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3 years ago

Please guys help. Thank you

Mathematics

1 answer:

Zolol [24]3 years ago

3 0

Answer:

<h3>72 is the answer </h3>

Step-by-step explanation:

<h3>I hope I helped you ❤️❤️</h3>

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Which of the following is equal to 5/8÷2/7?

natta225 [31]

Answer:

5/8 * 7/2

Step-by-step explanation:

Whenever you divide fractions, you can use the 'keep change flip' method.

You keep 5/8, change the division sign to multiplication, and then flip the second fraction.

1. 5/8 / 7/2 - Keep 5/8

2. 5/8 * 7/2 - Change division sign

3. 5/8 * 2/7 - Flip the second fraction, 7/2 -> 2/7

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3 years ago

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Which transformation from the graph of a function f(x) describes the graph of g(x) =10f(x)

Elanso [62]

Multiplying a function by a constant factor (10 in this example) stretches the graph relative to the x-axis. If f(x) would be a sine, you would blow up the amplitude from -1..+1 to -10..+10.

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3 years ago

Please answer this as fast as you can In the graphic above, ΔABC ≅ ΔEDC by: HL. AAS. AAA. SSS.

Masja [62]

Answer:

Step-by-step explanation:

Triangles by definition have 3 sides. If the sides are corresponding then it is beneficial to us if they are the same length as well. If all 3 sides in one triangle are equal in length to the corresponding sides in another triangle, then the triangles are congruent by SSS (side-side-side). This is the case for us. Side EC is corresponding and congruent to side AC; side CD is corresponding and congruent to side CB; side ED is corresponding and congruent to side AB.

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3 years ago

find the coordinates of the point P on the parabola y=1-x^2 with domain 0≤x≤1 that minimize the area of the triangle enclosed by

coldgirl [10]

Let point P be with coordinates $(x_0,y_0).$ Find the equation of the tangent line.

1. If $y=1-x^2,$ then $y'=-2x.$

2. The equation of the tangent line at point P is

$y-y_0=-2x_0(x-x_0).$

Find x-intercept and y-intercept of this line:

when x=0, then $y=y_0+2x_0^2;$
when y=0, then $x=\dfrac{y_0}{2x_0}+x_0=\dfrac{y_0+2x_0^2}{2x_0}.$

The area of the triangle enclosed by the tangent line at P, the x-axis, and y-axis is

$A=\dfrac{1}{2}\cdot (2x_0^2+y_0)\cdot \left(\dfrac{y_0+2x_0^2}{2x_0}\right)=\dfrac{(y_0+2x_0^2)^2}{4x_0}.$

Since point P is on the parabola, then $y_0=1-x_0^2$ and

$A=\dfrac{(1-x_0^2+2x_0^2)^2}{4x_0}=\dfrac{(1+x_0^2)^2}{4x_0}.$

Find the derivative A':

$A'=\dfrac{2(1+x_0^2)\cdot 2x_0\cdot 4x_0-4(1+x_0^2)^2}{16x_0^2}=\dfrac{12x_0^4+8x_0^2-4}{16x_0^2}.$

Equate this derivative to 0, then

$12x_0^4+8x_0^2-4=0,\\ \\3x_0^4+2x_0^2-1=0,\\ \\D=2^2-4\cdot 3\cdot (-1)=16,\ \sqrt{D}=4,\\ \\x_0^2_{1,2}=\dfrac{-2\pm4}{6}=-1,\dfrac{1}{3},\\ \\x_0^2=\dfrac{1}{3}\Rightarrow x_0_{1,2}=\pm\dfrac{1}{\sqrt{3}}.$