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DIA [1.3K]
3 years ago
5

Which statement is true? A. 1.2<-6.9 B. 6.9<1.2 C. -6.9<-1.2 D. -1.2>6.9

Mathematics
2 answers:
slega [8]3 years ago
6 0
C because if you look at it on a number line -2.9 is greater than -6.9.
Delvig [45]3 years ago
5 0
The answer is C because of when you look at the negative numbers on the number line
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Which of the following is the vertical asymptote for the graph below?
forsale [732]

Answer:

C

Step-by-step explanation:

Vertical asymptotes are always in the form x = ?

If you look at the dotted line, it lands on 2.  Because it's a vertical line, the asymptote is going to be x = 2

4 0
2 years ago
After being rearranged and simplified, which of the following equations could
masha68 [24]

Answer:

C and D

Step-by-step explanation:

The quadratic formula is

x= (-b±√b²-4ac)/2a

The formula uses the numerical coefficients in the quadratic equation.

The general quadratic equation is ax²+bx+c where a, b and c are the numerical coefficients

So, lets try and see;

A.

5x+4=3x^4-2\\\\=3x^4-5x-2-4\\=3x^4-5x-6\\a=3,b=-5,c=-6

But due to the fact that  in this equation you have x⁴, the equation is not a quadratic equation thus can not be solved using this formula

B

-x^2+4x+7=-x^2-9\\\\\\=-x^2+x^2+4x+7+9\\=4x+16

C

9x+3x^2=14+x-1\\\\\\=3x^2+9x-x-14+1\\\\=3x^2+8x-13\\\\\\a=3,b=8,c=-13\\

D.

2x^2+x^2+x=30\\\\\\=3x^2+x-30\\\\\\a=3,b=1,c=-30

From the checking above, the equations will be C and D

7 0
3 years ago
Read 2 more answers
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
3 years ago
I dont understand the math answer can you explain better please.
Lunna [17]

Answer:

Can you elaborate, please?

Step-by-step explanation:

7 0
3 years ago
I need help please i'll give brainly​
Readme [11.4K]

Answer:

60 +24 = 84

Step-by-step explanation:jjjjjhjhjhj

6 x 4 +5 x 12

7 0
3 years ago
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