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4 years ago

Let u = <-5, -9>, v = <6, 8>. Find -8u - 2v.

2 answers:

8 0

$\vec{u}=[-5;-9]\\\vec{v}=[6;\ 8]\\\\-8\vec{u}-2\vec{v}=-8\cdot[-5;-9]-2\cdot[6;\ 8]\\\\=[(-9)(-5);(-8)(-9)]-[(2)(6);\ (2)(8)]\\\\=[40;\ 72]-[12;\ 16]=[40-12;\ 72-16]\\\\=\boxed{[28;\ 56]}$

brilliants [131]4 years ago

5 0

If u and v are indeed vectors,
$- 8u - 2v = - 8( - 5, - 9) - 2(6,8) \\ = (40,72) - (12,16) \\ = (28,56)$
,

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5x - 7(2- 4x) = 8 - 3(6 - 6x) = 9a - 4(2a + 5) =

kati45 [8]

Answer:

the answer to 5x-7(2-4x)= =33x−14

the answer to 8 - 3(6 - 6x) ==18x−10

the answer to 9a - 4(2a + 5) ==a−20

Step-by-step explanation:

8 0

4 years ago

What is the mean for this set of data?

oee [108]

The mean is 14

add all the numbers up and divide the sum by 6 because that’s how many numbers you’re adding together.

12+10+14+18+14+16
= 84

84/6 = 14

5 0

3 years ago

Image attached below. Please help.

goldfiish [28.3K]

Answer:

T,F,T,F,T

Step-by-step explanation:

1. It crosses it there becuase 4(0)-15=-15 for y

2: both top and middle answers are -23, so continuous

3: it increases over that period because the exponent keeps getting bigger

4: No, its not decreasing because -12 plugged in is far less than -23 when -2 is plugged in, meaning its increasing

5: There is no number that would cause an imaginary number or make you divide by zero, meaning all real numbers are fine.

4 0

4 years ago

Delta makes 12-volt car batteries. These batteries are known to be normally

blondinia [14]

Answer:

The probability that Delta car batteries last between three and four years

P(36≤X≤48) = 0.5188

The percentage of that Delta car batteries last between three and four years

P(3≤X≤4) = 52%

Step-by-step explanation:

Step(i):-

Given that the sample size n =12 -volt car batteries

Let 'X' be a Random variable in a normal distribution

Given that mean of the normal distribution = 45 months

Given that the Standard deviation of the normal distribution = 8months

Step(ii):-

Let X₁ = 3 years = 12 × 3 = 36 months

$Z_{1} = \frac{x_{1} -mean}{S.D} = \frac{36-45}{8} = -1.125$

Let X₂ = 4 years = 12 × 4 = 48 months

$Z_{2} = \frac{x_{2} -mean}{S.D} = \frac{48-45}{8} = 0.375$

Step(iii):-

The probability that Delta car batteries last between three and four years

P(36≤X≤48) = P(-1.125≤Z≤0.375)

= P(Z≤0.375) - P(Z≤-1.125)

= 0.5 +A(0.375) - (0.5-A(1.125)

= 0.5 + 0.1480 - (0.5 -0.3708)

= 0.1480 + 0.3708

= 0.5188

Final answer:-

The probability that Delta car batteries last between three and four years

P(36≤X≤48) = 0.5188

The percentage of that Delta car batteries last between three and four years

P(3≤X≤4) = 52%

5 0

3 years ago

I need help with this problem 2(w+7)+6(3w−1)

Lubov Fominskaja [6]

Use distributive property: a(b + c) = ab + ac.

2(w + 7) + 6(3w - 1) = (2)(w) + (2)(7) + (6)(3w) + (6)(-1)

= 2w + 14 + 18w - 6

combine like terms

= (2w + 18w) + (14 - 6) = 20w + 8

<h3>Answer: 2(w + 7) + 6(3w - 1) = 20w + 8</h3>

5 0

4 years ago

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