Helppp I’m on a math test

Mathematics

1 answer:

Inessa [10]3 years ago

6 0

option (D) 28 √6 is the answer

Explanation :

p =3√6 + 2√24 + 7√54

3√6 + 2√2×2×6+ 7√3×3×6

3√6 +2 ×2√6 + 7 ×3√6

3√6 +4√6 + 21√6

28√6

You might be interested in

In the Apex Republic, 8% of the population was unemployed in May of 2006. If there were 176,000,000 people willing and able to w

dolphi86 [110]

176,000,000*0.08=14,080,000 people

8 0

4 years ago

What is the lenght when height is 36 and width is 28 and the area is 6112

vazorg [7]

A = l*w
6112 = l*28
l = 6112/28
l = 218.2857143

8 0

3 years ago

Write an equation in slope-intercept form of the line perpendicular to y = 3x + 4 that passes through the point (3, 4).

adell [148]

Ahh I remember doing this but I don't remember clearly. Lemme look for my old notes

6 0

3 years ago

Find an equation of the tangent line to the curve 2(x^2+y^2)2=25(x^2−y^2) (a lemniscate) at the point (3,1)

Sholpan [36]

$\bf 2[x^2+y^2]^2=25(x^2-y^2)\qquad \qquad 
\begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ 3}}\quad ,&{{ 1}})\quad 
\end{array}\\\\
-----------------------------\\\\
2\left[ x^4+2x^2y^2+y^4 \right]=25(x^2-y^2)\qquad thus
\\\\\\
2\left[ 4x^3+2\left[ 2xy^2+x^22y\frac{dy}{dx} \right]+4y^3\frac{dy}{dx} \right]=25\left[2x-2y\frac{dy}{dx} \right]
\\\\\\
2\left[ 4x^3+2\left[ 2xy^2+x^22y\frac{dy}{dx} \right]+4y^3\frac{dy}{dx} \right]=50\left[x-y\frac{dy}{dx} \right]
\\\\\\
$

$\bf \left[ 4x^3+2\left[ 2xy^2+x^22y\frac{dy}{dx} \right]+4y^3\frac{dy}{dx} \right]=25\left[x-y\frac{dy}{dx} \right]
\\\\\\
4x^3+4xy^2+4x^2y\frac{dy}{dx}+4y^3\frac{dy}{dx}+25y\frac{dy}{dx}=25x
\\\\\\
\cfrac{dy}{dx}[4x^2y+4y^3+25y]=25x-4x^3+4xy^2
\\\\\\
\cfrac{dy}{dx}=\cfrac{25x-4x^3+4xy^2}{4x^2y+4y^3+25y}\impliedby m=slope$

notice... a derivative is just the function for the slope

now, you're given the point 3,1, namely x = 3 and y = 1

to find the "m" or slope, use that derivative, namely $f'(3,1)=\cfrac{25x-4x^3+4xy^2}{4x^2y+4y^3+25y}$

that'd give you a value for the slope

to get the tangent line at that point, simply plug in the provided values
in the point-slope form

$\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad
\begin{cases}
x_1=3\\
y_1=1\\
m=slope
\end{cases}\\ \qquad \uparrow\\
\textit{point-slope form}$

and then you solve it for "y", I gather you don't have to, but that'd be the equation of the tangent line at 3,1

6 0

4 years ago

What is the discount if the price is 55 and the sale price is 46.75

SVEN [57.7K]

Starting: 46.75
After 55% discount: 21.04
Total: $21.04

5 0

4 years ago