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Mariana [72]
3 years ago
5

Fumio performed the following operations on vectors u, v, and w, where u = ⟨6, –8⟩, v = ⟨1, 2⟩, w = ⟨3, –3⟩ 2u – 3(v + w) 2 ⟨6,

–8⟩ – 3[⟨1, 2⟩ + ⟨3, –3⟩] ⟨12, –8⟩ – 3⟨4, –1⟩ ⟨12, –8⟩ + ⟨–12, 3⟩ ⟨0, –5⟩ What mistake did Fumio make? He did not distribute the 2 to both components of vector u. He added vectors v and w before distributing the –3 to their components. He subtracted vector w from vector v. He did not use the associative property correctly when rearranging the signs.
Mathematics
2 answers:
user100 [1]3 years ago
8 0

Answer:

(A) on edge

He did not distribute the 2 to both components of vector u.

Step-by-step explanation:

The -8 in his work did not change to -16 as it should if it was multiplied by 2

Correct on edge

uranmaximum [27]3 years ago
8 0

Answer:

A

Step-by-step explanation:

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Answer:

16

Step-by-step explanation:

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3 years ago
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3 0
3 years ago
That are equal to 4⋅10-3:
kozerog [31]
Answer: 37

Explanation:

Order of operations says you do multiplication first, you get 4 • 10 = 40

Then do subtraction, so 40 - 3 = 37

Hope this helps!
7 0
3 years ago
Evaluate tan45/sin30 . In your final answer, include all calculations.
Maslowich
From the identity \displaystyle{ \tan x= \frac{\sin x}{\cos x}, and from the values:

                 \sin45^{\circ}=\cos45^{\circ}= \frac{ \sqrt{2}}{2},

we have: \displaystyle{ \tan 45^{\circ}= \frac{\sin 45^{\circ}}{\cos 45^{\circ}}= \frac{ \frac{ \sqrt{2}}{2}}{\frac{ \sqrt{2}}{2}} =1.


We also know that \sin30^{\circ}= \frac{1}{2}.


Thus, \displaystyle{  \frac{\tan 45^{\circ}}{\sin30^{\circ}} = \frac{1}{\frac{1}{2}}=2.



Answer: 2
8 0
3 years ago
Find the area of the surface. The part of the surface z = xy that lies within the cylinder x2 + y2 = 36.
rewona [7]

Answer:

Step-by-step explanation:

From the given information:

The domain D of integration in polar coordinates can be represented by:

D = {(r,θ)| 0 ≤ r ≤ 6, 0 ≤ θ ≤ 2π) &;

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y =\dfrac{\partial z}{\partial x} , x = \dfrac{\partial z}{\partial y}

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= \iint_D \sqrt{x^2 +y^2 +1 } \ dA

= \int^{2 \pi}_{0} \int^{6}_{0} \ r  \sqrt{r^2 +1 } \ dr \ d \theta

=2 \pi \int^{6}_{0} \ r  \sqrt{r^2 +1 } \ dr

= 2 \pi \begin {bmatrix} \dfrac{1}{3}(r^2 +1) ^{^\dfrac{3}{2}} \end {bmatrix}^6_0

= 2 \pi \times \dfrac{1}{3}  \Bigg [ (37)^{3/2} - 1 \Bigg]

= \dfrac{2 \pi}{3} \Bigg [37 \sqrt{37} -1 \Bigg ]

3 0
2 years ago
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