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Nadya [2.5K]
3 years ago
10

explain how (Cos^2 / sin^2)-1 equals to (cos^2 - sin^2) / sin^2. (answer comes up in photomath but I need to know how to do it)

Mathematics
1 answer:
NemiM [27]3 years ago
8 0

Answer:

Check the photo tagged on the answer

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There are 150 students and 3/5 of these students are girls. 2/3 play school activities. how many play activities?
Murrr4er [49]

Answer:

100 or 60

Step-by-step explanation:

Since the question does not specify whether \frac{2}{3} refers to only the girls or all students, both will be calculated for your convenience.

If 2/3 refers to all students:

150*\frac{2}{3}

=100

If 2/3 refers to the girls:

150*\frac{3}{5}*\frac{2}{3}

=150*\frac{2}{5}

=60

8 0
3 years ago
Read 2 more answers
Evaluate <br> f(12)=3/2x+14<br> f(-4)=3/2x+14<br> f(0)=3/2x+14
Illusion [34]

When f(x) is substituted with a number, you just plug the number into wherever you see x:

_____________

f(12) = 3/2x + 14

f(12) = 3/2(12) + 14

f(12) = 18 + 14

f(12) = 32

f(-4) = 3/2x + 14

f(-4) = 3/2(-4) + 14

f(-4) = -6 + 14

f(-4) = 8

f(0) = 3/2(0) + 14

f(0) = 0 + 14

f(0) = 14

7 0
3 years ago
A circle has a center of (5, 2) and a radius of 10. Which the fallowing points will the circle pass through ?
Sergeeva-Olga [200]

Answer:

centre of the circle(h,k)=(5,2)

radius(r)=10

Step-by-step explanation:

Now,

equation of the circle is giveb by (x-h)^2+(y-k)^2=r^2

or,(x-5)^2+(y-2)^2=(10)^2

or,x^2-10x+25+y^2-4y+4=100

or,x^2+y^2-10x-4y+29-100=0

or,x^2+y^2-10x-4y-71=0

5 0
3 years ago
How many positive integers between 5 and 31
Anestetic [448]

Answer:

Part (A): There are 9 integers between 5 and 31 which are divisible by 3.

Part (B): There are 6 integers between 5 and 31 which are divisible by 4.

Part (C): There are 2 integers between 5 and 31 which are divisible by 3 and by 4.

Step-by-step explanation:

Consider the provided information.

Part (A) we need to find how many integers between 5 and 31 are divisible by 3.

Between 5 and 31 there are 25 integers.

According to quotient rule: \frac{25}{3} \approx8.33

That means either 8 or 9 integers are divisible by 3 as 8.33 lies between 8 and 9.

The integers are: 6, 9, 12, 15, 18, 21, 24, 27, 30

Hence, there are 9 integers between 5 and 31 which are divisible by 3.

Part (B) we need to find how many integers between 5 and 31  are divisible by 4.

Between 5 and 31 there are 25 integers.

According to quotient rule: \frac{25}{4} \approx6.25

That means either 6 or 7 integers are divisible by 4, as 6.25 lies between 6 and 7.

The integers are: 8, 12, 16, 20, 24, 28

Hence, there are 6 integers between 5 and 31 which are divisible by 4.

Part (C) we need to find how many integers between 5 and 31 are divisible by 3 and by 4

Between 5 and 31 there are 25 integers.

Integers should be divisible by 3 and by 4, that means integers should be divisible by 3×4=12.

According to quotient rule: \frac{25}{12} \approx2.08

That means either 2 or 3 integers are divisible by 3 and by 4 or 12, as 2.08 lies between 2 and 3.

The integers are: 12, 24,

Hence, there are 2 integers between 5 and 31 which are divisible by 3 and by 4.

7 0
3 years ago
EXAMPLE 3 A wooden cube has edges measuring 5 centimeters each. Find the
Serjik [45]
5(edge)•5(edge)=25
25(side) •6 total sides

150 cm
3 0
3 years ago
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