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3 years ago

Which points are on the graph of g(x) = (1/5)^x

Mathematics

1 answer:

11111nata11111 [884]3 years ago

7 0

Answer:

A) The point( -1 , 5) is satisfies the given graph $g(x) = (\frac{1}{5} )^{x}$

B) The point( 3 , 1/125) is satisfies the given graph $g(x) = (\frac{1}{5} )^{x}$

Step-by-step explanation:

Explanation:-

Given graph

y = $g(x) = (\frac{1}{5} )^{x}$

Put the point ( -1 , 5)

Put y =5 and x =-1

$5 = (\frac{1}{5} )^{-1} = 5$

The point( -1 , 5) is satisfies the given graph

ii)

Given graph

y = $g(x) = (\frac{1}{5} )^{x}$

$\frac{1}{125} = (\frac{1}{5} )^{3} = \frac{1}{125}$

The point( 3 , 1/125) is satisfies the given graph $g(x) = (\frac{1}{5} )^{x}$

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The volume of a box without a lid is given by the formula v = 4x(10-x)^2, where x is a length in inches and v is the volume in c

Aloiza [94]

The volume of the box without the lid is 500 cubic inches.

<h3>What is volume?</h3>

The capacity of an object is measured by its volume. For instance, a cup's capacity is stated to be 100 ml if it can hold 100 ml of water in its brim.

Another way to think of volume is as the amount of space a three-dimensional object takes up.

Calculation for the volume of the box-

The volume of a box without a lid is given by the formula v = 4x(10-x)².

Where,

'x' is a length in inches, and

'v' is the volume in cubic inches.

The volume when x = 5 inches.

Substitute the value of 'x' in the formula of volume;

v = 4×5(10-5)²

= 20×5²

v = 500

Therefore, the volume of a box without a lid is 500 cubic inches.

To know more about the volume of a cardboard shoe box, here

brainly.com/question/623789

#SPJ4

8 0

2 years ago

I need help with this and fast if possible

11Alexandr11 [23.1K]

Answer:

The answer is Hence Proved

Step-by-step explanation:

Given that CB║ED , CB ≅ ED

To prove Δ CBF ≅ Δ EDF

CB ≅ ED ( Given )

This means that the length of CB is equal to ED

As CB║ED The following conditions satisfies when a transversal cut

two parallel lines

∠ EDF = ∠ FBC ( Alternate interior points )
∠ DEF = ∠ FCB ( Alternate interior points )

∴ Δ CBF ≅ Δ EDF ( By ASA criterion)

The Δ CBF is congruent to Δ EDF By ASA criterion .

Hence proved

6 0

3 years ago

Explain how you would find the height of a rectangular prism if you know the volume is 60 centimeters and that the area of the b

ZanzabumX [31]

U have to add 60+10 then 60-10=50+20=70 and that is the answer that u u get addition subtraction those are the only to that u use.

4 0

3 years ago

DIRECTIONS: Use this information to answer Parts A, B, and C.

Talja [164]

Answer:

V=706.85

Step-by-step explanation:

Well, first of when you want to find the volume of the cone you need to have a radius and hieght. And the radius is half the diameter, so divide the diameter into half: 10/2=5 in. and you already have the height so just plug in the numbers.

Formula: V= $\frac{1}{3}$ $\pi$ $r^{2}$ h ---> V= $\frac{1}{3}$ $\pi$ ( $5^{2}$ )(27)

And put the plugged in formula in your calculator and you'll find the answer!

V=706.85

I hope this helped.

3 0

3 years ago

wo point charges lie on the $x$ axis. A charge of +6.24 $\mu C$ is at the origin, and a charge of -9.55 $\mu C$ is at $x$ = 12.0

Andreyy89

Answer:

$E_n=34,467,075.42\ N/C$

Step-by-step explanation:

Electric Field

The electric field produced by a point charge Q at a distance d is given by

$\displaystyle E=K\cdot \frac{Q}{d^2}$

Where

$K = 9\cdot 10^9\ Nw.m^2/c^2$

The net electric field is the vector addition of the individual electric fields produced by each charge. The direction is given by the rule: If the charge is positive, the electric field points outward, if negative, it points inward.

Let's calculate the electric fields of each charge at the given point. The first charge $q_1=+6.24\mu C=6.24\cdot 10^{-6}C$ is at the origin. We'll calculate its electric field at the point x=-3.85 cm. The distance between the charge and the point is d=3.85 cm = 0.0385 m, and the electric field points to the left:

$\displaystyle E_1=9\cdot 10^9\cdot \frac{6.24\cdot 10^{-6}}{0.0385^2}$

$E_1=37,888,345.42\ N/C$

Similarly, for $q_2=-9.55\mu C=-9.55\cdot 10^{-6}C$ , the distance to the point is 12 cm + 3.85 cm = 15.85 cm = 0.1585 m. The electric field points to the right:

$\displaystyle E_2=9\cdot 10^9\cdot \frac{9.55\cdot 10^{-6}}{0.1585^2}$

$E_2=3,421,270\ N/C$

Since E1 and E2 are opposite, the net field is the subtraction of both

$E_n=37,888,345.42\ N/C-3,421,270\ N/C$

$\boxed{E_n=34,467,075.42\ N/C}$

6 0

3 years ago