What is 1955 divide by 85

2 answers:

5 0

When you divide 1955 by 85 you get 23. Hope I helped. You wouldn't mind helping me out by giving me the brainliest answer :P

SpyIntel [72]3 years ago

3 0

It's 23!!!!!! I already said it

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What is 2/3, 7/6, 3 1/9, and 5 5/18 written as a decimal?

Paul [167]

Step-by-step explanation:

$\frac{2}{3} = 2 \times \frac{1}{3} = 2 \times 0.333 = 0.666 \\ \\ \frac{7}{6} = 7 \times \frac{1}{6} = 7 \times \frac{1}{2} \times \frac{1}{3} = 7 \times 0.5 \times 0.333 = 1.1655 \\ \\ 3 \frac{1}{9} = 3 + \frac{1}{9} = 3 + ( \frac{1}{3} \times \frac{1}{ 3} ) = 3 + (0.333 \times 0.333) = 3 \times 0.11088 = 3.11088 \\ \\ 5 \frac{5}{18} = 5 + (5 \times \frac{1}{18} ) = 5 + (5 \times \frac{1}{3} \times \frac{1}{3} \times \frac{1}{2} ) = 5 +(5 \times 0.333 \times 0.333 \times 0.5) = 5 + 0.27722 = 5.27722$

4 0

2 years ago

Read 2 more answers

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

3 years ago

Frn3n ionfrnffn kfrk jfrnfn

kvasek [131]

I tried to help but couldn’t :)

5 0

2 years ago

What is the slope of the line?

Ad libitum [116K]

Answer:

m= 4/3

Step-by-step explanation: