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3 years ago

I'm super confused. Can anybody answer while writing the steps?

Mathematics

1 answer:

kvv77 [185]3 years ago

5 0

The formula for desnity is d= m/v (mass over volume), so

1) 0.18 = 240/v Plug in the numbers you have (0.18 being the density and 240 being the volume)

2) 0.18(240) = 240/v(240) multiply 240 on each side to get rid of the 240 on the right side of the equation.

3) 43.2 = m

So, the formula would be d=m/v and m= 43.2

You might be interested in

9x - 3y = -2<br> -3x + y = -3

morpeh [17]

Multiply 2nd equation by 3 and add to first
9x-3y=-2
<u>-9x+3y=-9 +</u>
0x+0y=-7

0=-7
false
no solution
these lines aer paraalell so no solution

6 0

3 years ago

If 20 out of 25 students in a P.E. class are boys, what percent of the class is made up of boys?

Ivanshal [37]

80% !
hope this helps

4 0

3 years ago

The production of what number and 7 is less than 49

MArishka [77]

Many numbers, when multiplied by $7$ , are less than $49$ . We can find an inequality for numbers $x$ that are sufficient.

Clearly, $7x < 49$ is the desired inequality. We divide by 7 to find that $x < 7$ determines working numbers.

4 0

3 years ago

f(x) is a quadratic function with x-intercepts at (−1, 0) and (−3, 0). If the range of f(x) is [−4, [infinity]) and g(x) = 2x^2

iren2701 [21]

Answer with Step-by-step explanation:

We are given that function f(x) which is quadratic function.

x -intercept of function f(x) at (-1,0) and (-3,0)

x-Intercept of f means zeroes of f

x=-1 and x=-3

Range of f =[-4, $\infty$ )

g(x)= $2x^2+8x+6=2(x^2+4x+3)$

$g(x)=0$

$2(x^2+4x+3)=0$

$x^2+4x+3=0$

$x^2+3x+x+3=0$

$x(x+3)+1(x+3)=0$

$(x+1)(x+3)=0$

$x+1=0\implies x=-1$

$x+3=0\implies x=-3$

Therefore, x-intercept of g(x) at (-1,0) and (-3,0).

Substitute x=-2

$g(-2)=2(-2)^2+8(-2)+6=8-16+6=-2$

$g(x)=2(x^2+4x)+6$

$g(x)=2(x^2+2\times x\times 2+4-4)+6=2(x^2+2\times x\times 2+4)-8+6$

$g(x)=2(x+2)^2-2$

By comparing with the equation of parabola

$y=a(x-h)^2+k$

Where vertex=(h,k)

We get vertex of g(x)=(-2,-2)

Range of g(x)=[-2, $\infty$ )

Zeroes of f and g are same .

But range of f and g are different.

Range of f contains -3 and -4 but range of g does not contain -3 and -4.

f and g are both quadratic functions.

3 0

3 years ago

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.

Vera_Pavlovna [14]

Split up the integration interval into 4 subintervals:

$\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]$

The left and right endpoints of the $i$ -th subinterval, respectively, are

$\ell_i=\dfrac{i-1}4\left(\dfrac\pi2-0\right)=\dfrac{(i-1)\pi}8$

$r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8$

for $1\le i\le4$ , and the respective midpoints are

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8$

Trapezoidal rule

We approximate the (signed) area under the curve over each subinterval by

$T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4T_i\approx\boxed{3.038078}$

Midpoint rule

We approximate the area for each subinterval by

$M_i=f(m_i)(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}$

Simpson's rule

We first interpolate the integrand over each subinterval by a quadratic polynomial $p_i(x)$ , where

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It so happens that the integral of $p_i(x)$ reduces nicely to the form you're probably more familiar with,

$S_i=\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))$

Then the integral is approximately

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}$

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.

3 0

3 years ago