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svet-max [94.6K]
3 years ago
15

What makes a right trapezoid unique from all the other trapezoids

Mathematics
1 answer:
Tems11 [23]3 years ago
3 0

Answer:

A right trapezoid is a trapezoid in which one of the sides is perpendicular to the two bases: In this special case, if you know the length of the perpendicular side, that's the same as the altitude of the trapezoid.

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Find the area of the kite !!!! PLEASE HELP!!!!
Andrew [12]

ANSWER

D. 18 m^2

EXPLANATION

The area of a kite is half the product of the diagonals.

The diagonals of the kite are

3+3=6m

and

2+4=6m

The area of the kite

=  \frac{1}{2}  \times 6 \times 6

= 3 \times 6

= 18 {m}^{2}

The correct answer is D.

7 0
3 years ago
Solve the problem.
julsineya [31]

Answer:

2 dogs

Step-by-step explanation:

2 dogs at 1st plus 7 dogs added equals 9

3 0
2 years ago
What slope would make the lines<br> parallel?<br> y = -3x + 9<br> y = [?]x + 4
Vladimir79 [104]

Answer:

- 3

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = - 3x + 9 ← is in slope- intercept form

with slope m = - 3

Parallel lines have equal slopes, thus

y = - 3x + 4 ← is the equation of a parallel line

4 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
What are the factors of 3/4 xy​
Eva8 [605]
The answer for this is 3 and 4
4 0
3 years ago
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