What is the slope m in the equation y= -2x + 5

(4-5x)+43=3-32(-82) please answer!!

kotykmax [81]

Answer:

x = -516

Step-by-step explanation:

(4 - 5x) + 43 = 3 - 32(-82) {Multiply 32 and (-82)}

4 - 5x + 43 = 3 - (-2624)

4 - 5x + 43 = 3 + 2624 {add 4 and 43 in LHS}

-5x + 47 = 2627 {Subtract 47 from both sides}

-5x = 2627 - 47

-5x = 2580 {Divide both sides by (-5)}

x = 2580/(-5)

x = -516

7 0

3 years ago

On the day after my birthday this year, I can truthfully say: "The day after tomorrow is a Monday". On which day is my birthday?

Dvinal [7]

Answer: Friday

Step-by-step explanation:

3 0

4 years ago

Read 2 more answers

Nina can stitch 2/3 of a dress in 4 hours. If d represents the number of dresses and h represents the number of hours, which equ

Ronch [10]

Answer:

The answer is A.

Explanation:

A is true because 2/3 of a dress takes a full four hours. That means 1/3 is finished every 2 hours. So if you add both 4 and 2 together you get 6 hours :)

Hope this helped

3 0

2 years ago

If g(x)=3/4x+2,find g(-12)

Triss [41]

Answer:

g(-12) = -7

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Find a decomposition of a=⟨−5,−1,1⟩ into a vector c parallel to b=⟨−6,0,6⟩ and a vector d perpendicular to b such that c+d=a.

dezoksy [38]

The projection of vector A parallel to vector B is $\langle -3, 0, 3\rangle$ and the projection of vector A perpendicular to vector B is $\langle -2, -1, -2\rangle$ .

In this question, we need to determine all projections of a vector with respect to another vector. In this case, the projection of vector A parallel to vector B is defined by this formula:

$\vec a_{\parallel , \vec b} = \frac{\vec a \,\bullet\,\vec b}{\|\vec b\|^{2}}\cdot \vec b$ (1)

Where $\|\vec b\|$ is the norm of vector B.

And the projection of vector A perpendicular to vector B is:

$\vec a_{\perp, \vec b} = \vec a - \vec a_{\parallel, \vec b}$ (2)

If we know that $a = \langle -5, -1, 1 \rangle$ and $\vec b = \langle -6, 0, 6 \rangle$ , then the projections are now calculated:

$\vec a_{\parallel, \vec b} = \frac{(-5)\cdot (-6)+(-1)\cdot (0)+(1)\cdot (6)}{(-6)^{2}+0^{2}+6^{2}} \cdot \langle -6, 0, 6 \rangle$

$\vec a_{\parallel, \vec b} = \frac{1}{2}\cdot \langle -6, 0, 6 \rangle$

$\vec a_{\parallel, \vec b} = \langle -3, 0, 3\rangle$

$\vec a_{\perp, \vec b} = \langle -5, -1, 1 \rangle - \langle -3, 0, 3 \rangle$

$\vec a_{\perp, \vec b} = \langle -2, -1, -2\rangle$

The projection of vector A parallel to vector B is $\langle -3, 0, 3\rangle$ and the projection of vector A perpendicular to vector B is $\langle -2, -1, -2\rangle$ .

We kindly invite to check this question on projection of vectors: brainly.com/question/24160729

7 0

3 years ago