} \right )$

Step 3:

The given directional derivative of f at A $\vec{u}_{AB}$ is 9,

$D\vec{u}_{AB}f=f_{x} \cdot 1 + f_{y}\cdot 0\\f_{x} =9$

The given directional derivative of f at A $\vec{u}_{AC}$ is 2,

$D\vec{u}_{AB}f=f_{x} \cdot 0 + f_{y}\cdot 1\\f_{y} =2$

The given directional derivative of f at A $\vec{u}_{AD}$ is

$D\vec{u}_{AD}f=f_{x} \cdot \frac{3}{5} + f_{y}\cdot \frac{4}{5}$

$D\vec{u}_{AD}f=9 \cdot \frac{3}{5} + 2\cdot \frac{4}{5}$

$D\vec{u}_{AD}f= \frac{27+8}{5} =7$

The directional derivative of f at A in the direction of $\vec{u}_{AD}$ is 7.

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