Line m has an equation of y = 3x + 2

Use the graph below to find f(-2) =

Vedmedyk [2.9K]

F(-2) is when x = -2
On the graph it’s (-2, 2)
Solution: f(-2) = 2

7 0

3 years ago

Find the area of an equilateral triangle (regular 3 Gon) with the given measurement 9 in perimeter a=SQ.in

blsea [12.9K]

Answer:

3.897 $in^{2}$

Step-by-step explanation:

equilateral triangles are also equiangular, meaning the have equal angles.

Triangle sum theory says that angles of a triangle add up to 180.

That means each angle is 60.

A = bh/2

You need the (h). The base of is 3. Perimeter = 9, so each side is 3

Draw a perpendicular line for the height. The line cuts the base in half (1.5)

Using trigonometry you can find the height.

tan 60° = h/1.5

h = height, 1.5 is half of 3, 60° is the base angle.

multiply each side by 1.5

1.5(tan 60°) = h

h=2.598

then substitute h into formula

A= (2.598)(3)

2

A = 3.897 rounded

4 0

2 years ago

The probability that a can of paint contains contamination is 3.23%, and the probability of a mixing error is 2.4%. The probabil

stich3 [128]

Answer:

4.6%.

Step-by-step explanation:

The probability that a can of paint contains contamination(C) is 3.23%

P(C)=3.23%

The probability of a mixing(M) error is 2.4%.

P(M)=2.4%

The probability of both is 1.03%.

$P(C \cap M)=1.03\%$

We want to determine the probability that a randomly selected can has contamination or a mixing error. i.e. $P(C \cup M)$

In probability theory:

$P(C \cup M) = P(C)+P(M)-P(C \cap M)\\P(C \cup M)=3.23+2.4-1.03\\P(C \cup M)=4.6\%$

The probability that a randomly selected can has contamination or a mixing error is 4.6%.

3 0

3 years ago

The base of a solid is the region enclosed by the graphs of y=e^x, y=0, x=0, and x = 1. If each cross-section perpendicular to t

yanalaym [24]

Answer:

A

Step-by-step explanation:

The base of a solid in the region enclosed by the graphs of y = eˣ, y = 0, x = 0, and x = 1. Each cross-section perpendicular to the x-axis is an equilateral triangle. We want to find the volume of the solid.

Please refer to the graph below. We are concerned with the red region.

In order to find the volume, we essentially sum up the area of the figure at each x value. So, we integrate from x = 0 to x = 1.

The area for an equilateral triangle is given by:

$\displaystyle A=\frac{\sqrt{3}}{4}s^2$

Where s is the side length of the triangle.

Since the triangle lies perpendicular on the region, the side length of the triangle at x is simply y, which is eˣ.

Therefore, our volume is:

$\displaystyle V=\int_0^1\frac{\sqrt3}{4}(y)^2\, dx$

Substitute:

$\displaystyle V=\int_0^1\frac{\sqrt3}{4}(e^x)^2\, dx$

Evaluate the integral. Simplify:

$\displaystyle V=\frac{\sqrt3}{4}\int_0^1e^{2x}\, dx$

Integrate using u-substitution:

$\displaystyle V=\frac{\sqrt3}{8}\left(e^{2x}\Big|_0^1\right)$

Evaluate:

$\displaystyle V=\frac{\sqrt3}{8}\left(e^{2(1)}-e^{2(0)} \right)$

Therefore, the volume of the solid is:

$\displaystyle V=\frac{\sqrt3}{8}\left(e^2-1\right)$

Our answer is A.

3 0

2 years ago

Read 2 more answers

Find the measure of /_ MFZ

marshall27 [118]

Given:

$MZ\cong PZ$

$m\angle MFZ=(x+9)^\circ$

$m\angle PFZ=(2x-1)^\circ$

To find:

The measure of angle MFZ.

Solution:

In triangles MFZ and PFZ,

$m\angle FMZ\cong \angle FPZ$ (Right angle)

$MZ\cong PZ$ (Given)

$FZ\cong FZ$ (Common side)

$\Delta MFZ\cong \Delta PFZ$ (HL congruent postulate)

$\angle MFZ\cong \angle PFZ$ (CPCTC)

$m\angle MFZ=m\angle PFZ$

$x+9=2x-1$

$9+1=2x-x$

$10=x$

The value of x is 10.

The measure of angle MFZ is:

$m\angle MFZ=(x+9)^\circ$

$m\angle MFZ=(10+9)^\circ$

$m\angle MFZ=19^\circ$

Therefore, the measure of angle MFZ is 19 degrees.

7 0

2 years ago