The measure of angle A is 144.3 degrees and the angle to cut the molding is 54.3 degrees
<h3>How to solve for angle A?</h3>
Start by solving the acute part of angle A using the following sine function
sin(Ax) = (30 - 4)/32
Evaluate the quotient
sin(Ax) = 0.8125
Take the arc sin of both sides
Ax = 54.3
The measure of angle A is then calculated as:
A = 90 + Ax
This gives
A = 90 + 54.3
Evaluate
A = 144.3
Hence, the measure of angle A is 144.3 degrees
<h3>The angle to cut the molding</h3>
In (a), we have:
Ax = 54.3
This represents the angle where the molding would be cut
Hence, the angle to cut the molding is 54.3 degrees
Read more about angles at:
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Answer:
log(x^7·y^2)
Step-by-step explanation:
The applicable rules are ...
... log(a^b) = b·log(a)
... log(ab) = log(a) +log(b)
_____
The first term, 7log(x) can be rewritten as log(x^7). Note that an exponentiation operator is needed when this is written as text.
The second term 2log(y) can be rewritten as log(y^2). These two rewrites make use of the first rule above.
Now, you have the sum ...
... log(x^7) +log(y^2)
The second rule tells you this can be rewritten as ...
... log(x^7·y^2) . . . . . seems to match the intent of the 3rd selection
Answer:
D
Step-by-step explanation:
observe, grouping by gcd not 1
28r² + 35ry – 4xr – 5xy
= 28r² – 4xr + 35ry – 5xy
= 4r(7r – x) + 5y(7r –x)
= (4r + 5y)(7r – x)
The answer to the above question can be explained as under -
We know that, the sum of angles of triangle is 180°.
So, vertex angle plus base angles are equal are equal to 180°.
Let the vertex angle be represented by "v" and base angles be represented by "b".
Thus, v + b + b = 180°
So, v + 2b = 180°
Next, the question says, the vertex angle is 20° less than the sum of base angles.
Thus, 2b - 20° = v
<u>Thus, we can conclude that the correct option is A) v + 2b = 180°, 2b - 20° = v</u>
Answer:
The answer is "It would decrease, but not necessarily by 8%".
Step-by-step explanation:
They know that width of the confidence level is proportional to a confidence level. As just a result, reducing the confidence level decreases the width of a normal distribution, but not with the amount of variance in the confidence level. As just a result, when a person teaches a 90% standard deviation rather than a 98 percent normal distribution, the width of the duration narrows.
