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4 years ago

HELP put in order and find the solution- are these correct?

Mathematics

1 answer:

Tasya [4]4 years ago

6 0

The steps need to go in 2, 1, 3, 4, 5 order (just flip your 1 and 2!) and the answer is correct on your second problem! nice job! :)

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Help me please and thank you

ollegr [7]

C) 51 is the answer

3 0

3 years ago

Determine whether the sequence converges or diverges. If it converges, find the limit. (If an answer does not exist, enter DNE.)

In-s [12.5K]

Answer:

The sequence diverges.

Step-by-step explanation:

A sequence $a_{n}$ converges when $\lim_{n \rightarrow \infty} a_{n}$ is a real number.

In this question, the sequence given is:

$a_{n} = 4n\cos{(7n\pi)}$

The cosine is always going to be between -1 and 1, so for the convergence of the sequence, we look it as: $a_{n} = 4n$ . So

$\lim_{n \rightarrow \infty} a_{n} = \lim_{n \rightarrow \infty} 4n = \infty$

Since the limit is not a real number, the sequence diverges.

5 0

3 years ago

What is the focus of the parabola? <img src="https://tex.z-dn.net/?f=y%3D%20%5Cfrac%7B1%7D%7B4%7D%20x%5E%7B2%7D%20-x%2B3" id="Te

OlgaM077 [116]

We know that
the equation of the parabola is of the form
y=ax²+bx+c
in this problem
y=1/4x²−x+3
where
a=1/4
b=-1
c=3
the coordinates of the focus are
(-b/2a,(1-D)/4a)
where D is the discriminant b²-4ac
D=(-1)²-4*(1/4)*3-----> D=1-3---> D=-2
therefore
x coordinate of the focus
-b/2a----> 1/[2*(-1/4)]----> 2

y coordinate of the focus
(1-D)/4a------> (1+2)/(4/4)---> 3

the coordinates of the focus are (2,3)

3 0

3 years ago

Read 2 more answers

Plzzz help me with this I’ll give brainliest

scoray [572]

Answer18:

The quadrilateral ABCD is not a parallelogram

Answer19:

The quadrilateral ABCD is a parallelogram

Step-by-step explanation:

For question 18:

Given that vertices of a quadrilateral are A(-4,-1), B(-4,6), C(2,6) and D(2,-4)

The slope of a line is given m= $\frac{Y2-Y1}{X2-X1}$

Now,

The slope of a line AB:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{6-(-1)}{(-4)-(-4)}$

m= $\frac{7}{0}$

The slope is 90 degree

The slope of a line BC:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{6-6}{(-4)-(-1)}$

m= $\frac{0}{(-3)}$

The slope is zero degree

The slope of a line CD:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{(-4)-6}{2-2}$

m= $\frac{-10}{0}$

The slope is 90 degree

The slope of a line DA:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{(-1)-(-4)}{(-4)-(2)}$

m= $\frac{3}{-6}$

m= $\frac{-1}{2}$

The slope of the only line AB and CD are the same.

Thus, The quadrilateral ABCD is not a parallelogram

For question 19:

Given that vertices of a quadrilateral are A(-2,3), B(3,2), C(2,-1) and D(-3,0)

The slope of a line is given m= $\frac{Y2-Y1}{X2-X1}$

Now,

The slope of a line AB:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{2-3}{3-(-2)}$

m= $\frac{-1}{5}$

The slope of a line BC:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{(-1)-2}{2-3}$

m= $\frac{-3}{-1}$

m=3

The slope of a line CD:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{0-(-1)}{(-3)-2}$

m= $\frac{-1}{5}$

The slope of a line DA:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{3-0}{(-2)-(-3)}$

m=3

The slope of the line AB and CD are the same

The slope of the line BC and DA are the same

Thus, The quadrilateral ABCD is a parallelogram

3 0

3 years ago

Last month, when Joe took his puppy to the veterinarian, the puppy weighed 4.6 pounds. This month, the puppy weighed 5.98 pounds

Karo-lina-s [1.5K]

Percent increase : (new number - original number) / (original number)....then multiply that by 100 to get ur percent.

(5.98 - 4.6) / (4.6) = 1.38/4.6 = 0.3......0.3 x 100 = 30% <==

6 0

4 years ago