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3 years ago

What is the value of x? enter your answer in the box

Mathematics

2 answers:

Leya [2.2K]3 years ago

6 0

Answer:

Step-by-step explanation:

Remark

Every triangle has 180 degrees. The three angles must add to 180

Solution

x + 102 + 39 = 180

x + 141 = 180 Subtract 141 from both sides.

x + 141-141 = 180-141

x = 39

Lelechka [254]3 years ago

5 0

Answer:

Step-by-step explanation:

If i'm right, a triangle should add up to 180, so you add 39 and 102 to get 141. Then you subtract 141 from 180 and get 39

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PLEASE ANSWER! Given the functions f(x) = x2 + 6x - 1, g(x) = -x2 + 2, and h(x) = 2x2 - 4x + 3, rank them from least to greatest

Ratling [72]

Answer:

he rank from least to great based on their axis of symmetry:

0, 1, -3 ⇒ g(x), h(x), f(x)

So, option C is correct.

Step-by-step explanation:

A quadratic equation is given by:

$ax^2+bx+c =0$

Here, a, b and c are termed as coefficients and x being the variable.

Axis of symmetry can be obtained using the formula

$x = \frac{-b}{2a}$

Identification of a, b and c in f(x), g(x) and h(x) can be obtained as follows:

$f(x) = x^2 + 6x - 1$

⇒ a = 1, b = 6 and c = -1

$g(x) = -x^2 + 2$

⇒ a = -1, b = 0 and c = 2

$h(x) = 2^2 - 4x + 3$

⇒ a = 2, b = -4 and c = 3

So, axis of symmetry in $f(x) = x^2 + 6x - 1$ will be:

$x = \frac{-b}{2a}$

x = -6/2(1) = -3

and axis of symmetry in $g(x) = -x^2 + 2$ will be:

$x = \frac{-b}{2a}$

x = -(0)/2(-1) = 0

and axis of symmetry in $h(x) = 2^2 - 4x + 3$ will be:

$x = \frac{-b}{2a}$

x = -(-4)/2(2) = 1

So, the rank from least to great based on their axis of symmetry:

0, 1, -3 ⇒ g(x), h(x), f(x)

So, option C is correct.

Keywords: axis of symmetry, functions

Learn more about axis of symmetry from brainly.com/question/11800108

#learnwithBrainly

7 0

4 years ago

How do you solve this limit of a function math problem?

hram777 [196]

If you know that

$e=\displaystyle\lim_{x\to\pm\infty}\left(1+\frac1x\right)^x$

then it's possible to rewrite the given limit so that it resembles the one above. Then the limit itself would be some expression involving $e$ .

For starters, we have

$\dfrac{3x-1}{3x+3}=\dfrac{3x+3-4}{3x+3}=1-\dfrac4{3x+3}=1-\dfrac1{\frac34(x+1)}$

Let $y=\dfrac34(x+1)$ . Then as $x\to\infty$ , we also have $y\to\infty$ , and

$2x-1=2\left(\dfrac43y-1\right)=\dfrac83y-2$

So in terms of $y$ , the limit is equivalent to

$\displaystyle\lim_{y\to\infty}\left(1-\frac1y\right)^{\frac83y-2}$

Now use some of the properties of limits: the above is the same as

$\displaystyle\left(\lim_{y\to\infty}\left(1-\frac1y\right)^{-2}\right)\left(\lim_{y\to\infty}\left(1-\frac1y\right)^y\right)^{8/3}$

The first limit is trivial; $\dfrac1y\to0$ , so its value is 1. The second limit comes out to

$\displaystyle\lim_{y\to\infty}\left(1-\frac1y\right)^y=e^{-1}$

To see why this is the case, replace $y=-z$ , so that $z\to-\infty$ as $y\to\infty$ , and

$\displaystyle\lim_{z\to-\infty}\left(1+\frac1z\right)^{-z}=\frac1{\lim\limits_{z\to-\infty}\left(1+\frac1z\right)^z}=\frac1e$

Then the limit we're talking about has a value of

$\left(e^{-1}\right)^{8/3}=\boxed{e^{-8/3}}$

# # #

Another way to do this without knowing the definition of $e$ as given above is to take apply exponentials and logarithms, but you need to know about L'Hopital's rule. In particular, write

$\left(\dfrac{3x-1}{3x+3}\right)^{2x-1}=\exp\left(\ln\left(\frac{3x-1}{3x+3}\right)^{2x-1}\right)=\exp\left((2x-1)\ln\frac{3x-1}{3x+3}\right)$

(where the notation means $\exp(x)=e^x$ , just to get everything on one line).

Recall that

$\displaystyle\lim_{x\to c}f(g(x))=f\left(\lim_{x\to c}g(x)\right)$

if $f$ is continuous at $x=c$ . $\exp(x)$ is continuous everywhere, so we have

$\displaystyle\lim_{x\to\infty}\left(\frac{3x-1}{3x+3}\right)^{2x-1}=\exp\left(\lim_{x\to\infty}(2x-1)\ln\frac{3x-1}{3x+3}\right)$

For the remaining limit, write

$\displaystyle\lim_{x\to\infty}(2x-1)\ln\frac{3x-1}{3x+3}=\lim_{x\to\infty}\frac{\ln\frac{3x-1}{3x+3}}{\frac1{2x-1}}$

Now as $x\to\infty$ , both the numerator and denominator approach 0, so we can try L'Hopital's rule. If the limit exists, it's equal to

$\displaystyle\lim_{x\to\infty}\frac{\frac{\mathrm d}{\mathrm dx}\left[\ln\frac{3x-1}{3x+3}\right]}{\frac{\mathrm d}{\mathrm dx}\left[\frac1{2x-1}\right]}=\lim_{x\to\infty}\frac{\frac4{(x+1)(3x-1)}}{-\frac2{(2x-1)^2}}=-2\lim_{x\to\infty}\frac{(2x-1)^2}{(x+1)(3x-1)}=-\frac83$

and our original limit comes out to the same value as before, $\exp\left(-\frac83\right)=\boxed{e^{-8/3}}$ .

3 0

3 years ago

A cubical water tank can contain 1000/125 cubic meters of water. Find the length of a side of the water tank.

Annette [7]

Given:

Volume of cubical tank = $\dfrac{1000}{125}$ cubic meters.

To find:

The length of a side of the water tank.

Solution:

The volume of a cubical tank is:

$V=a^3$

Where, a is the side length.

It is given that the volume of cubical tank is $\dfrac{1000}{125}$ cubic meters. So,

$a^3=\dfrac{1000}{125}$

$a^3=\dfrac{(10)^3}{5^3}$

Taking cube root on both sides, we get

$a=\dfrac{10}{5}$

$a=2$

Therefore, the length of a side of the water tank is 2 meters.

3 0

3 years ago

Which West African kingdom is responsible for the introduction of Islam to the region?

natita [175]

Answer:

Ghana this is because they are very religious

4 0

3 years ago

I need help pleaseee, someoneeee

Arisa [49]

Answer:

Step-by-step explanation:

Use the exponents to guide your answer. When we examine -∞, if the exponent is even, it will be a positive value. If it is odd, it will be a negative value. For positive ∞, it doesn't matter whether odd or even, the value will be positive.

So we automatically know as x approaches ∞, y has to be ∞.

But for -∞ it is trickier. You have 3x^6 and 75x^4 together that is positive, and 30x^5 that is negative. The first two is greater than 30x^5 overall, so as x approaches -∞, y also approaches ∞.

If you need a visual, this is what it looks like:

6 0

3 years ago

Read 2 more answers