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2 years ago

Which statement must be true to prove | || m?

Mathematics

1 answer:

Yakvenalex [24]2 years ago

6 0

Answer C is correct. If angle 1 + angle 2 = 180 degrees then L is parallel to M.
Co interior angles are only supplementary in parallel lines

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at a certain conference of 100 people there are 29 indian women and 23 indian men. out of these indian people, 4 are doctors and

tangare [24]

Answer:

4 women doctors

Step-by-step explanation:

because if 24 are men or doctor the 4 gotta be women doctor

3 0

2 years ago

The vertices of ABC are A(-3,5), B(-2,3), and C(-5,3). If ABC is reflected across the y-axis to produce the image A'B'C', find t

Nat2105 [25]

Answer:

$C' =(5,3)$

Step-by-step explanation:

Given

$A = (-3,5)$

$B = (-2,3)$

$C = (-5,3)$

Reflection across y-axis

Required

Find C'

For reflection across y-axis, the rule is:

$(x,y) \to (-x,y)$

So, the image of C is:

$(-5,3) \to (-(-5),3)$

$(-5,3) \to (5,3)$

Hence:

$C' =(5,3)$

7 0

2 years ago

Which 2 expressions are equivalent?

LuckyWell [14K]

Answer:

i had this question it's right

6 0

2 years ago

Read 2 more answers

Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer

agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming: the function is $f(x)=x^{2}$ in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

$g'(0)=g'(1)$

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

$g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0$

$g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2$

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

$f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]$

This is what the Bilateral Theorem says:

$\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L$

4 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