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3 years ago

What is the value of x?

Mathematics

2 answers:

julsineya [31]3 years ago

7 0

For this case we have that the sum of the interior angles of a triangle is 180 degrees.

So:

$G + 45 + 53 = 180\\G = 180-45-53\\G = 82 \ degrees$

Thus, the angle of the vertex G of the triangle is 82 degrees.

Now, we must find the exterior angle x.

We know that, by definition:

$x + 82 = 180\\x = 180-82\\x = 98$

So, the angle $x = 98$

Answer:

kykrilka [37]3 years ago

4 0

Answer:

Step-by-step explanation:

triangle adds upto 180 degrees

180 - (53+45) = 98

Then two lines adapt to 180 degrees

therefore, 180 -98 = 82

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PLEASE HEL MUST BE DONE URGENTLY

LekaFEV [45]

Y = x + 2 is closest to what I believe is the solution.

If the club begins with 2 members, then +2 MUST appear in the formula.

If the club adds 2 members every week, then the total number of members would be

f(x) = y = 2 + 2(x-1), where x is the number of weeks.

Test this: If x = 1, then y = 2+2(1-1) = 2 + 0 = 2. This is correct. The club began with 2 members in week 1

8 0

3 years ago

Read 2 more answers

What’s the correct answer for this?

Nitella [24]

Answer:

6 pounds and .125 ounces

Step-by-step explanation:

6 0

2 years ago

What is the equation of the graph below?

IgorC [24]

ANSWER

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

EXPLANATION

The vertex of the graph is at

$(-3,-1)$

The equation of the graph in vertex form is given by the formula,

$f(x) = a{(x - h)}^{2} + k$

The above graph has a minimum point, hence
$a > \: 0$

When we put these values into the equation we obtain,

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

This implies that,

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

The point (-2,0) lies on the line.

$a( - 2 + 3)^{2} - 1 = 0$

$a = 1$

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

The correct answer is D.

6 0

3 years ago

2. G(x) = 2x² + 3x<br>Is the function even, odd, or neither?

Bingel [31]

The function appears to be neither odd nor even

8 0

3 years ago

CALCULUS - Find the values of in the interval (0,2pi) where the tangent line to the graph of y = sinxcosx is

Rufina [12.5K]

Answer:

$\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}$

Step-by-step explanation:

We want to find the values between the interval (0, 2π) where the tangent line to the graph of y=sin(x)cos(x) is horizontal.

Since the tangent line is horizontal, this means that our derivative at those points are 0.

So, first, let's find the derivative of our function.

$y=\sin(x)\cos(x)$

Take the derivative of both sides with respect to x:

$\frac{d}{dx}[y]=\frac{d}{dx}[\sin(x)\cos(x)]$

We need to use the product rule:

$(uv)'=u'v+uv'$

So, differentiate:

$y'=\frac{d}{dx}[\sin(x)]\cos(x)+\sin(x)\frac{d}{dx}[\cos(x)]$

Evaluate:

$y'=(\cos(x))(\cos(x))+\sin(x)(-\sin(x))$

Simplify:

$y'=\cos^2(x)-\sin^2(x)$

Since our tangent line is horizontal, the slope is 0. So, substitute 0 for y':

$0=\cos^2(x)-\sin^2(x)$

Now, let's solve for x. First, we can use the difference of two squares to obtain:

$0=(\cos(x)-\sin(x))(\cos(x)+\sin(x))$

Zero Product Property:

$0=\cos(x)-\sin(x)\text{ or } 0=\cos(x)+\sin(x)$

Solve for each case.

Case 1:

$0=\cos(x)-\sin(x)$

Add sin(x) to both sides:

$\cos(x)=\sin(x)$

To solve this, we can use the unit circle.

Recall at what points cosine equals sine.

This only happens twice: at π/4 (45°) and at 5π/4 (225°).

At both of these points, both cosine and sine equals √2/2 and -√2/2.

And between the intervals 0 and 2π, these are the only two times that happens.

Case II:

We have:

$0=\cos(x)+\sin(x)$

Subtract sine from both sides:

$\cos(x)=-\sin(x)$

Again, we can use the unit circle. Recall when cosine is the opposite of sine.

Like the previous one, this also happens at the 45°. However, this times, it happens at 3π/4 and 7π/4.

At 3π/4, cosine is -√2/2, and sine is √2/2. If we divide by a negative, we will see that cos(x)=-sin(x).

At 7π/4, cosine is √2/2, and sine is -√2/2, thus making our equation true.

Therefore, our solution set is:

$\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}$

And we're done!

Edit: Small Mistake :)

5 0

2 years ago