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AnnZ [28]
3 years ago
12

What is the vaule of ( f o g)(-5) ?

Mathematics
1 answer:
saul85 [17]3 years ago
3 0

Answer:

It's C

Step-by-step explanation:

g(-5) = (-5)^2 = 25

so (f o g)(-5)

= 2(25) + 1

= 51  

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Which number will make the fractions equal? 5/10=?/100
Mashcka [7]

50 hopefully this will help you

4 0
3 years ago
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Quadrilateral BCDE is inscribed inside a circle as shown below. Write a proof showing that angles C and E are supplementary.
erma4kov [3.2K]

Each of those angles are Inscribed angles. A useful fact is that the measure of the intercepted arc of an inscribed angle is twice the measure of the angle.

The intercepted arc of ∠C is arc BED. The intercepted arc of ∠E is arc BCD.

m(arc BED) + m(arc BCD) = 2( m∠C + m∠E)

The two arcs combined make u p the entire circle, so the sum of their measures is 360.

360 = 2(m∠C + m∠E)
180 = m∠C + m∠E

The angles are supplementary.

4 0
3 years ago
-8 is a term.<br> Yes <br> No
pantera1 [17]

Answer:

yes

Step-by-step explanation:

3 0
3 years ago
An amusement park offers two options. Option 1 involves a $10 admission fee plus $0.50 per ride. Option 2 involves a $6 admissio
Art [367]

Answer:

16 rides

Step-by-step explanation:

Option 1 . Admission fee = $10

                Each ride = $0.50

Option 2 . Admission fee = $6

                Each ride = $0.75

Let no. of rides be x

So, cost of ride according to option 1 = 0.50x

So, total cost after having x rides according to option 1 :

= 10+0.50x  ---1

Cost of ride according to option 2 = 0.75x  

So, total cost after having x rides according to option 2 :

= 6+0.75x  --2

Now to find the beak even point i.e. having the same cost

Equate 1 and 2

10+0.50x=6+ 0.75x

10-6= 0.75x-0.50x

4= 0.25x

\frac{4}{0.25} =x

16=x

Thus for 16 rides , the two options have the same cost .

Hence the break even point is 16 rides

5 0
4 years ago
In a group of 60 students, 15 liked maths only, 20 liked science only and 5 did not liked any of two subjects. How many of them
evablogger [386]

Answer:

35

Step-by-step explanation:

Given

n (A) = 15

n (B) = 20

Students who do not like any subject = 5

Hence, number of students who would like either both or either of the two subjects = 60-5 = 55

n (A or B) = n (A) + n (B) - n (A and B)

Number of students linking both the subjects

55 - 15-20

= 55-35 = 20

Number of students linking only one subject = 60-20-5 = 35

4 0
3 years ago
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