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3 years ago

Bart, Lisa, and Maggle are pooling their money to buy a snack from the comer store. Bart has

Mathematics

1 answer:

zaharov [31]3 years ago

8 0

Answer:

Subtract $4.60 from 0.05

Step-by-step explanation:

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A student is solving the equation. 2x – 4 – (x + 2) = 6x If the student solves the equation correctly, what is the value of x ?

ANTONII [103]

Answer:

x = $-\frac{6}{5}$

Step-by-step explanation:

2x – 4 – (x + 2) = 6x is the given equation

Now simplifying both sides, we get

2x - 4 - x - 2 = 6x

taking like terms on each side of the equality, we get

2x - x - 6x = 4 + 2

or, -5x = 6

or x = $\frac{6}{-5}$

hence, the value of x = $-\frac{6}{5}$

7 0

3 years ago

What is solve a^3 = 11

IgorC [24]

Answer:

Step-by-step explanation:

√11

4 0

3 years ago

Simplify the expression. -(10)^-2

valentinak56 [21]

Answer:

-1/100

Step-by-step explanation:

-(10)^-2

Exponents first

We need to make the negative in the exponent positive, which means move it to the denominator

-1

--------

10^2

-1

---------

100

6 0

3 years ago

Read 2 more answers

A triangular lamina has vertices (0, 0), (0, 1) and (c, 0) for some positive constant c. Assuming constant mass density, show th

a_sh-v [17]

The equation of the line through (0, 1) and (c, 0) is

y - 0 = (0 - 1)/(c - 0) (x - c) ==> y = 1 - x/c

Let L denote the given lamina,

L = {(x, y) : 0 ≤ x ≤ c and 0 ≤ y ≤ 1 - x/c}

Then the center of mass of L is the point $(\bar x,\bar y)$ with coordinates given by

$\bar x = \dfrac{M_x}m \text{ and } \bar y = \dfrac{M_y}m$

where $M_x$ is the first moment of L about the x-axis, $M_y$ is the first moment about the y-axis, and m is the mass of L. We only care about the y-coordinate, of course.

Let ρ be the mass density of L. Then L has a mass of

$\displaystyle m = \iint_L \rho \,\mathrm dA = \rho\int_0^c\int_0^{1-\frac xc}\mathrm dy\,\mathrm dx = \frac{\rho c}2$

Now we compute the first moment about the y-axis:

$\displaystyle M_y = \iint_L x\rho\,\mathrm dA = \rho \int_0^c\int_0^{1-\frac xc}x\,\mathrm dy\,\mathrm dx = \frac{\rho c^2}6$

Then

$\bar y = \dfrac{M_y}m = \dfrac{\dfrac{\rho c^2}6}{\dfrac{\rho c}2} = \dfrac c3$

but this clearly isn't independent of c ...

Maybe the x-coordinate was intended? Because we would have had

$\displaystyle M_x = \iint_L y\rho\,\mathrm dA = \rho \int_0^c\int_0^{1-\frac xc}y\,\mathrm dy\,\mathrm dx = \frac{\rho c}6$

and we get

$\bar x = \dfrac{M_x}m = \dfrac{\dfrac{\rho c}6}{\dfrac{\rho c}2} = \dfrac13$

8 0

3 years ago

Identify the scale factor of 8 and 8/3

amm1812

Answer:

scale factor = $\frac{1}{3}$

Step-by-step explanation:

note

8 × $\frac{1}{3}$ = $\frac{8}{3}$ , so

scale factor = $\frac{1}{3}$

8 0

3 years ago