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kodGreya [7K]
2 years ago
6

Select the correct answer.

Mathematics
2 answers:
Tamiku [17]2 years ago
8 0

Answer:

B

Step-by-step explanation:

SVEN [57.7K]2 years ago
3 0
Select the correct answer.
If a = 0.3 and b =0.5, what is the value of a+b?
A.8/10
B.8/9
C.80/99
D.88/100

Hey Love!<3 It would be

A.8/10
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The diameter of a round dinner plate is 10 inches. What's the area of the plate?
topjm [15]
A= pi times r*r so if you are using 3.14 as pi the answer is 78.5 because 5x5=25
25x3.14=78.5
5 0
3 years ago
Prove that 1/sec0-tan0= sec0+tan0
Leviafan [203]

Answer:

Step-by-step explanation:

\text{LHS}=\frac{1}{\sec \theta-\tan \theta}\\=\frac{\sec \theta+\tan \theta}{(\sec \theta-\tan \theta)(\sec \theta+\tan \theta)}\\=\frac{\sec \theta+\tan \theta}{\sec^{2} \theta-\tan^{2} \theta}\\=\frac{\sec \theta+\tan \theta}{1}\\=\sec \theta+\tan \theta\\=\text{RHS}

7 0
2 years ago
If the side length is increased by a factor of 3, how much larger is the perimeter of a face of the new cube?
Luden [163]
The <span>perimeter of a face of the new cube is 8 times larger than the previous cube.
</span>Let us say, the side of cube before increase is "a"
Then perimeter of the cube = 12a ,

because the cube has 12 sides.
And perimeter of a face of cube = 4a
Now, the side length is increased by factor of 3 i.e. new side is 3a
Thus, perimeter of new face of the cube = 4*3a = 12a

Therefore, the new perimeter is larger by = 12a - 4a = 8a 
6 0
3 years ago
If cos Θ = negative 4 over 7, what are the values of sin Θ and tan Θ?
FinnZ [79.3K]
Given cos theta is equal to - 4/ 7 then we can conclude that theta is in the second and third quadrants. In this case, the other leg is equal to square root of (7^2 - 4^2 ) equal to square root of 33. In this case, sin theta can be equal to +- square root of 33 / 7 and tan theta is equal to +-square root of 33 / 4.
7 0
3 years ago
What are the first three terms of the Arithmetic Sequence an=11-3(n-1)?
svlad2 [7]

Answer:

11,8,5

Step-by-step explanation:

an=11-3(n-1)

Let n=1

a1 = 11 - 3(1-1)

    = 11 -0

    =11

Let n=2

a2 = 11-3(2-1)

     = 11 -3(1)

     = 8

Let n=3

a3 = 11-3(3-1)

    = 11 -3(2)

    = 11 -6

    =5

5 0
3 years ago
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