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Maslowich
3 years ago
12

I need helps can someone help me please!!!

Mathematics
2 answers:
expeople1 [14]3 years ago
8 0

Answer:

JK=27

LM=27

KL=16

MJ=16

Step-by-step explanation:

JK=4x+7

=27

LM=5x+2

=27

KL=2y-2

=16

MJ=y+7

=16

so, JK= 27 and KL=16

(I hope u got it)

notsponge [240]3 years ago
4 0

Answer:

Jk= 27

LM= 22

KL= 16

MJ=16

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Given:

The graph of a function y=h(x).

To find:

The interval where h(x)>0.

Solution:

From the given graph graph it is clear that, the function before x=0 and after x=3.6 lies above the x-axis. So, h(x)>0 for x and x>3.6.

The function between x=0 and x=3.6 lies below the x-axis. So, h(x) for 0.

Now,

For -1, the graph of h(x) is above the x-axis. So, h(x)>0.

For 0, the graph of h(x) is below the x-axis. So, h(x).

For 1, the graph of h(x) is below the x-axis. So, h(x).

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STALIN [3.7K]
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2 years ago
If g(x)= square root of 3-5x, find the domain of g'(x).
BaLLatris [955]

Answer:

domain: x>3/5

Step-by-step explanation:

First we need to derive our function g(x) to get a new function g'(x)

To do this we will have to apply chain rule because we have an inner and outer functions.

Our G(x) = square root(3-5x)

Chain rule formula states that: d/dx(g(f(x)) = g'(f(x))f'(x)

where d/dx(g(f(x)) = g'(x)

g(x) is the outer function which is x^1/2

f(x) is our inner function which is 3-5x

therefore f'(x)= 1/2x^(-1/2) and f'(x) = -5

g'(f(x)) = -1/2(3-5x)^(-1/2)

Applying chain rule then g'(x) = 1/2 (3-5x)^(-/1/2)*(-5)

But the domain is the values of x where the function g'(x) is not defined

In this case it will be 3-5x > 0, because 3-5x is a denominator and anything divide by zero is infinity/undefined

which gives us x >3/5

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3 years ago
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Answer:

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Step-by-step explanation:

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3 years ago
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Answer:

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Step-by-step explanation:

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