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zaharov [31]
2 years ago
12

Are the two figures congruent?

Mathematics
1 answer:
geniusboy [140]2 years ago
6 0

Answer:

No, because the reflection of ABC is not congruent to A’B’C’.

Step-by-step explanation:

we know that

The rule of the reflection across the line y=x is equal to

(x,y) -------> (y,x)

so

A(-6,6) -------> A'(6,-6) ----> is ok

B(-3,3) ------> B'(3,-3) ----> is ok

C(-8,2) ------> C'(8,-2) ----> is not ok ( is not a reflection acros the line y=x)

therefore

The triangles are no t congruent, because the reflection of ABC is not congruent to A’B’C’

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WARRIOR [948]

Answer:

the graph one is G and the second one is C hope this helps

4 0
3 years ago
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Man help plzzz ive been struggling
SpyIntel [72]

Answer:

a

Step-by-step explanation:

add all together

4 0
2 years ago
(5) Find the Laplace transform of the following time functions: (a) f(t) = 20.5 + 10t + t 2 + δ(t), where δ(t) is the unit impul
Aloiza [94]

Answer

(a) F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}

(b) F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}

Step-by-step explanation:

(a) f(t) = 20.5 + 10t + t^2 + δ(t)

where δ(t) = unit impulse function

The Laplace transform of function f(t) is given as:

F(s) = \int\limits^a_0 f(s)e^{-st} \, dt

where a = ∞

=>  F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt

where d(t) = δ(t)

=> F(s) = \int\limits^a_0 {(20.5e^{-st} + 10te^{-st} + t^2e^{-st} + d(t)e^{-st}) \, dt

Integrating, we have:

=> F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3}  )\left \{ {{a} \atop {0}} \right.

Inputting the boundary conditions t = a = ∞, t = 0:

F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}

(b) f(t) = e^{-t} + 4e^{-4t} + te^{-3t}

The Laplace transform of function f(t) is given as:

F(s) = \int\limits^a_0 (e^{-t} + 4e^{-4t} + te^{-3t} )e^{-st} \, dt

F(s) = \int\limits^a_0 (e^{-t}e^{-st} + 4e^{-4t}e^{-st} + te^{-3t}e^{-st} ) \, dt

F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt

Integrating, we have:

F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.

Inputting the boundary condition, t = a = ∞, t = 0:

F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}

3 0
3 years ago
Which expression helps you find the length x of a side of a rectangle that has a diagonal of 15 units and a width of 9 units?
aliina [53]
15²-9²= x² use Pythagoras' theorum and square the two smaller sides to get the square of the hypotenuse (the diagonal) then rearrange this to get the above calculation
6 0
3 years ago
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please answer this question fast and please show all your work because I am locking for how can you do it please ​
inysia [295]

Option C is the correct values of the relationship between the number of cakes the baker makes and the number of bags of flour uses.

Solution:

Option A: Ratio of cakes baked to bags of flour used

$\frac{9}{4}, \frac{18}{6}=\frac{3}{1} \text { and } \frac{27}{9}=\frac{3}{1}

Here the ratios are not same.

So, this option is not true.

Option  B: Ratio of cakes baked to bags of flour used

$\frac{6}{2}=\frac{3}{1}, \frac{12}{4}=\frac{3}{1} \text { and } \frac{18}{6}=\frac{3}{1}

Here the ratios are same.

So, this option is true.

Option C: Ratio of cakes baked to bags of flour used

$\frac{4}{9}, \frac{6}{18}=\frac{1}{3} \text { and } \frac{9}{17}

Here the ratios are not same.

So, this option is not true.

Option D: In this table cakes baked is 6 and the bags of flour is 18.

But a baker made 18 cakes using 6 bags of flour.

So, this option is not true.

Hence option C is the correct values of relationship between the number of cakes the baker makes and the number of bags of flour uses.

6 0
3 years ago
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