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3 years ago

Solve for k. m=−8j−9k+9

Mathematics

2 answers:

Svet_ta [14]3 years ago

8 0

(m + 8j - 9) / -9 ............

inn [45]3 years ago

7 0

Answer: k = -8/9j + -1/9m + 1

Step 1: Flip the equation.−8j − 9k + 9 = m

Step 2: Add -9 to both sides.−8j − 9k + 9 + −9 = m + −9−8j − 9k = m − 9

Step 3: Add 8j to both sides.−8j −9k + 8j = m −9 + 8j−9k = 8j + m −9

Step 4: Divide both sides by -9.
-9k/-9 = 8j + m - 9/-9
k = -8/9j + -1/9m + 1

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A recipe requires ¾ cup of nuts for 1 batch of muffins. Write the number of batches of muffins that can be made using 7 ½ cups o

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Answer:

10 batches

Step-by-step explanation:

3/4 cups = 1 muffins

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For x muffins,

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3 years ago

The ratio of the length 2.0 meters to the length 1.2 meters is

photoshop1234 [79]

Answer:

The Ration of length 2.0 meters to the length 1.2 meters = 5 : 3

Step-by-step explanation:

The given two length are 2.0 m and 1.2 m

Length L1 = 2.0 m And

Length L2 = 1.2 m

SO, The ratio of L1 : L2 = $\frac{2.0}{1.2}$

Or, The ratio of L1 : L2 = $\frac{20}{12}$

∴ The ratio of L1 : L2 = $\frac{5}{3}$

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3 years ago

What is 0.004% of 6800 round to the nearest hundredth

Elis [28]

Just remember that the word "of" means "multiply" in mathematical terms
Now lets get started working out this problem of yours
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Also, remember: $5$ and down is rounded to $0$ | $6$ and up is going closer to $10$

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3 years ago

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Anna [14]

Answer:

a) 0.4452

b) 0.0548

c) 0.0501

d) 0.9145

e) 6.08 minutes or greater

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.7 minutes

Standard Deviation, σ = 0.50 minutes.

We are given that the distribution of length of the calls is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(calls last between 4.7 and 5.5 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{4.7 - 4.7}{0.50} \leq z \leq \displaystyle\frac{5.5-4.7}{0.50}) = P(0 \leq z \leq 1.6)\\\\= P(z \leq 1.6) - P(z$

$P(4.7 \leq x \leq 5.5) = 44.52\%$

b) P(calls last more than 5.5 minutes)

$P(x > 5.5) = P(z > \displaystyle\frac{5.5-4.7}{0.50}) = P(z > 1.6)\\\\P( z > 1.6) = 1 - P(z \leq 1.6)$

Calculating the value from the standard normal table we have,

$1 - 0.9452 = 0.0548 = 5.48\%\\P( x > 5.5) = 5.48\%$

c) P( calls last between 5.5 and 6 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{5.5 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(1.6 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$

$P(5.5 \leq x \leq 6) = 5.01\%$

d) P( calls last between 4 and 6 minutes)

$P(4 \leq x \leq 6) = P(\displaystyle\frac{4 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(-1.4 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$