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3 years ago

Ranking the correct answer brainliest !

Mathematics

2 answers:

nikklg [1K]3 years ago

6 0

Answer:

angle 3 =68°+47°=115°

Hope it helps you

AnnZ [28]3 years ago

4 0

Answer:

∠ 3 = 115°

Step-by-step explanation:

The exterior angle of a triangle is equal to the sum of the 2 opposite interior angles.

∠ 3 is an exterior angle of the triangle, then

∠ 3 = ∠ 1 + ∠ 2 = 68° + 47° = 115°

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Avary wanted to buy a gold necklace that cost $578. She pain an initial amount of $128 and agreed to pay the rest in monthly ins

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Answer:

75 each month

Step-by-step explanation:

578-128=450

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2 years ago

How do I find QS here?

juin [17]

Answer: 12.
To solve this problem, let's first solve for x. Thi is easiest done by figuring what QR is in terms of x using two equations, both from different lines.
In the first line: QR = 15 - 4x.
In the third line:QR = 13x - 1 - x = 12x - 1.
Now, we have to set these equations equal to each other. 15 - 4x = 12x - 115 = 16x - 116x = 16x = 1
Next, we take line three, 13x - 1, and substitute x as 1. The answer is 12.

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3 years ago

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Step-by-step explanation: sloppy or clean?

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3 years ago

A junk box in your room contains a dozen old batteries, five of which are totally dead. You start picking batteries one at a tim

MArishka [77]

Answer:

7/12 of a chance

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6 0

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

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

And

$y_0=1-\left(\pm\dfrac{1}{\sqrt{3}}\right)^2=\dfrac{2}{3}.$

Answer: two points: $P_1\left(-\dfrac{1}{\sqrt{3}},\dfrac{2}{3}\right), P_2\left(\dfrac{1}{\sqrt{3}},\dfrac{2}{3}\right).$

6 0

3 years ago