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3 years ago

Pls help!! Find the value of x.

Mathematics

1 answer:

Gelneren [198K]3 years ago

6 0

I think it’s b but I’m not to sure sorry if it’s wrong!

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Given that r = ( 7, 3, 9) and v = ( 3, 7, -9), evaluate r + v

uysha [10]

Answer:

b. (10,10,0)

Step-by-step explanation:

r+v can be evaluated if the vectors/matrices have the same dimensions.

These do. They are both 1 by 3 vectors.

Just add first to first in each.

Just add second to second in each.

Just add third to third in each.

Example:

(5,-5,6)+(1,2,3)

=(5+1,-5+2,6+3)

=(6,-3,9)

Done!

In general, (a,b,c)+(r,s,t)=(a+r,b+s,c+t).

r+v

=(7,3,9)+(3,7,-9)

=(7+3,3+7,9+-9)

=(10,10,0)

Done!

5 0

3 years ago

What is 2+2 I need help and how to explain.

Ket [755]

Lol its 4 put up two fingers then add two more

7 0

4 years ago

Read 2 more answers

Find the distance and displacement for the following figures :

Vinil7 [7]

The figure (i) have a distance of 15.996 meters and a displacement of 13.892 meters and (ii) a distance of 480 centimeters and a displacement of 339.411 centimeters

<h3>How to find the distance and displacement in each trajectory</h3>

The distance is the sum of the lengths that form a trajectory and the displacement is the distance between the <em>initial</em> and <em>final</em> point of a trajectory. Then, we have the following results for each case:

Case I

Distance

d = 5 m + 0.5π · (7 m)

d ≈ 15.996 m

Displacement

$D = \sqrt{(12\,m)^{2}+(7 m)^{2}}$

D ≈ 13.892 m

Case II

Distance

d = 8 · (60 cm)

d = 480 cm

Displacement

$D =\sqrt{(240\,cm)^{2}+(240\,cm)^{2}}$

D ≈ 339.411 cm

To learn more on displacement: brainly.com/question/11934397

#SPJ1

4 0

2 years ago

if <img src="https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7B1%7D%7B3%7D%20-%20%5Cfrac%7B1%7D%7B2%7Dx%20" id="TexFormula1" title="f

Dimas [21]

Answer:

$\large\boxed{(f+g)(x)=2x^2+\dfrac{1}{2}x+4\dfrac{1}{3}}$

Step-by-step explanation:

$(f+g)(x)=f(x)+g(x)\\\\\text{We have}\\\\f(x)=\dfrac{1}{3}-\dfrac{1}{2}x,\ g(x)=2x^2+x+4.\\\\\text{Substitute:}\\\\(f+g)(x)=\left(\dfrac{1}{3}-\dfrac{1}{2}x\right)+(2x^2+x+4)\\\\=\dfrac{1}{3}-\dfrac{1}{2}x+2x^2+x+4\qquad\text{combine like terms}\\\\=2x^2+\left(-\dfrac{1}{2}x+x\right)+\left(\dfrac{1}{3}+4\right)\\\\=2x^2+\dfrac{1}{2}x+4\dfrac{1}{3}$

6 0

3 years ago

........................................................................

eduard

Answer:

Step-by-step explanation:

8 0

1 year ago