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dimulka [17.4K]
3 years ago
7

1÷1+cosa+1÷1_cos a please solve question​

Mathematics
1 answer:
zloy xaker [14]3 years ago
4 0

ok so THIS IS UR ANSWER I THINK PLEASE MAKE ME BRAINLIEST

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Any one good with proportion
olasank [31]

Answer:

4. No

5. No

6. No

7. No

8. No

9a. 2 : 3 = 8 : 12 Extremes: 2, 12. Means: 3, 8

9b. 90 : 36 = 135 : 54 Extremes: 90, 54. Means: 36, 135

Step-by-step explanation:

4. 12 : 25 and 6 : 13 are NOT equal. NO.

5. 64 : 120 = 8 : 15 and 60 : 100 = 3 : 5. NOT equal. NO

6. 37 : 115 and 296 : 966 =  148 : 483. NOT equal. NO

7. 300 : 200 = 3 : 2 and 6 : 5 are NOT equal. NO.

8. 8 : 10 = 4 : 5 and 11 : 14 are NOT equal. NO.

9a. 2 : 3 = 8 : 12 Extremes: 2, 12. Means: 3, 8

9b. 90 : 36 = 135 : 54 Extremes: 90, 54. Means: 36, 1135

6 0
3 years ago
Solve the equation. Check your solutions when necessary.<br><br><br>2x² − 7x = 4
prisoha [69]

Answer:

x= -1/2 = -0.500

x=4

Step-by-step explanation:

2x^2-7x-4=0

2x^2-8x+1x-4=0

2x(x-4)+1(x-4)=0

(2x+1)(x-4)=0

2x+1=0...2x= -1.. x= -1/2         x-4=0 .. x=4

checking the solution

2(-1/2)^2-7(-1/2)=4.....  4=4

2(4)^2-7(4)=4....... 4=4

5 0
3 years ago
Triangles PIC Included
ZanzabumX [31]

Answer:

sin70° = x/6

Step-by-step explanation:

When we draw out our picture and triangle, we should see that our angle is on the ground 70°, the height <em>x</em> is across from the angle and is the vertical leg up, and that our hypotenuse is equal to 6:

sin70° = x/6

6 0
3 years ago
Read 2 more answers
Grace wants to find the probability of a family of 3 children having 2 boys and 1 girl. Use a tree diagram to list the possibili
pychu [463]

Answer:

Step-by-step explanation:

My approach was to draw out the probabilities, since we have 3 children, and we are looking for 2 boys and 1 girl, the probabilities can be Boy-Boy-Girl, Boy-Girl-Boy, and Girl-Boy-Boy. So a 2/3 chance if you think about it, your answer 2/3 can't be correct. If we assume that boys and girls are born with equal probability, then the probability to have two girls (and one boy) should be the same as the probability to have two boys and one girl. So you would have two cases with probability 2/3, giving an impossible 4/3 probability for both cases. Also, your list "Boy-Boy-Girl, Boy-Girl-Boy, and Girl-Boy-Boy" seems strange. All of those are 2 boys and 1 girl, so based on that list, you should get a 100 percent chance. But what about Boy-Girl-Girl, or Girl-Girl-Girl? You get 2/3 if you assume that adjacencies in the (ordered) list are important, i.e., "2 boys and a girl" means that the girl was not born between the boys.

5 0
2 years ago
I am a two digit square number. the sum of my digits is 13. what Square number am I
skelet666 [1.2K]
49 Because 7x7 is 49 & 4+9=13
5 0
3 years ago
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