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FrozenT [24]
3 years ago
6

118° D B angle C= ? can anybody help pls?

Mathematics
2 answers:
uysha [10]3 years ago
6 0

Answer:

118

Step-by-step explanation:

there is a property that i can't remember the name of that is dealing with angles across a flat line.

another way, is that 118+D=180, so D is 62

so that means D + C = 180, where D is 62 so C must be 118

Phoenix [80]3 years ago
4 0

Answer:

∡ c = 118°

Step-by-step explanation:

angle c is 118 degree (being vertically opposite angle)

∡ c = 118°

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STatiana [176]

Answer:

With 800 extra kg load: speed = 66.667 km/h

With 2400 extra kg load: speed = 40 km/h

Step-by-step explanation:

Let's call the weight of the car W = 1600 kg, and its speed after accelerating for 8 seconds with this weight S = 100 km/h

If we load the car with 800 extra kg, the total weight will be:

W2 = W + 800 = 2400 kg

The relation between the weights is:

W2 / W = 2400 / 1600 = 1.5

If the weight increased 1.5 times, the relation between the speeds will be 1/1.5 (as they are inversely proportional), so:

S2 / S = 1 / 1.5

S2 = S / 1.5 = 66.667 km/h

If we load the car with 2400 extra kg, the total weight will be:

W3 = W + 2400 = 4000 kg

The relation between the weights is:

W3 / W = 4000 / 1600 = 2.5

If the weight increased 2.5 times, the relation between the speeds will be 1/2.5, so:

S3 / S = 1 / 2.5

S3 = S / 2.5 = 40 km/h

3 0
3 years ago
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soldi70 [24.7K]
-4d + 2(3+d)= -14
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If f(x) = 2x2^2 + 1 and g(x) = x^2 - 7, find (f+g) (x)
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(f+g)(x) = f(x) + g(x)

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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;
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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

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Find the derivative A':

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Equate this derivative to 0, then

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

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