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3 years ago

12

Graph the functions f(x)=−6x+14 and g(x)=−2x+6 on the same coordinate plane. What are the solutions of the equation −6x+14=−2x+6

? Select each correct answer. 0 1 2 3 4

1 answer:

asambeis [7]3 years ago

8 0

Answer:

2 AND 4

Step-by-step explanation:

Brainliest please

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I ONLY need 8c<br> Please show ALL STEPS

viktelen [127]

Answer:

8c

f(g(x)) = x^4 + 2x^3 - x

g(f(x)) = x^4 + 2x^3 + 2x^2 - x

Step-by-step explanation:

f(x) = x^2 - x ; g(x) = x^2 + x

f(g(x)) = (x^2 + x)^2 - (x^2 + x)

f(g(x)) = (x^2 + x)^2 - x^2 - x

f(g(x)) = (x^2 + x)(x^2 + x) - x^2 - x

f(g(x)) = x^4 + x^3 + x^3 + x^2 - x^2 - x

f(g(x)) = x^4 + 2x^3 - x

g(f(x)) = (x^2 - x)^2 + x^2 - x

g(f(x)) = (x^2 + x)(x^2 + x) + x^2 - x

g(f(x)) = x^4 + x^3 + x^3 + x^2 + x^2 - x

g(f(x)) = x^4 + 2x^3 + 2x^2 - x

3 0

2 years ago

A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 10.4

vlabodo [156]

Answer:

Therefore the diameter of the hole is $1.94 \times 10^{-3}$ m.

Step-by-step explanation:

Bernoulli's equation,

$P_1+\frac12 \rho v^2_1+\rho g h_1= P_2+\frac12 \rho v^2_2+\rho g h_2$

P₁ = P₂= atmospheric presser

$\rho$ = density

$\frac12 \rho v^2_1+\rho g h_1= \frac12 \rho v^2_2+\rho g h_2$ [since P₁ = P₂]

$\Rightarrow\rho (\frac12 v^2_1+ g h_1)= \rho(\frac12 v^2_2+ g h_2)$

$\Rightarrow\frac12 v^2_1+ g h_1= \frac12 v^2_2+ g h_2$

$\Rightarrow\frac12 v^2_2-\frac12 v^2_1=g h_1- g h_2$

$\Rightarrow v^2_2- v^2_1=2g h$ [ $h_1-h_2=h$ ]

Here $v_1\approx 0$

$\Rightarrow v^2_2=2g h$

$\therefore v_2=\sqrt {2gh$

Here g= 9.8 m/s² , h = 10.4 m

The velocity of water that leaves from the hole $v_2$ = $\sqrt {2\times 9.8\times 10.4}$ m/s

=14.28 m/s.

Given, the rate of flow from the leak is $2.53\times 10^{-3}$ $m^3/min$

$=\frac{2.53\times 10^{-3}}{60}$ $m^3/s$

Let the diameter of the hole be d.

Then the cross section area of the hole is $=\pi (\frac d2)^2$

We know that,

The rate of flow = Cross section area × speed

$\Rightarrow \frac{2.53\times 10^{-3}}{60} =\pi (\frac d2)^2\times 14.28$

$\Rightarrow (\frac d2)^2=\frac{2.53\times 10^{-3}}{60\times 14.28\times \pi}$

$\Rightarrow d= 1.94 \times 10^{-3}$

Therefore the diameter of the hole is $1.94 \times 10^{-3}$ m.

4 0

3 years ago