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4 years ago

Need this geometry answer quick!(I think its answer c but not sure)

Mathematics

1 answer:

Vsevolod [243]4 years ago

7 0

180 is the answer....

angle B =50
angle C = 130

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Help Please, anyone good at domain values of composite functions.

Delicious77 [7]

Answer:

second choice

Step-by-step explanation:

$(f \circ g)(x) =$

$= f(g(x))$

$= f(-7x + 7)$

$= \sqrt{-7x + 7 + 3}$

$= \sqrt{-7x + 10}$

Let x = 5:

$(f \circ g)(5) = \sqrt{-7(5) + 10}$

$(f \circ g)(5) = \sqrt{-35 + 10}$

$(f \circ g)(5) = \sqrt{-25}$

Since using x = 5 gives the square root of a negative number, 5 is not in the domain of the composition of the functions.

Answer: second choice

3 0

3 years ago

4. From a faculty of six professors, six associate professors, ten assistant professors, and twelve instructors, a committee of

Degger [83]

Answer:

P=0.228

Step-by-step explanation:

We know that from a faculty of six professors, six associate professors, ten assistant professors, and twelve instructors, a committee of size six is to be selected.

Therefore, we have 34 people.

We calculate the number of possible combinations:

$C^{34}_6=\frac{34!}{6!(34-6)!}=1344904\\$

Of the 6 professors we choose 2, and of the other 28 people we choose 4.

We calculate the number of favorable combinations:

$C_2^6\cdot C_4^{28}=\frac{6!}{2!(6-2)!}\cdot \frac{28!}{4!(28-4)!}=15\cdot 20475=307125\\$

Therefore, the probability is:

$P=\frac{307125}{1344904}=0.228$

8 0

3 years ago

490 animals increase at an annual rate of 20%

Radda [10]

My final answer is automatic 588

5 0

3 years ago

Read 2 more answers

PLEASE HELP PLEASE HELP

Yakvenalex [24]

Answer:

Option D - Line P

Step-by-step explanation:

Look at the coordinates in the (x, y) column of Max's table and see which line has the same points.

8 0

2 years ago

Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE

EleoNora [17]

Answer:

0, 10

Step-by-step explanation:

The given function is:

$g(y) = \frac{y-5}{y^2-3y+15}$

According to the quotient rule:

$d(\frac{f(y)}{h(y)}) = \frac{f(y)*h'(y)-h(y)*f'(y)}{h^2(y)}$

Applying the quotient rule:

$g(y) = \frac{y-5}{y^2-3y+15}\\g'(y)=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}$

The values for which g'(y) are zero are the critical points:

$g'(y)=0=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}\\(y-5)*(2y-3)-(y^2-3y+15)=0\\2y^2-3y-10y+15-y^2+3y-15\\y^2-10y=0\\y=\frac{10\pm \sqrt 100}{2}\\y_1=\frac{10-10}{2}= 0\\y_2=\frac{10+10}{2}=10$

The critical values are y = 0 and y = 10.

5 0

3 years ago