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3 years ago

Which of the following statements about this is true??

Mathematics

2 answers:

stepan [7]3 years ago

8 0

Answer:

lines tx and vs are perpendicular

fenix001 [56]3 years ago

8 0

Answer:

I think it is the 2 one don't blame me if I am wrong

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maria [59]

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3 years ago

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QUESTION 8 The orthocenter of a triangle is formed by the ____________of a triangle. perpendicular bisectors angle bisectors med

Otrada [13]

<span>The orthocenter of a triangle is formed by the ALTITUDES of a triangle.</span>

Source:
http://www.1728.org/trictr.htm

4 0

3 years ago

-x/14+2>4?? I’m confused

Lelechka [254]

Answer: x<−28

Step-by-step explanation:

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3 years ago

An empty can weighs 30 grams. The weight of the same can filled with liquid is 47.3 grams. Which equation could be used to find

Alika [10]

Answer:

w = y - x is the equation to find w

Where,

w = Weight of the liquid

x = Weight of the empty can

y = Weight of the can with liquid

Step-by-step explanation:

Let

Weight of the empty can = x

Weight of the liquid = w

Weight of the can with liquid = x + w = y

x= 30 grams

w = ?

y= 47.3 grams

y = x + w

w = y - x

= 47.3 - 30

= 17.3

w= 17.3 grams

Therefore, the weight of the liquid is 17 .3 grams

3 0

3 years ago

Question 8 Find the unit vector in the direction of (2,-3). Write your answer in component form. Do not approximate any numbers

slamgirl [31]

Answer:

The unit vector in component form is $\hat{u} = \left(\frac{2}{\sqrt{13} },-\frac{3}{\sqrt{13}} \right)$ or $\hat{u} = \frac{2}{\sqrt{13}}\,i-\frac{3}{13}\,j$ .

Step-by-step explanation:

Let be $\vec u = (2,-3)$ , its unit vector is determined by following expression:

$\hat {u} = \frac{\vec u}{\|\vec u \|}$

Where $\|\vec u \|$ is the norm of $\vec u$ , which is found by Pythagorean Theorem:

$\|\vec u\|=\sqrt{2^{2}+(-3)^{2}}$

$\|\vec u\| = \sqrt{13}$

Then, the unit vector is:

$\hat{u} = \frac{1}{\sqrt{13}} \cdot (2,-3)$

$\hat{u} = \left(\frac{2}{\sqrt{13} },-\frac{3}{\sqrt{13}} \right)$

The unit vector in component form is $\hat{u} = \left(\frac{2}{\sqrt{13} },-\frac{3}{\sqrt{13}} \right)$ or $\hat{u} = \frac{2}{\sqrt{13}}\,i-\frac{3}{13}\,j$ .

6 0

3 years ago