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ipn [44]
2 years ago
13

A proportional relationship can be represented in the form of a

Mathematics
1 answer:
lidiya [134]2 years ago
4 0

Answer:

a graph

Step-by-step explanation:

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Can someone help me i will give brainliest to whom ever right 28=x+13
frozen [14]
The answer is that x equals 15
6 0
2 years ago
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M is between points C and Q. CM=17 and CQ=21. What is MQ?
katrin2010 [14]
Sometimes it helps to draw it out...

     (17)
C_______M_________Q
|___________________|
             21

therefore, CM + MQ = CQ
17 + MQ = 21
MQ = 21 - 17
MQ = 4 <===
8 0
3 years ago
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Please answer this question now
Alex787 [66]

Answer:

541.7 (m2)

Step-by-step explanation:

Applying the sine theorem:

WV/sin(X) = XV/sin(W)

=> WV = XV*sin(X)/sin(W) = 37*sin(50)/sin(63) = 31.81

Angle V = 180 - X - W = 180 - 50 - 63 = 67

Denote WH is a height of the triangle VWX, H lies on XV

=> WH = WV*sin(V) = 31.81*sin(67) =  29.28

=> Area of triangle VWX is calculated by:

S = side*height/2 = XV*WH/2 = 37*29.28/2 = 541.7 (m2)

8 0
3 years ago
Find the number b such that the line y = b divides the region bounded by the curves y = 36x2 and y = 25 into two regions with eq
Gemiola [76]

Answer:

b = 15.75

Step-by-step explanation:

Lets find the interception points of the curves

36 x² = 25

x² = 25/36 = 0.69444

|x| = √(25/36) = 5/6

thus the interception points are 5/6 and -5/6. By evaluating in 0, we can conclude that the curve y=25 is above the other curve and b should be between 0 and 25 (note that 0 is the smallest value of 36 x²).

The area of the bounded region is given by the integral

\int\limits^{5/6}_{-5/6} {(25-36 \, x^2)} \, dx = (25x - 12 \, x^3)\, |_{x=-5/6}^{x=5/6} = 25*5/6 - 12*(5/6)^3 - (25*(-5/6) - 12*(-5/6)^3) = 250/9

The whole region has an area of 250/9. We need b such as the area of the region below the curve y =b and above y=36x^2 is 125/9. The region would be bounded by the points z and -z, for certain z (this is for the symmetry). Also for the symmetry, this region can be splitted into 2 regions with equal area: between -z and 0, and between 0 and z. The area between 0 and z should be 125/18. Note that 36 z² = b, then z = √b/6.

125/18 = \int\limits^{\sqrt{b}/6}_0 {(b - 36 \, x^2)} \, dx = (bx - 12 \, x^3)\, |_{x = 0}^{x=\sqrt{b}/6} = b^{1.5}/6 - b^{1.5}/18 = b^{1.5}/9

125/18 = b^{1.5}/9

b = (62.5²)^{1/3} = 15.75

8 0
3 years ago
The 4 - pack is priced at £2.04
Archy [21]
The 4 Pack is the better value ;)
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2 years ago
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