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2 years ago

Will mark brainlyest

Mathematics

2 answers:

larisa86 [58]2 years ago

4 0

Answer:

i believe that its angle 4 and 5 since they would sum up to 180 degrees

Step-by-step explanation:

ArbitrLikvidat [17]2 years ago

3 0

Answer: Angles 4 and 5

Step-by-step explanation:

Supplementary angles are angles that add up to 180. The only angle pairs the add up to 180 (you can tell by seeing if two angles are the only two on a straight line) are angles 4 and 5, and angles 3 and 4. Angles 3 and 4 isn’t an option, so it’s 4 and 5.

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Simplify the expression 35e^9/5e^8

erastovalidia [21]

$\frac{35e {}^{9} }{5 {e}^{8} } \ \\ \\ \frac{7e {}^{9} }{e {}^{8} } \\ \\ \\ = 7e$

Step By Step Explanation:

Reduce: Reduce the fraction with 5
Simplify: Simplify the expression

Alternate Forms:

19.02797

<h3>☆彡Hanna</h3>

5 0

2 years ago

The width x of your iris decreases from 4 millimeters to 2 millimeters when you enter a dark room. How many times greater is the

inysia [295]

Answer:

The area of the Iris is 4 times greater.

Step-by-step explanation:

Firstly, we have to realize that we cannot solve this problem without knowing the radius of the iris. because from the information given, the radius of the iris could well be the size of the galaxy and contract from width of 4mm to 2mm and we wouldn't know!

The average radius of the iris is 6mm, so we take this value.

Now, initially the width of the iris is 4mm, that means the radius of the pupil is:

$6mm-4mm=2mm$

Therefore it's area $A$ is:

$A=\pi r^2=\pi (2)^2=4\pi$

When the iris contracts to 2mm, the radius of the pupil becomes:

$6mm-2mm=4mm$

Then it's area is $A$ :

$A=\pi r^2=\pi (4)^2=16\pi$

To find how many times greater this final area is than the initial area, we just divide it by the initial area:

$\frac{16\pi }{4\pi } =4$

This is 4 times greater than the initial area.

3 0

3 years ago

For the function y=3x2: (a) Find the average rate of change of y with respect to x over the interval [3,6]. (b) Find the instant

nirvana33 [79]

Answer:

The instantaneous rate of change of $y$ with respect to $x$ at the value $x = 3$ is 18.

Step-by-step explanation:

a) Geometrically speaking, the average rate of change of $y$ with respect to $x$ over the interval by definition of secant line:

$r = \frac{y(b) -y(a)}{b-a}$ (1)

Where:

$a$ , $b$ - Lower and upper bounds of the interval.

$y(a)$ , $y(b)$ - Function exaluated at lower and upper bounds of the interval.

If we know that $y = 3\cdot x^{2}$ , $a = 3$ and $b = 6$ , then the average rate of change of $y$ with respect to $x$ over the interval is:

$r = \frac{3\cdot (6)^{2}-3\cdot (3)^{2}}{6-3}$

$r = 27$

The average rate of change of $y$ with respect to $x$ over the interval $[3,6]$ is 27.

b) The instantaneous rate of change can be determined by the following definition:

$y' = \lim_{h \to 0}\frac{y(x+h)-y(x)}{h}$ (2)

Where:

$h$ - Change rate.

$y(x)$ , $y(x+h)$ - Function evaluated at $x$ and $x+h$ .

If we know that $x = 3$ and $y = 3\cdot x^{2}$ , then the instantaneous rate of change of $y$ with respect to $x$ is:

$y' = \lim_{h \to 0} \frac{3\cdot (x+h)^{2}-3\cdot x^{2}}{h}$

$y' = 3\cdot \lim_{h \to 0} \frac{(x+h)^{2}-x^{2}}{h}$

$y' = 3\cdot \lim_{h \to 0} \frac{2\cdot h\cdot x +h^{2}}{h}$

$y' = 6\cdot \lim_{h \to 0} x +3\cdot \lim_{h \to 0} h$

$y' = 6\cdot x$

$y' = 6\cdot (3)$

$y' = 18$

The instantaneous rate of change of $y$ with respect to $x$ at the value $x = 3$ is 18.

5 0

2 years ago

How do I solve 5x+1=7x+19

notsponge [240]

Answer:

x=-9

Step-by-step explanation:

5x+1=7x+19

1 = 7x-5x +19

1-19 = 2x

-18/2 = x

x = -9

Hope this helps!

Please mark brainliest if you think I helped! Would really appreciate!

5 0

2 years ago

Explain why a 3 dimensional figure with a curved surface is not a polyhedron

nekit [7.7K]

Polyhedron, by definition, is formed by intersecting planes, so each part of its surface is either a vertex or an edge, or a flat piece of a plane.
Explanation:
The definition of polyhedron implies that each side is a piece of a flat surface.

5 0

2 years ago