All categories

1 year ago

UV

Mathematics

1 answer:

icang [17]1 year ago

8 0

I think the answer would be SSS. but i haven’t done proofs and shi for like 2 years

You might be interested in

randy read a book in the afternoon. He looked at his watch when he started and finishing reading. How long did randy read

mr Goodwill [35]

The length of the time that randy read is 45 minutes.

Time is known to be an element that is often measured or that is measurable.

From the image attached, Randy started reading at 4:10 and Randy finished reading at 4:55.

We therefore need to subtract:

4:55 - 4:10 = 45 minutes

So, Randy is said to read for 45 minutes.

It is a period via which an action, process, etc. is alive or continues such as duration. One can tell the time by counting the minutes and seconds.

Learn more about time from

brainly.com/question/25607827

6 0

1 year ago

(b) dy/dx = (x - y+ 1)^2

Elanso [62]

Substitute $v(x)=x-y(x)+1$ , so that

$\dfrac{\mathrm dv}{\mathrm dx}=1-\dfrac{\mathrm dy}{\mathrm dx}$

Then the resulting ODE in $v(x)$ is separable, with

$1-\dfrac{\mathrm dv}{\mathrm dx}=v^2\implies\dfrac{\mathrm dv}{1-v^2}=\mathrm dx$

On the left, we can split into partial fractions:

$\dfrac12\left(\dfrac1{1-v}+\dfrac1{1+v}\right)\mathrm dv=\mathrm dx$

Integrating both sides gives

$\dfrac{\ln|1-v|+\ln|1+v|}2=x+C$

$\dfrac12\ln|1-v^2|=x+C$

$1-v^2=e^{2x+C}$

$v=\pm\sqrt{1-Ce^{2x}}$

Now solve for $y(x)$ :

$x-y+1=\pm\sqrt{1-Ce^{2x}}$

$\boxed{y=x+1\pm\sqrt{1-Ce^{2x}}}$

3 0

2 years ago

The black graph is the graph of

bija089 [108]

Answer:

Who knows ask goku

Step-by-step explanation:

5 0

2 years ago

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

2 years ago

Jordana was shown this diagram of an isosceles trapezoid and its reduction. She was asked to find the perimeter of the

PilotLPTM [1.2K]

The question is unclear to me, 30x3=10 WHAT, 30x3=90 not 10

And also it seems you already know the answer Perimeter=28cm

3 0

2 years ago

Read 2 more answers