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3 years ago

Write and solve an equation to find the complement of the angle that measure 42 degree

2 answers:

7 0

Complementary angles add up to 90 degrees.

So we can write:

42 + x = 90

To solve, subtract 42 to both sides:

x = 48

So the complement of a 42 degree angle is a 48 degree angle.

Butoxors [25]3 years ago

3 0

Complementary angles measures to the sum of 90 degrees. Since there is already an angle of 42 degrees, simply subtract 90 from 42, and you get 48 degrees...

angle A + angle B= 90 degrees, so

42 +x=90
x=48

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Otis constructed a rectangle with a compass and straightedge.

trapecia [35]

B. square
C. Draw arcs on the circle with the compass set at r units.

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3 years ago

Solve the following equation by factoring. 2x2−7x+3=0

slava [35]

Answer:

x=3

Step-by-step explanation:

Step 1: Factor left side of equation.

(2x−1)(x−3)=0

Step 2: Set factors equal to 0.

x−3=0

ANSWER:

x=3

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3 years ago

Solve x2 + 2x + 6 = 0 using the quadratic formula.

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No solution for the quadratic equation because of the √-20

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3 years ago

Consider the equation below. f(x) = 2x3 + 3x2 − 336x (a) Find the interval on which f is increasing. (Enter your answer in inter

Vaselesa [24]

We have been given a function $f(x)=2x^3+3x^2-336x$ . We are asked to find the interval on which function is increasing and decreasing.

(a). First of all, we will find the critical points of function by equating derivative with 0.

$f'(x)=2(3)x^{2}+3(2)x^1-336$

$f'(x)=6x^{2}+6x-336$

$6x^{2}+6x-336=0$

$x^{2}+x-56=0$

$x^{2}+8x-7x-56=0$

$(x+8)-7(x+8)=0$

$(x+8)(x-7)=0$

$x=-8,x=7$

So $x=-8,7$ are critical points and these will divide our function in 3 intervals $(-\infty,-8)U(-8,7)U(7,\infty)$ .

Now we will find derivative over each interval as:

$f'(x)=(x+8)(x-7)$

$f'(-9)=(-9+8)(-9-7)=(-1)(-16)=16$

Since $f'(9)>0$ , therefore, function is increasing on interval $(-\infty,-8)$ .

$f'(x)=(x+8)(x-7)$

$f'(1)=(1+8)(1-7)=(9)(-6)=-54$

Since $f'(1)$ , therefore, function is decreasing on interval $(-8,7)$ .

Let us check for the derivative at $x=8$ .

$f'(x)=(x+8)(x-7)$

$f'(8)=(8+8)(8-7)=(16)(1)=16$

Since $f'(8)>0$ , therefore, function is increasing on interval $(7,\infty)$ .

(b) Since $x=-8,7$ are critical points, so these will be either a maximum or minimum.

Let us find values of f(x) on these two points.

$f(-8)=2(-8)^3+3(-8)^2-336(-8)$

$f(-8)=1856$

$f(7)=2(7)^3+3(7)^2-336(7)$

$f(7)=-1519$

Therefore, $(-8,1856)$ is a local maximum and $(7,-1519)$ is a local minimum.

Let us find 2nd derivative.

$f''(x)=6(2)x^{1}+6$

$f''(x)=12x+6$

$12x+6=0$

$12x=-6$

$\frac{12x}{12}=-\frac{6}{12}$

$x=-\frac{1}{2}$

Therefore, $x=-\frac{1}{2}$ is an inflection point of given function.

3 0

3 years ago

Points M, N, and P are respectively the midpoints of sides AC , BC , and AB of △ABC. Prove that the area of △MNP is on fourth of

Hunter-Best [27]

Answer:

The area of △MNP is one fourth of the area of △ABC.

Step-by-step explanation:

It is given that the points M, N, and P are the midpoints of sides AC, BC and AB respectively. It means AC, BC and AB are median of the triangle ABC.

Median divides the area of a triangle in two equal parts.

Since the points M, N, and P are the midpoints of sides AC, BC and AB respectively, therefore MN, NP and MP are midsegments of the triangle.

Midsegments are the line segment which are connecting the midpoints of tro sides and parallel to third side. According to midpoint theorem the length of midsegment is half of length of third side.

Since MN, NP and MP are midsegments of the triangle, therefore the length of these sides are half of AB, AC and BC respectively. In triangle ABC and MNP corresponding side are proportional.

$\triangle ABC \sim \triangle NMP$