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3 years ago

A B 21 48° 6x-24 D [?] << <[ ] 11 C

Mathematics

1 answer:

r-ruslan [8.4K]3 years ago

3 0

The shape in the figure is a composite shape with two triangles, and the inequality is 4 < x < 12

<h3>How to determine the range of value of x?</h3>

In the given figure, the side lengths AB and AD are equivalent.

This means that, we have the following angle relationship:

CAD < CAB

Substitute known values

6x - 24 < 48

Add 24 to both sides

6x < 72

Divide both sides by 6

x < 12

Also, we have the following angle relationship:

CAD > 0

Substitute known values

6x - 24 > 0

Add 24 to both sides

6x < 24

Divide both sides by 6

x < 4

This means that:

x < 4 and x < 12

Combine both inequalities

4 < x < 12

Hence, the inequality is 4 < x < 12

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Answer this pls..................

Sindrei [870]

Answer:

so we know:

total of 20 students

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

chemistry has a total of 14 students

1 2 3 4 5 6 7 8 9 10 11 12 13 14

8 students are in physics

1 2 3 4 5 6 7 8

3 are in all of them

1 2 3

1) underline the main 3

total of 20 students

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

chemistry has a total of 14 students

1 2 3 4 5 6 7 8 9 10 11 12 13 14

8 students are in physics

1 2 3 4 5 6 7 8

2) lets bold the physics students, if you are in physics you have to do chem so that means 8 of the chem students also take physics.

total

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

chemistry

1 2 3 4 5 6 7 8 9 10 11 12 13 14

physics

1 2 3 4 5 6 7 8

3) now we bring biology into the "equation"

off the bat we know 3 are definitely in Biology.

so if 4 study ONLY biology we can cross them off our list see the {}

total

1 2 3 4 5 6 7 8 {9 10 11 12} 13 14 15 16 17 18 19 20

chemistry

1 2 3 4 5 6 7 8 {9 10 11 12} 13 14

physics

1 2 3 4 5 6 7 8

biology

1 2 3 

4) so we have a total of 20 students, 3 in all of the classes (underlined), 8 in physics (bolded). If there 14 in chemistry, 4 take ONLY chemistry {}, 8 take chemistry as well (bolded), that leaves two students, so they must belong in biology. lets italicise them.

total

1 2 3 4 5 6 7 8 {9 10 11 12} 13 14 15 16 17 18 19 20

chemistry

1 2 3 4 5 6 7 8 {9 10 11 12} 13 14

physics

1 2 3 4 5 6 7 8

biology

1 2 3  13 14

5) out of our total of 20 students we know 14 of the students classes, so the remaining six must be in biology.

1 2 3 4 5 6 7 8 {9 10 11 12} 13 14 15 16 17 18 19 20

chemistry

1 2 3 4 5 6 7 8 {9 10 11 12} 13 14

physics

1 2 3 4 5 6 7 8

biology

1 2 3 13 14 15 16 17 18 19 20

6) so that was our "diagram" so now lets count how many are in chemistry and biology but NOT physics.

1 2 3 4 5 6 7 8 {9 10 11 12} 13 14 15 16 17 18 19 20

chemistry

-1 2 3 4 5 6 7 8- {9 10 11 12} 13 14

physics

-1 2 3 4 5 6 7 8-

biology

-1 2 3 -13 14 15 16 17 18 19 20

there are 14 students in chemistry and 11 in biology, 5 of them are in both classes so 14-5= 9 and 11-5= 6 9+6=16. BUT 8 of those sixteen are also in -physics- so we arent counting them, we put those students in dashes. so lets count the students out of the dashes. so twelve of them are in chemistry and biology but not physics.

7) so how many are in just biology but not the other classes? lets look at the graph.

1 2 3 4 5 6 7 8 {9 10 11 12} 13 14 15 16 17 18 19 20

chemistry

1 2 3 4 5 6 7 8 {9 10 11 12} 13 14

physics

1 2 3 4 5 6 7 8

biology

1 2 3 13 14 15 16 17 18 19 20

in the class of biology students 1 2 3 13 and 14 take other classes that leaves students 15 16 17 18 19 and 20. so 6 students take ONLY biology.

sorry its long but i hope this helps!

3 0

3 years ago

4. Simplify the expression. (6 points)

Lunna [17]

4. Simplify the expression.
sine of x to the second power minus one divided by cosine of negative x

(1−sin2(x))/(sin(x)−csc(x))

sin2x+cos2x=1 1−sin2x=cos2x

cos2(x)/(sin(x)−csc(x)) csc(x)=1/sin(x) cos2(x)/(sin(x)− 1/sin(x))= cos2(x)/((sin2(x)− 1)/sin(x)) sin2(x)− 1=-cos2(x) cos2(x)/(( -cos2(x))/sin(x))
=-sin(x)
the answer is the letter a) -sin x

5. Find all solutions in the interval [0, 2π). (6 points)
sin2x + sin x = 0
 using a graphical tool
the solutions x1=0 x2=pi x3=3pi/2
the answer is the letter D) x = 0, π, three pi divided by two