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3 years ago

5/8 and 1/4 is subtracted from the product of 2 3/4 and 1 3/5 what is the difference

Mathematics

1 answer:

olga55 [171]3 years ago

3 0

Answer:

-233/40

Step-by-step explanation:

5/8 + 1/4 - 2(3/4 + 13/5)

5/8 + 2/8 - 6/4 + 26/5

7/8 - 30/20 + 104/20

7/8 - 134/20

35/40 - 268/40

-233/40

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Finding Derivatives Implicity In Exercise,Find dy/dx implicity.<br> x2e - x + 2y2 - xy = 0

Klio2033 [76]

Answer:

the question is incomplete, the complete question is

"Finding Derivatives Implicity In Exercise,Find dy/dx implicity . $x^{2}e^{-x}+2y^{2}-xy$ "

Answer : $\frac{dy}{dx}=\frac{y-(2-x)xe^{-x}}{(4y-x)}$

Step-by-step explanation:

From the expression $x^{2}e^{-x}+2y^{2}-xy$ " y is define as an implicit function of x, hence we differentiate each term of the equation with respect to x.

we arrive at

$\frac{d}{dx}(x^{2}e^{-x )+\frac{d}{dx} (2y^{2})-\frac{d}{dx}xy=0\\$

for the expression $\frac{d}{dx}(x^{2}e^{-x})$ we differentiate using the product rule, also since y^2 is a function of y which itself is a function of x, we have

$(2xe^{-x}-x^{2}e^{-x})+4y\frac{dy}{dx}-x\frac{dy}{dx} -y=0\\\\(2-x)xe^{-x}+(4y-x)\frac{dy}{dx}-y=0 \\$ .

if we make dy/dx subject of formula we arrive at

$(4y-x)\frac{dy}{dx}=y-(2-x)xe^{-x}\\\frac{dy}{dx}=\frac{y-(2-x)xe^{-x}}{(4y-x)}$

5 0

3 years ago

BRAINLIEST ASAP! PLEASE HELP ME :)<br> thxx

olga nikolaevna [1]

Answer:

2nd point

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Marcus looked out in the audience and saw 40% of the chairs were filled. There were 42 people sitting in the chairs. Enter the t

puteri [66]

Answer:

I say it is 16.8

Step-by-step explanation:

40 x 42

100

= 16.8

4 0

2 years ago

Make n the subject of m= 4n-gn+1

Misha Larkins [42]

Answer:

n = (m-1)/(4-g)

Step-by-step explanation:

m= 4n-gn+1

Subtract 1 from each side

m-1= 4n-gn+1-1

m-1 = 4n -gn

Factor out n

m-1 = n(4-g)

Divide each side by (4-g)

(m-1)/(4-g) = n(4-g)/(4-g)

(m-1)/(4-g) = n

n = (m-1)/(4-g)

6 0

2 years ago

Read 2 more answers

Help me with question a please ! With full workings !

frosja888 [35]

A)

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
Q&({{ 0}}\quad ,&{{ 2}})\quad 
% (c,d)
P&({{ 0.5}}\quad ,&{{ 0}})
\end{array}\qquad 
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}$

$\bf QP=\sqrt{(0.5-0)^2+(0-2)^2}\implies QP=\sqrt{0.5^2+2^2}
\\\\\\
QP=\sqrt{\left( \frac{1}{2} \right)^2+4}\implies QP=\sqrt{ \frac{1^2}{2^2}+4}\implies QP=\sqrt{\frac{1}{4}+4}
\\\\\\
QP=\sqrt{\frac{17}{4}}\implies QP=\cfrac{\sqrt{17}}{\sqrt{4}}\implies QP=\cfrac{\sqrt{17}}{2}$

b)

since QR=QP, that means that QO is an angle bisector, and thus the segments it makes at the bottom of RO and OP, are also equal, thus RO=OP

thus, since the point P is 0.5 units away from the 0, point R is also 0.5 units away from 0 as well, however, is on the negative side, thus R (-0.5, 0)

c)

what's the equation of a line that passes through the points (-0.5, 0) and (0,2)?

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
Q&({{ 0}}\quad ,&{{ 2}})\quad 
% (c,d)
R&({{ -0.5}}\quad ,&{{ 0}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{0-2}{-0.5-0}\implies \cfrac{-2}{-0.5}$

$\bf m=\cfrac{\frac{-2}{1}}{-\frac{1}{2}}\implies \cfrac{-2}{1}\cdot \cfrac{2}{-1}\implies 4
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-2=4(x-0)\implies y=4x+2\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}$

7 0

3 years ago