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3 years ago

11

I need help someone plz help me

2 answers:

meriva3 years ago

8 0

Answer:

If the number gets bigger to the right.

The number gets less if it moves to the left.

Step-by-step explanation:

Andrej [43]3 years ago

3 0

Answer:

You move the decimal points the amount at which the number that is next to the 10 is, or you could find 10^3, which is 1000, and move the decimal to the right 3 times (or places)... Resulting in 4,580... Do you understand?

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Pets Plus and Pet Planet are having a sale on the same aquarium. At Pets Plus the aquarium is on sale for 30% off the original p

SOVA2 [1]

Answer:

Pets Plus

Step-by-step explanation:

A 30% discount is greater than a 25% discount

4 0

3 years ago

Find the volume of each of the following cubes. a. s = 7 kilometers b. s = 11 yards

nadya68 [22]

Answer:

Part 1) $V=343\ km^{3}$

Part 2) $V=1,331\ yd^{3}$

Step-by-step explanation:

we know that

The volume of a cube is equal to

$V=s^{3}$

where

s is the length of the cube

Part 1)

we have

$s=7\ km$

substitute in the formula

$V=7^{3}$

$V=343\ km^{3}$

Part 2)

we have

$s=11\ yd$

substitute in the formula

$V=11^{3}$

$V=1,331\ yd^{3}$

6 0

4 years ago

Read 2 more answers

If you saved 2.00 on january 1, 4.00 on february 1, 6.00 on march 1 and so on how much money would you save in a year

Ronch [10]

The answer that I got:

$156

5 0

3 years ago

One of the tables shows a proportional relationship. Graph the line representing the proportional relationship from this table i

Mkey [24]

Answer:

Table C

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $y/x=k$ or $y=kx$

Find the value of the constant of proportionality in each table

Table A

For $x=-2, y=-4$ ------> $k=-4/-2=2$

For $x=-1, y=-3$ ------> $k=-3/-1=3$

This table has different values of k

therefore

the table A does not represent a proportional relationship

Table B

For $x=-2, y=-4$ ------> $k=-4/-2=2$

For $x=-1, y=-2$ ------> $k=-2/-1=2$

For $x=1, y=3$ ------> $k=3/1=3$

This table has different values of k

therefore

the table B does not represent a proportional relationship

Table C

For $x=-2, y=-4$ ------> $k=-4/-2=2$

For $x=-1, y=-2$ ------> $k=-2/-1=2$

For $x=1, y=2$ ------> $k=2/1=2$

For $x=2, y=4$ ------> $k=4/2=2$

This table has the same value of k

therefore

the table C represent a proportional relationship

Table D

For $x=-2, y=-4$ ------> $k=-4/-2=2$

For $x=-1, y=-2$ ------> $k=-2/-1=2$

For $x=1, y=2$ ------> $k=2/1=2$

For $x=2, y=6$ ------> $k=6/2=3$

This table has different values of k

therefore

the table D does not represent a proportional relationship

3 0

3 years ago

Read 2 more answers

Find the solution of the differential equation that satisfies the given initial condition. dy/dx=x/y , y(0)=-1

Charra [1.4K]

Separating variables, we have

$\dfrac{dy}{dx} = \dfrac xy \implies y\,dy = x\,dx$

Integrate both sides.

$\displaystyle \int y\,dy = \int x\,dx$

$\dfrac12 y^2 = \dfrac12 x^2 + C$

Given that $y(0)=-1$ , we find

$\dfrac12 (-1)^2 = \dfrac12 0^2 + C \implies C = \dfrac12$

Then the particular solution is

$\dfrac12 y^2 = \dfrac12 x^2 + \dfrac12$

$y^2 = x^2 + 1$

$y = \pm\sqrt{x^2 + 1}$

and because $y(0)=-1$ , we take the negative solution to accommodate this initial value.

$\boxed{y(x) = -\sqrt{x^2+1}}$

7 0

2 years ago