Ask question

Ask question

All categories

Svetradugi [14.3K]

3 years ago

8

Find the length of line segment GF.

1 answer:

amm18123 years ago

8 0

Pythagorean theorem
c2 = a2 + b2
71^2 = 43^2 + b2
5041 = 1849 + b2
3192 = b2
b = 56.5 which is answer choice d

You might be interested in

Find the measure of each side of this triangles. Answer in exact form. Find x and y

Firdavs [7]

Answer:

x = 24/√3

4√3 = y

Step-by-step explanation:

We are given that this is a right triangle as well as two angles and a side.

Given that the angles in a triangle add up to 180 degrees, and a right angle is 90 degrees, we can say that the missing angle is equal to

(180-90-30) = 60 degrees

Next, given sohcahtoa, we can say that

sin(30) = opposite/hypotenuse = y/x

cos(30) = adjacent/hypotenuse = 12/x

tan(30) = opposite/adjacent = y/12

We want to only have one variable per equation so we can solve for the variable, so we can use

cos(30) = 12/x and tan(30) = y/12

Starting with cos(30) = 12/x, we can multiply both sides by x to remove the denominator to get

cos(30) * x = 12, and divide both sides by cos(30) to isolate the x and get

x = 12/cos(30). Next, cos(30) = √3/2, so we have

x = 12/(√3/2)

x = (12/1) / (√3/2)

x = (12/1) * (2/√3)

x = 24/√3

For y, we have

tan(30) = y/12

multiply both sides by 12 to isolate the y and remove the denominator

tan(30) * 12 = y

√3/3 * 12 = y

√3 * 4 = y

4√3 = y

4 0

3 years ago

What is the remainder when (x3 + 1) is divided by (x2 – x + 1)? x + 1 x 2 0

mariarad [96]

Answer:

x+1

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Find the missing length indicated

miskamm [114]

Check the picture below.

5 0

3 years ago

Decimal equivalent of 143%

daser333 [38]

Answer:

1.43 as a decimal

Step-by-step explanation:

%143 = 1.43

7 0

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.

(c) To find inflection points, we need to check where 2nd derivative is equal to 0.

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