All categories

3 years ago

What similarity statement can you write relating the three triangles in the diagram?

Mathematics

1 answer:

kvv77 [185]3 years ago

7 0

From the diagram, ΔRST, ΔRUS and ΔSUT are right triangles. This means that the corresponding legs of the three triangles are proportional, then we can say that the triangles are similar by side-angle-side similarity.
RS/RU=RU/SU
In the figure we can also notice that the acute angles of the corresponding angles are similar, then by angle-angle similarity the triangles are similar.
∠TRS~∠URS~TSU
This implies that: ΔSUT~ΔSRU~ΔRTS

You might be interested in

When h has the value 4, calculate 16-3h

azamat

h=4

calculate 16-3h

16-(3*4 )

16-12 =4

I hope that's help !

3 0

3 years ago

Quadrilateral KLMN is a rectangle. The coordinates of L are L(1,-4), and the coordinates of Mare M(3,-2). Find the slopes of sid

lozanna [386]

L(1, -4)=(xL, yL)→xL=1, yL=-4
M(3, -2)=(xM, yM)→xM=3, yM=-2

Slope of side LM: m LM = (yM-yL) / (xM-xL)
m LM = ( -2 - (-4) ) / (3-1)
m LM = ( -2+4) / (2)
m LM = (2) / (2)
m LM = 1

The quadrilateral is the rectangle KLMN
The oposite sides are: LM with NK, and KL with NK
In a rectangle the opposite sides are parallel, and parallel lines have the same slope, then:
Slope of side LM = m LM = 1 = m NK = Slope of side NK
Slope of side NK = m NK = 1

Slope of side KL = m KL = m MN = Slope of side MN

The sides KL and LM (consecutive sides) are perpendicular (form an angle of 90°), then the product of their slopes is equal to -1:
(m KL) (m LM) = -1
Replacing m LM = 1
(m KL) (1) = -1
m KL = -1 = m MN

Answer:
Slope of side LM =1
Slope of side NK =1
Slope of side KL = -1
Slope of side MN = -1

3 0

3 years ago

Find the derivative of<br> <img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%5Cfrac%7B6%7D%7Bx%7D%20" id="TexFormula1" titl

aev [14]

$f'(x_0)=\lim\limits_{h\to0}\dfrac{f(x_0+h)-f(h)}{h}=\lim\limits_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}\\\\f(x)=\dfrac{6}{x};\ x_0=2\\\\subtitute\\\\f'(2)=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-\frac{6}{2}}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-3}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6-3x}{x}}{x-2}=\lim\limits_{x\to 2}\dfrac{6-3x}{x(x-2)}\\\\=\lim\limits_{x\to2}\dfrac{-3(x-2)}{x(x-2)}=\lim\limits_{x\to2}\dfrac{-3}{x}=-\dfrac{3}{2}=-1.5\\\\\\An swer:\boxed{f'(2)=-1.5}$

8 0

3 years ago

Given the following observed and expected data (total of 1000), using chi-squared calculate a p-value that corresponds with this

finlep [7]

Answer:

The answer is " $0.90>p>0.75.$ "

Step-by-step explanation:

$\text{Cinnabar vestigial} \ \ \ \ \ \ \ \ \ \ \ 384 \ \ \ \ \ \ \ \ \ \ \ 390 \ \ \ \ \ \ \ \ \ \ \ -6 36 \ \ \ \ \ \ \ \ \ \ \ 0.092308\\\\$

$roof \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 408 \ \ \ \ \ \ \ \ \ \ \ 390 \ \ \ \ \ \ \ \ \ \ \ 18 \ \ \ \ \ \ \ \ \ \ \ 324 \ \ \ \ \ \ \ \ \ \ \ 0.830769\\\\\text{Cinnabar vestigial roof} \ \ \ \ \ \ \ \ \ \ \ \ \ 63 \ \ \ \ \ \ \ \ \ \ \ \ \ 70\ \ \ \ \ \ \ \ \ \ \ \ -7 \ \ \ \ \ \ \ \ \ \ \ 49 \ \ \ \ \ \ \ \ \ \ \ 0.7\\\\\text{wild type} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 72 \ \ \ \ \ \ \ \ \ \ \ 70 \ \ \ \ \ \ \ \ \ \ \ 2 \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ 0.057143\\\\$

$vestigial \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 32 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 35 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.257143\\\\$

$\text{Cinnabar roof} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 34\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 35 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.028571\\\\\text{roof vestigial} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5 \ \ \ \ \ \ \ \ \ \ \ \ \ \ -1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.2\\\\$

$\text{cinnabar} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.8 \\\\$

$Total \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1000 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.965934$

Eight phenotypes were present.

Df is provided also by a number of phenotypes -1 The degree of freedom

$\to df = 8-1= 7$

For p-value 0,9, Chi-square is 2.83;

The p-value of 0.75 is 4.5. Chi-square

Chi-sqaure value is observed at 2.965.

That means 0.90>p-value>0.75.

7 0

3 years ago

Which is true about characterization?

Vanyuwa [196]

C.Playwrights can reveal characterization both directly and indirectly

6 0

3 years ago