All categories

3 years ago

Express the area of the figure as a monomial.

Mathematics

1 answer:

Reil [10]3 years ago

6 0

Answer:

Step-by-step explanation:

Area of circle=πr²

=3.14*(5x³)² = 3.14 * 5x³*5x³

= 3.14*25*x⁶

=78.5x⁶

You might be interested in

The Graphs With Angles And Lines

Arada [10]

Answer:

Step-by-step explanation:

Soo sorry if i am worng :(

8 0

2 years ago

Which undefined geometric term is described as a location on a coordinate plane that is designated by an ordered pair, (x, y)?

jolli1 [7]

Answer: D. Point

A point is a location. On the 2D coordinate plane, it's described by (x,y). Think of it like an address or like latitude/longitude. A point is an undefined term in geometry as it doesn't build off of smaller elements.

8 0

2 years ago

Read 2 more answers

A line with a slope of 3 passes through (2,5). Which choice is an equation of this line? Please help

jek_recluse [69]

Answer:

Top right.

Step-by-step explanation:

So we want a line with a slope of 3 and passes through (2,5).

To do so, we can use the point-slope form.

The point-slope form is:

$y-y_1=m(x-x_1)$

m is the slope and x₁ and y₁ is an ordered pair.

Thus, let m be 3, y₁ be 5, and x₁ be 2. Thus:

$y-5=3(x-2)$

Our answer is the top right :)

3 0

3 years ago

Read 2 more answers

I dont know i need help please

Minchanka [31]

Answer:

Step-by-step explanation:

8 0

2 years ago

Find the x- and y- intercepts of parabola y=5x^2-16x+10

____ [38]

Y-INTERCEPT

$y = 5x^2 - 16x + 10$

The y-intercept is where the equation/curve/parabola cosses the y-axis.

The y-axis is where x = 0. (The x-axis is where y = 0)

To find the y-intercept:

$\text{y-axis} \rightarrow \text{x = 0} \rightarrow y = 5(0)^2 -16(0) + 10 = 10$

The y-intercept must be at (0, 10)

X-INTERCEPT (ROOTS/SOLUTIONS)

$y = 5x^2 - 16x + 10\\\text{make it equal 0}\\y = 0\\\therefore 5x^2 - 16x + 10 = 0$

We need to use the quadratic formula

The quadratic formula helps us find what values of $x$ make the equation = 0

Quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-16) + \sqrt{(-16)^2-4(5)(10)}}{2(5)}\\\\x = \frac{16 + \sqrt{256-200}}{10}\\x = \frac{16 + \sqrt{56}}{10}\\x = \frac{16 + 2\sqrt{14}}{10}\\x = \frac{8 + \sqrt{14}}{5}\\\\\\x=\frac{-(-16) - \sqrt{(-16)^2-4(5)(10)}}{2(5)}\\\text{doing the same thing...}\\\text{end up with...}\\x = \frac{8 - \sqrt{14}}{5}\\$

The x-intercepts are at:

$(\frac{8 + \sqrt{14}}{5}, 0)\\(\frac{8 - \sqrt{14}}{5}, 0)$

5 0

1 year ago