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3 years ago

Need help ASAP giving 30 points

Mathematics

1 answer:

HACTEHA [7]3 years ago

6 0

Answer:

Step-by-step explanation:

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Use the alternative form of the dot product to find u · v.

vodka [1.7K]

Answer:

Step-by-step explanation:

u = 45

v = 30

Angle between them, θ = 5π / 6 = 150°

(a) The formula for the dot product is given by

$\overrightarrow{u}.\overrightarrow{v}= u v Cos\theta$

$\overrightarrow{u}.\overrightarrow{v}= 45 \times 30 \times Cos150$

$\overrightarrow{u}.\overrightarrow{v}= -1169.13$

(b) $\overrightarrow{u}=Cos\frac{\pi }{3}\widehat{i}+Sin\frac{\pi }{3}\widehat{j}$

$\overrightarrow{u}=0.5\widehat{i}+0.866\widehat{j}$

$\overrightarrow{v}=Cos\frac{2\pi }{3}\widehat{i}+Sin\frac{2\pi }{3}\widehat{j}$

$\overrightarrow{v}=-0.707\widehat{i}+0.707\widehat{j}$

Let the angle between them is θ

The formula in terms of the dot product is given by

$Cos\theta =\frac{\overrightarrow{u}.\overrightarrow{v}}{u v}$

$u=\sqrt{0.5^{2}+0.866^{2}}=1$

$v=\sqrt{-0.707^{2}+0.707^{2}}=1$

$Cos\theta =\frac{0.5 \times (-0.707) + 0.866 \times 0.707)}{1\times 1}$

$Cos\theta=0.2587$

θ = 75°

$u=\sqrt{1^{2}+4^{2}+8^{2}}=9$

$Cos\alpha =\frac{1}{9}$

$Cos\beta =\frac{4}{9}$

$Cos\gamma =\frac{8}{9}$

7 0

4 years ago

Matthew is 7 years older then twice the age of his sister, if matthew is 13, how old is his sister

Irina18 [472]

This is subtraction an the answer would be 6

5 0

4 years ago

Read 2 more answers

Look at the planes ABCD and EFGH shown below.. Which statement is true about the two planes?. A.There are more than two segments

scZoUnD [109]

According to your image which has a diagram where 2 planes are both perpendicularly attached to each other. So base on your diagram I conclude that the answer is C. There are two segments along the planes intersect. I hope i answered your question

7 0

3 years ago

HELP PLEASE AND THANK YOU 4JAAB

Maslowich

Answer:

$\bold{Q13}\ \angle W\\\\\bold{Q14}\ \angle S\\\\\bold{Q15}\ side\ WA\\\\\bold{Q16}\ \triangle PTA$

Step-by-step explanation:

For Q13, Q14, Q15

$\text{for}\ \overline{WA}\ \text{and}\ \overline{AS}\ -\ A\ \text{is common}.\ \text{Therefore inclu}\text{ded angle is}\ \angle A\\\\\text{for}\ \overline{WS}\ \text{and}\ \overline{WA}\to\angle W\\\\\text{for}\ \overline{WS}\ \text{and}\ \overline{SA}\to\angle S\\\\============================$

$\text{Inc}\text{lude side between}\ \angle A\ \text{and}\ \angle W\ \text{is}\ AW\ (WA).\\\\\text{Inc}\text{lude side between}\ \angle S\ \text{and}\ \angle W\ \text{is}\ SW\ (WS).\\\\\text{Inc}\text{lude side between}\ \angle A\ \text{and}\ \angle S\ \text{is}\ AS\ (SA).$