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3 years ago

Stuck on this please help!!

Mathematics

2 answers:

Dmitry [639]3 years ago

8 0

Answer:

welp i got -2x+89

Step-by-step explanation:

Serhud [2]3 years ago

7 0

You would have to sell them $79.62 i think

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Lara wrote the statements shown in the chart below:

Brums [2.3K]

Answer:

Option 4:

Statement 1 and Statement 2 are theorems because they can be proved.

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Tyler saved 1/3 as much as kara. tyler saved $55. write an equation to find how much kara saved?

siniylev [52]

X - 1/3 = 55

55x3= 165

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3 years ago

If k(x) = 5x – 6, which expression is equivalent to (k + k)(4)?. . 5(4 + 4) – 6. 5(5(4) – 6) – 6. 54 – 6 + 54 – 6. 5(4) – 6 + 5(

Dmitrij [34]

K(x) = 5x-6
(k+k)(x) = k(x) + k(x) = 5x-6 +5x-6
just plug in x = 4,
(k+k)(4) = 5(4) -6 + 5(4) -6
thats your answer, the last option.

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3 years ago

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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,

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

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

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3 years ago

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Nimfa-mama [501]

Answer:

256

Step-by-step explanation:

I judt know I don't know hoet to explain sorry hope it helps

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