A line passes through the point (-2,3) and has a slope of 5/2

Carlotta thinks that 3y + 5y3 is the same as 8y4. Which statement shows that it is NOT the same?

Vesna [10]

Answer:

C. 3(4) + 5(4)³ ≠ 8(4)⁴

Step-by-step explanation:

3y + 5y³ is NOT the same as 8y⁴

Assume y = 4

Not equal to sign ≠

3y + 5y³ ≠ 8y⁴

3(4) + 5(4)³ ≠ 8(4)⁴

12 + 20³ ≠ 32⁴

12 + 8000 ≠ 1,048,576

8012 ≠ 1,048,576

The statement which shows that 3y + 5y³ is NOT the same as 8y⁴ is 3(4) + 5(4)³ ≠ 8(4)⁴

4 0

2 years ago

Jane has 5 books.fred has 2 more books than jane.how many books do fred and jane have in all

jolli1 [7]

Step-by-step explanation:

5+7=12 books all together

7 0

3 years ago

Read 2 more answers

What is the value of | -5 | - (-5)?

musickatia [10]

|-5|-(-5)

The absolute value of any number is always positive

5-(-5)

When there is a in front of an expression in parentheses, change the sign of each term in the expression

5+5

Add the numbers

10

Answer=10

3 0

2 years ago

Convert from rectangular to polar coordinates: note: choose rr and θθ such that rr is nonnegative and 0≤θ<2π0≤θ<2π (a)(9,0

DochEvi [55]

Answer:

Step-by-step explanation:

Convert rectangle (x , y) to polar coordinates ( r , θ)

$x=r \cos \theta, y= r \sin \theta$

$r=\sqrt{x^2+y^2} , \theta =tan^-^1 (\frac{y}{x} )$

a) converts (9, 0) to polar coordinates ( r , θ)

$r=\sqrt{x^2+y^2} \\\\=\sqrt{9^2+0} \\\\=9$

$\theta= \tan^-^1 (\frac{0}{9} )\\\\=0$

b) Convert $(18,\frac{18}{\sqrt{3} } )$ to polar coordinates ( r, θ)

$r = \sqrt{18^2+(\frac{18}{\sqrt{3} })^2 } \\\\=\sqrt{324+108} \\\\=\sqrt{432}$

$\frac{x}{y} \theta = \tan^-^1(\frac{\frac{18}{\sqrt{3} } }{18} )\\\\= \tan ^-^1(\frac{1}{\sqrt{3} } )\\\\= \frac{\pi}{6}$

c) converts (-5, 5) to polar coordinates ( r , θ)

$r =\sqrt{(-5)^2+(5)^2} \\\\=\sqrt{50} \\\\=5\sqrt{2}$

$\theta=\tan^-^1(\frac{5}{-5} )\\\\= \tan^-^1(-1)\\\\=\frac{3\pi}{4}$

d) converts (-1, √3) to polar coordinates ( r , θ)

$r=\sqrt{(-1)^2+(\sqrt{3})^2 } \\\\= \sqrt{4} \\\\=2$

$\theta=\tan^-^1(\frac{\sqrt{3} }{-1} )\\\=\tan^-^1(-\sqrt{3} )\\\\=\frac{2\pi}{3}$

$= \frac{2\pi}{\sqrt{3} }$

6 0

3 years ago

Question 1: Find the equation of the line through point (5,4) and perpendicular to y=−43x−2. Use a forward slash (i.e. "/") for

Mariulka [41]

Answer:

Question 1: y = 3/4x + 1/4.

Question 2: y = 6/5x + 7/5.

Step-by-step explanation:

Question 1: A line perpendicular to another line would have a slope that is the negative reciprocal of the other line. If the slope of the first line is -4/3, the slope of a line perpendicular to the first would have a slope of 3/4.

Since the line goes through (5, 4), we can just put the points into the equation, y = 3/4x + b.

4 = 3/4(5) + b

b + 15/4 = 4

b = 16/4 - 15/4

b = 1/4

So, the equation of the line is y = 3/4x + 1/4.

Question 2: 5x + 6y = -6

6y = -5x - 6

y = -5/6x - 1

As stated before, a line perpendicular to another will have a slope that is the negative reciprocal of the other. So, the slope of the other line is 6/5.

The line goes through (-2, -1), so we can put the points into the equation, y = 6/5x + b.

-1 = 6/5(-2) + b

b - 12/5 = -5/5

b = -5/5 + 12/5

b = 7/5

So, the equation of the line is y = 6/5x + 7/5.

Hope this helps!

8 0

3 years ago