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kvv77 [185]
3 years ago
7

Does anyone know this?

Mathematics
1 answer:
NARA [144]3 years ago
6 0

Answer:

42 cm²

Step-by-step explanation:

Length of the shaded rectangle = 7 cm

Width of the shaded rectangle= 7 - (0.5 + 0.5) = 7 - 1

Width of the shaded rectangle = 6 cm

Area of the shaded rectangle = L × W

Area = 7 × 6 = 42 cm²

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Please help me with this, it’s really important
Ivanshal [37]

Answer:

RST is obtuse ABC is a right angle XYZ is an acute angle

Step-by-step explanation:

4 0
3 years ago
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For each of the following equations,give the centre and radius of the circle.
allsm [11]

Answer:

a) Center is (0,0) and radius r = 5

b) Center is (0,0) and radius r = 1

d) Center is (5,3) and radius r = 9

e Center is (-2,1) and radius r = 3

g) Center is (-1,-3) and radius r = 7

h) Center is (0,0) and radius r = 2.5

Step-by-step explanation:

<u><em>a)</em></u>

Given circle  x² + y² = 25

The standard form of circle equation is

                   ( x-h)² +(y-0)² = r²

Center is (0,0) and radius r = 5

b)

Given circle  x² + y² = 1

The standard form of circle equation is

                   ( x-h)² +(y-0)² = r²

Center is (0,0) and radius r = 1

d)

Given circle  (x-5)² + (y-3)² = 81

The standard form of circle equation is

                   ( x-h)² +(y-0)² = r²

                   (x-5)² + (y-3)² = 9²

Center is (5,3) and radius r = 9

e)

Given circle  (x+2)² + (y+1)² = 9

The standard form of circle equation is

                   ( x-h)² +(y-0)² = r²

                  (x+2)² + (y+1)² = 3²

Center is (-2,-1) and radius r = 3

g)

Given circle  (x+1)² + (y+3)² = 49

The standard form of circle equation is

                   ( x-h)² +(y-0)² = r²

                  (x+1)² + (y+3)² = 7²

Center is (-1,-3) and radius r = 7

h)

Given circle  x² + y² = 6.25

The standard form of circle equation is

                   ( x-h)² +(y-0)² = r²

   Given circle  x² + y² = (2.5)²

Center is (0,0) and radius r = 2.5

5 0
3 years ago
6. Find the sum or difference.<br> 146<br> 155<br> a. +<br> 8<br> 8
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5 0
3 years ago
You move down 3 units and up 2 units. You end at (-5,1). Where did you start?
-BARSIC- [3]

Answer:

(-5, 2).

Step-by-step explanation:

Working backwards

Move up 3 units (-5, 1)  --->  ( -5, 4)

Move down 2 units (-5, 4)  ----> (-5, 2).

7 0
3 years ago
If the differential equation t 2y 00 − 2y 0 + (3 + t)y = 0 has y1(t) and y2(t) as a fundamental set of solutions and if W[y1, y2
podryga [215]

In the ODE, solve for y'':

t^2y''-2y'+(3+t)y=0\implies y''=\dfrac{2y'+(3+t)y}{t^2}

The Wronskian is then

W=\begin{vmatrix}y_1&y_2\\{y_1}'&{y_2}'\end{vmatrix}=y_1{y_2}'-{y_1}'y_2

Differentiating the Wronskian gives

W'=({y_1}'{y_2}'+y_1{y_2}'')-({y_1}''y_2+{y_1}'{y_2}')=y_1{y_2}''-{y_1}''y_2

Substitute y_1,y_2 into the equation for y'', then substitute {y_1}'',{y_2}'' into W':

W'=y_1\dfrac{2{y_2}'+(3+t)y_2}{t^2}-y_2\dfrac{2{y_1}'+(3+t)y_1}{t^2}

\implies W'=\dfrac{2W}{t^2}

which is another separable ODE; we have

\dfrac{\mathrm dW}W=\dfrac2{t^2}\,\mathrm dt\implies \ln|W|=-\dfrac2t+C\implies W=Ce^{-2/t}

Given that W(y_1,y_2)(2)=3, we find

3=Ce^{-2/2}\implies C=3e

so that

W(y_1,y_2)(t)=3e^{1-2/t}

and so

W(y_1,y_2)(6)=\boxed{3e^{2/3}}

8 0
3 years ago
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