I need Help Asap Please Please

Please please help ! Thank you

miskamm [114]

Answer:

4/3c^4 -it wont let me put it as it shows on your screen but the 4 is on top and under the line is 3c^4,

Step-by-step explanation:

Cancel the common factor of c^3 and z^7 -im not sure if this is what you were trying to get but its a answer-

4 0

2 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 $\vec{v} = 2i+2j$

4.

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

Solve the problem by entering and solving an equation.

asambeis [7]

Answer: -26.60

Step-by-step explanation:

Heres -30 degrees as faherenheit

3 0

2 years ago

Read 2 more answers

Whats the answer to 3(q-7)=27? im looking for q.

VMariaS [17]

Answer:

Step-by-step explanation:

i hope i am right q=16 is my answer my bad if it is wrong

3 0

2 years ago

Read 2 more answers

An Archer shoots an arrow horizontally at 250 feet per second. The bullseye on the target and the arrow are initially at the sam

Genrish500 [490]

Answer:

Step-by-step explanation:

Using the formula for finding range in projectile, Since range is the distance covered in the horizontal direction;

Range $R = U\sqrt{\frac{H}{g} }$

U is the velocity of the arrow

H is the maximum height reached = distance below the bullseye reached by the arrow.

R is the horizontal distance covered i.e the distance of the target from the archer.

g is the acceleration due to gravity.

Given R = 60ft, U = 250ft/s, g = 32ft/s H = ?

On substitution,

$60 = 250\sqrt{\frac{H}{32}} \\\frac{60}{250} = \sqrt{\frac{H}{32}}\\\frac{6}{25} = \sqrt{\frac{H}{32}$

Squaring both sides we have;

$(\frac{6}{25} )^{2} = (\sqrt{\frac{H}{32} } )^{2} \\\frac{36}{625} = \frac{H}{32} \\625H = 36*32\\H = \frac{36*32}{625} \\H = 1.84feet$

The arrow will hit the target 1.84feet below the bullseye.

5 0

2 years ago