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2 years ago

0.3(3a-4)-0.05(8a)(-7a)

Mathematics

1 answer:

Anit [1.1K]2 years ago

3 0

Answer:

2.8a²+0.9a - 1.2

Step-by-step explanation:

Given the expression 0.3(3a-4)-0.05(8a)(-7a)

Expand using the distributive law

0.3(3a-4)-0.05(8a)(-7a)

0.3(3a)-0.3(4)-0.05(8a)(-7a)

0.9a-1.2 - 0.05(8a)(-7a)

0.9a - 1.2 + 2.8a²

Rearrange

2.8a²+0.9a - 1.2

Hence the required expression is 2.8a²+0.9a - 1.2

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What two variables are likely to have a negative correlation

kogti [31]

Answer:

Negative Correlation

The variable x - value increases as y- value decreases is called negative correlation.

Step-by-step explanation:

Explanation:-

Negative correlation:-

The negative correlation is a relation between two variables such that they are part of a function in which depends and independent variables move in different directions in terms of value.

Positive correlation

The variable x - value increases as y- value increases is called positive correlation.

Negative Correlation

The variable x - value increases as y- value decreases is called negative correlation.

No correlation

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3 years ago

ABC has side lengths that are each equal to

likoan [24]

Answer:

Step-by-step explanation:no

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3 years ago

Read 2 more answers

Suppose that documentation lists the average sales price of a single-family home in the metropolitan Dallas/Ft. Worth/Irving, Te

Julli [10]

Answer:

"0.0373" seems to be the appropriate solution.

Step-by-step explanation:

The given values are:

n₁ = 43

n₂ = 35

$\bar{x_1}=217800$

$\bar{x_2}=204700$

$\sigma_1=30300$

$\sigma_2=33800$

Now,

The test statistic will be:

⇒ $Z=\frac{\bar{x_1}-\bar{x_2}}\sqrt{\frac{\sigma_1^2}{n_1} +\frac{\sigma_2^2}{n_2} }$

On substituting the given values in the above formula, we get

⇒ $=\frac{217800-204400}{\sqrt{\frac{(30300)^2}{43} +\frac{(33800)^2}{35} } }$

⇒ $=\frac{217800-204400}{\sqrt{\frac{918090000}{43} +\frac{1142440000}{35} } }$

⇒ $=\frac{13400}{7347.93}$

⇒ $=1.7828$

then,

P-value will be:

= $P(Z>1.7828)$

= $0.0373$

8 0

2 years ago

Would appreciate the help !

aleksandr82 [10.1K]

This is one pathway to prove the identity.

Part 1

$\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 2

$\frac{\sin^2(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)-\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-(\cos(\theta)-\cos^2(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-\cos(\theta)+\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 3

$\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\$

As the steps above show, the goal is to get both sides be the same identical expression. You should only work with one side to transform it into the other. In this case, the left side transforms while the right side stays fixed the entire time. The general rule is that you should convert the more complicated expression into a simpler form.

We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity $\sin^2(\theta)+\cos^2(\theta) = 1$ in the second to last step. I broke the steps into three parts to hopefully make it more manageable.

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3 years ago

The fractions 3/4, 6/8, 9/12, and 12/16 are examples of _____.

Maru [420]

Answer:

Equivalent fractions

Step-by-step explanation:

All of these fractions can be reduced to 3/4, so they're examples of equivalent fractions.

8 0

3 years ago