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zysi [14]
3 years ago
14

There are 5 blue marbles, 11 green marbles and

Mathematics
2 answers:
Margarita [4]3 years ago
4 0

Step-by-step explanation:

step by step by solutions are given

check it

hope you like the answer

Vlad1618 [11]3 years ago
4 0
C. 55% is. The answer
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Solve this system of equations by graphing. First graph the equations, and then type the solution.
stealth61 [152]

Answer:

The solution is the point (1,-4)

Step-by-step explanation:

we have

y=-3x-1 ----> equation A

y=-2x-2 ----> equation B

we know that

The solution of the system of equations by graphing is the intersection point both graphs

using a graphing tool

The intersection point is (1,-4)

see the attached figure

therefore

The solution is the point (1,-4)

5 0
3 years ago
Help with this slopeee ASAPPP
anygoal [31]

Answer: 1 over y 2 over x

Step-by-step explanation:

8 0
2 years ago
If someone could help me it would be awesome
alex41 [277]
The kite is 1500 feet above the ground
3 0
3 years ago
To qualify for the championship a runner must complete the race in less than 55 minutes ....... Use "t" to represent the time in
Svet_ta [14]

The inequality is t < 55

<em><u>Solution</u></em><em><u>:</u></em>

Given that, To qualify for the championship a runner must complete the race in less than 55 minutes

Let "t" represent the time in minutes of a runner who qualifies for the championship

Here it is given that the value of t is less than 55 minutes

Therefore, "t" must be less than 55, so that the runner qualifies the championship

<em><u>This is represented by inequality:</u></em>

t

The above inequality means, that time taken to complete the race must be less than 55 for a runner to qualify

Hence the required inequality is t < 55

3 0
3 years ago
What is the derivative of the function below?
11111nata11111 [884]

g(x)=4x-\dfrac{1}{4x}=4x-(4x)^{-1}\\\\\\\\g'(x)=\left[4x-(4x)^{-1}\right]'=(4x)'-\left[(4x)^{-1}\right]'=4-[-1(4x)^{-2}]\cdot4\\\\=4+\dfrac{1}{(4x)^2}\cdot4=4+\dfrac{4}{4^2x^2}=\boxed{4+\dfrac{1}{4x^2}}\\\\Used:\\\\(f(x)-g(x))'=f'(x)-g'(x)\\\\.[f(g(x))]'=f'(g(x))\cdot g'(x)\\\\(x^n)'=nx^{n-1}

8 0
3 years ago
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