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Vedmedyk [2.9K]
3 years ago
11

Figure one and figure 2 are similar quadrilaterals. Which proportion must be true?

Mathematics
1 answer:
amm18123 years ago
5 0

Step-by-step explanation:

option C is the correct answer of this question

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Write 1/10 in hundredth,fraction,decimal
Ipatiy [6.2K]
0.10
1/10

Hope I helped!

Let me know if you need anything else!

<span>~ Zoe</span>
4 0
3 years ago
Are the point (-1, 1) and (3, 1/5) solutions to the equation 3x-5y=8
Sati [7]
The (3, 1/5) would be a solution while the (-1,1) wouldn’t because when you write this equation in the slope intercept form which is y = 3/5x - 8/5

1/5 = 3/5 (3) - 8/5

would end up with 1/5 = 1/5 which is write

While,

1 = 3/5 (-1) - 8/5

Would end up with
1 = 11/5 which is wrong



5 0
3 years ago
on a bulletin board the principal ms gomez put 115 photos of the fourth grade students in her school she put the photos in 5 equ
avanturin [10]
So 115 divided by 5 is 23, so the answer is 23.
5 0
3 years ago
Find the general solution of x'1 = 3x1 - x2 + et, x'2 = x1.
Ann [662]
Note that if {x_2}'=x_1, then {x_2}''={x_1}', and so we can collapse the system of ODEs into a linear ODE:

{x_2}''=3{x_2}'-x_2+e^t
{x_2}''-3{x_2}'+x_2=e^t

which is a pretty standard linear ODE with constant coefficients. We have characteristic equation

r^2-3r+1=\left(r-\dfrac{3+\sqrt5}2\right)\left(r+\dfrac{3+\sqrt5}2\right)=0

so that the characteristic solution is

{x_2}_C=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}

Now let's suppose the particular solution is {x_2}_p=ae^t. Then

{x_2}_p={{x_2}_p}'={{x_2}_p}''=ae^t

and so

ae^t-3ae^t+ae^t=-ae^t=e^t\implies a=-1

Thus the general solution for x_2 is

x_2=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}-e^t

and you can find the solution x_1 by simply differentiating x_2.
7 0
3 years ago
Find all solutions of the given system of equations and check your answer graphically. HINT [See Examples 1-4.] (If there is no
Likurg_2 [28]

Answer:

Infinitely\ many\ solutions\ exist.\\\\Solutions\ are\ (x,\frac{3}{4}x-1)

Step-by-step explanation:

Given\ equations\ are\\\\3x-4y=4.................eq(1)\\\\9x-12y=12..............eq(2)\\\\divide\ eq(2)\ by\ 3\\\\\frac{1}{3}(9x-12y=12)\\\\\Rightarrow 3x-4y=4\\\\Hence\ equations\ represent\ the\ same\ line.\\Hence\ Infinitely\ many\ solutions\ exist.\\\\3x-4y=4\\\\4y=3x-4\\\\y=\frac{3}{4}x-1\\\\Solutions\ are\ (x,\frac{3}{4}x-1)

3 0
2 years ago
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