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2 months ago

The law of cosines is a^2+b^2-2abcosC=c^2. Find the value of 2abcosC.

Mathematics

2 answers:

Zina [12.3K]2 months ago

5 0

Response:

(D) $7=2abcosC$

Step-by-step breakdown:

The given figure shows c=1, a=2, and b=2, so we apply the law of cosines and obtain

$c^2=a^2+b^2-2abcosC$

By substituting the provided figures, we solve for

$(1)^2=(2)^2+(2)^2-2abcosC$

$1=4+4-2abcosC$

$1=8-2abcosC$

$1-8=-2abcosC$

$-7=-2abcosc$

$7=2abcosC$

Therefore, the value of $2abcosC$ is $7$ .

Thus, option (D) is the right choice.

zzz [12.3K]2 months ago

4 0

2ab x cos(C) = 7 since

a^2 + b^2 - 2ab x cos(C) = c^2
2ab x cos(C) = a^2 + b^2 - c^2
2ab x cos(C) = 2^2 + 2^2 - 1^2
2ab x cos(C) = 7

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Suppose you earn $5.50 an hour working at the plant nursery on weekends. Given your table of values in part a, write an equation

Leona [12618]

Answer:

For the first year, the total amount comes to $266, and with an hourly wage of $5.50, you can express this as the equation 5.50x=266

Step-by-step explanation:

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1 month ago

The weights of certain machine components are normally distributed with a mean of 8.01 g and a standard deviation of 0.06 g. Fin

PIT_PIT [12445]

Answer:

Option D) 7.90 g and 8.12 g

Step-by-step explanation:

The details provided in the question are:

Mean, μ = 8.01 g

Standard Deviation, σ = 0.06 g

The weights are distributed in a bell-shaped normal distribution.

Formula:

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

We need to determine the value of x for which the probability is 0.03.

P(X > x)

$P( X > x) = P( z > \displaystyle\frac{x - 8.01}{0.06})=0.03$

$= 1 -P( z \leq \displaystyle\frac{x - 8.01}{0.06})=0.03$

$=P( z \leq \displaystyle\frac{x - 8.01}{0.06})=0.97$

Using the standard normal z table, we find that,

$\displaystyle\frac{x - 8.01}{0.06} = 1.881\\\\x = 8.12$

Thus, the value of 8.17 g separates the upper 3% of the weights.

P(X < x)

$P( X < x) = P( z < \displaystyle\frac{x - 8.01}{0.06})=0.03$

From the standard normal z table, we derive,

$\displaystyle\frac{x - 8.01}{0.06} = -1.881\\\\x = 7.90$

Consequently, 7.90 separates the lower 3% of the weights.

Option D) 7.90 g and 8.12 g

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7 0

2 months ago

Write a subtraction fact with the same difference as 16-7

PIT_PIT [12445]

Subtracting seven from sixteen results in nine
The solution is nine

8 0

2 months ago

Read 2 more answers

A survey asked students whether they have any siblings and pets. the survey data are shown in the relative frequency table.

PIT_PIT [12445]

Response:

The probability that a student has a pet, given they do not have any siblings is:

Option: D ( 60%)

Step-by-step breakdown:

Let A represent the situation where a student lacks a sibling.

Let B signify the occurrence that a student has a pet.

Consequently, A∩B refers to the event in which a student is without siblings but possesses a pet.

Let P denote the chance of an event happening.

We need to determine:

P(B|A)

From our knowledge:

$P(B|A)=\dfrac{P(A\bigcap B)}{P(A)}$

From the data provided:

P(A)=0.25

and P(A∩B)=0.15

Thus,

$P(B|A)=\dfrac{0.15}{0.25}\\\\P(B|A)=\dfrac{3}{5}\\\\P(B|A)=0.6$

which expressed as a percentage is:

$0.6\times 100=60\%$

Therefore, the probability is:

60%

5 0

2 months ago