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3 years ago

I WILL GIVE ALL MY POINTS AND BRAINIEST FOR THE RIGHT ANSWER !!!!!!

1 answer:

3 0

Since there is no image given, the perimeter is the sum of measurement of the sides of the triangle. Assuming that the triangle is equilateral. Then the perimeter of the triangle is equal to
P = x + x + x
P = 3x
P = 30 mm

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Two of the angles in a triangle measure 18° and 72°. What is the measure of the third angle?

Delvig [45]

Answer:

Step-by-step explanation:

the sum of angles in a triangle is 180 degrees

18 + 72 + X = 180

90 + X = 180

X = 180 -90

X = 90

6 0

4 years ago

I'm super confused on parts g and h because no formula is given, just the limit values. These are the values:

Mrrafil [7]

Answer:

See below.

Step-by-step explanation:

So we have the limit:

$\lim_{x\to 4} (f\circ g)(x)$

This is equivalent to:

$\lim_{x\to 4} f(g(x))$

To solve, we can use the following property:

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

Therefore, our limit is the same as:

$\lim_{x\to 4} f(g(x))\\=f(\lim_{x\to 4} g(x))$

We are already given that the limit as x approaches 4 of g(x) is -2. Therefore:

$=f(-2)$

You simply need to evaluate this to solve the limit. I don't have enough information to solve this, so maybe you have something more.

Similarly, for the second one:

We have:

$\lim_{x\to 4} (g\circ f)(x)\\=\lim_{x\to 4}g(f(x))$

Using the same property:

$=g(\lim_{x\to 4} f(x))$

And we are told that this is 16, so:

$=g(16)$

And we just need to evaluate this to find the limit.

I hope you find this helpful!

8 0

3 years ago

Sin(x)^4+ cos(x)^4=1/2

posledela

I'll assume that what was meant was $\sin ^4 x + \cos ^4 x = \dfrac{1}{2}$ .

The exponent in the funny place is just an abbreviation: $\sin ^4 x = (\sin x)^4$ .

I hope that's what you meant. Let me know if I'm wrong.

Let's start from the old saw

$\cos^2 x + \sin ^2x = 1$

Squaring both sides,

$(\cos^2 x + \sin ^2x)^2 = 1^2$

$\cos^4 x + 2 \cos ^2 x \sin ^2x +\sin ^4x = 1$

$\cos^4 x + \sin ^4x = 1 - 2 \cos ^2 x \sin ^2x$

So now the original question

$\sin ^4 x + \cos ^4 x = \dfrac{1}{2}$

becomes
$1 - 2 \cos ^2 x \sin ^2x = \dfrac{1}{2}$

$4 \cos ^2 x \sin ^2x = 1$

Now we use the sine double angle formula

$\sin 2x = 2 \sin x \cos x$

We square it to see

$\sin^2 2x = 4\sin^2 x \cos^2 x = 1$

Taking the square root,

$\sin 2x = \pm 1$

Not sure how you want it; we'll do it in degrees.

When we know the sine of an angle, there's usually two angles on the unit circle that have that sine. They're supplementary angles which add to $180^\circ$ . But when the sine is 1 or -1 like it is here, we're looking at $90^\circ$ and $-90^\circ$ , which are essentially their own supplements, slightly less messy.

That means we have two equations:

$\sin 2x = 1 = \sin 90^\circ$

$2x = 90^\circ + 360^\circ k \quad$ integer $k$

$x = 45^\circ + 180^\circ k$

or

$\sin 2x = -1 = \sin -90^\circ$

$2x = -90^\circ+ 360^\circ k$

$x = - 45^\circ + 180^\circ k$

We can combine those for a final answer,

$x = \pm 45^\circ + 180^\circ k \quad$ integer $k$

Check. Let's just check one, how about

$x=-45^\circ + 180^\circ = 135^\circ$

$\sin(135)= 1/\sqrt{2}$

$\sin ^4(135)=(1/\sqrt{2})^4 = 1/4$

$\cos ^4(135)=(-1/\sqrt{2})^4 = 1/4$

$\sin ^4(135^\circ) +\cos ^4(135^\circ) = 1/2 \quad\checkmark$

6 0

3 years ago

Ninety two timesthe sum of 31 +60

shusha [124]

The answer is 8,372 for the final.

8 0

3 years ago

Read 2 more answers

Just realized this is a trapezoid and I don't know much about trapezoids and area so I have no clue how to attempt a three dimen

Ivanshal [37]

To find the area og a trapezoid: (base 1+ Base 2)h divided by 2

3 0

4 years ago