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3 years ago

Helpppp!!!! ASAP!! Please???!

Mathematics

2 answers:

faltersainse [42]3 years ago

7 0

Answer:

c 111 mark me

Step-by-step explanation:

and ur welcome lol

neonofarm [45]3 years ago

3 0

Answer:

105

Step-by-step explanation:

if it true i add 65 +40 =105 so yeah

You might be interested in

Which point is located in Quadrant III ?

Firdavs [7]

Answer:

here you go :)

Quadrant lll is always both negative so the answer is (-1/2, -1/4)

Step-by-step explanation:

3 0

2 years ago

Which equation’s graph is a horizontal line? *<br> x = 9<br> 3x + 2y = 8<br> y = - 5

Natasha2012 [34]

Answer:

Your answer would by y = -5

Step-by-step explanation:

This is horizontal because it does not have a slope. It is a horizontal straight line at the -5 point on the y-axis

3 0

2 years ago

Solve for the measure of angle G in degrees.....

7nadin3 [17]

Answer:

Angle G = 76 degrees

Step-by-step explanation:

66 + 2x + x = 180

66 + 3x = 180

3x = 114

x = 38

2(38) = 76

7 0

2 years ago

Read 2 more answers

Help help help help help help

Vsevolod [243]

Answer: 19 mins

Step-by-step explanation:

hope this helps

7 0

3 years ago

NO LINKS OR ANSWERING QUESTIONS YOU DON'T KNOW!!! THIS IS NOT A TEST OR AN ASSESSMENT!! Please help me with these math questions

Art [367]

Answer:

See below

Step-by-step explanation:

3. What are two ways that a vector can be represented?

Considering a vector $\vec{v}$ in some vector space $\mathbb R^n$ we have

$\vec{v} = \langle a,b\rangle$

This is the component form. I don't like that way. It is probably used in high school, but

$\vec{v} = \begin{pmatrix} a\\ b\\ \end{pmatrix}$

is preferable because the inner product on $\mathbb R^n$ is defined to be

$\langle a,b\rangle := \sum_{i = 1}^n a_i b_i$

You can also write it using linear form such as $\vec{v} = 2i+2j$

For this question, I think you meant

vectors

$\vec{u_1} = (-8, 12)$

$\vec{u_2} = (13, 15)$

Once

$\cos(\theta)=\dfrac{\vec{u_1} \cdot\vec{u_2}}{||\vec{u_1}||||\vec{u_2}||}$

Considering that the dot product is

$\vec{u_1}\cdot \vec{u_2} = (-8)\cdot 13 + 12\cdot 15 = -104+180= 76$

and the norm of $\vec{u_1}$ is $||\vec{u_1}|| = \sqrt{(-8)^2 + 12^2} = \sqrt{64 + 144}= \sqrt{208}$

and the norm of $\vec{u_2}$ is $||\vec{u_2}|| = \sqrt{13^2 + 15^2} = \sqrt{169 + 225}= \sqrt{394}$

Thus,

$\cos(\theta)=\dfrac{76}{\sqrt{208} \sqrt{394}} = \dfrac{19}{\sqrt{13}\sqrt{394}}=\dfrac{19}{\sqrt{5122}}$

$\therefore \theta = \arccos \left(\dfrac{19}{\sqrt{5122}} \right)$

3 0

2 years ago