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2 years ago

F(x) = -6(x + 2) - 3 Find the inverse relation

Mathematics

2 answers:

oksano4ka [1.4K]2 years ago

7 0

Answer:

inverse relation of x is f⁻(X) = (x+15)/-6

Step-by-step explanation:

Sphinxa [80]2 years ago

3 0

Hi i got y=(-x-15)/6 hahah

here’s my work:

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Given m LMN = 145°, what is m XMN?

astra-53 [7]

Answer:

m∠XMN = 80°

Step-by-step explanation:

you know that:

4x + 5 + 6x - 10 = 145

solve for x:

10x - 5 = 145

10x = 150

x = 15

Use this information to solve for m∠XMN

6(15) - 10

90 - 10

m∠XMN = 80°

7 0

2 years ago

A bucket that weighs 6 lb and a rope of negligible weight are used to draw water from a well that is 80 ft deep. The bucket is f

zaharov [31]

Answer:

3200 ft-lb

Step-by-step explanation:

To answer this question, we need to find the force applied by the rope on the bucket at time $t$

At $t=0, the weight of the bucket is 6+36=42 \mathrm{lb}$

After $t$ seconds, the weight of the bucket is $42-0.15 t \mathrm{lb}$

Since the acceleration of the bucket is the force on the bucket by the rope is equal to the weight of the bucket.

If the upward direction is positive, the displacement after $t$ seconds is $x=1.5 t$

Since the well is 80 ft deep, the time to pull out the bucket is $\frac{80}{2}=40 \mathrm{~s}$

We are now ready to calculate the work done by the rope on the bucket.

Since the displacement and the force are in the same direction, we can write

$W=\int_{t=0}^{t=36} F d x$

Use $x=1.5 t$ and $F=42-0.15 t$

$W=\int_{0}^{36}(42-0.15 t)(1.5 d t)$

$=\int_{0}^{36} 63-0.225 t d t$

$=63 \cdot 36-0.2 \cdot 36^{2}-0=3200 \mathrm{ft} \cdot \mathrm{lb}$

$=\left[63 t-0.2 t^{2}\right]_{0}^{36}$

$W=3200 \mathrm{ft} \cdot \mathrm{lb}$

4 0

2 years ago

Porfavor se los agradeceria mucho y de corazon :D

Zanzabum

Answer:

1=2p+3

Step-by-step explanation:

5 0

1 year ago

Please hepl with math

Ivenika [448]

Answer:

$\dfrac{5^{n+2}-6\times 5^{n+1}}{13 \times5^{n}-2\times5^{n+1}} = -\dfrac{5}{3}$

Step-by-step explanation:

We are given the expression to be simplified:

$\dfrac{5^{n+2}-6\times 5^{n+1}}{13 \times5^{n}-2\times5^{n+1}}$

Let us take common a term with a power of 5 from the numerator and the denominator of the given expression.

We know that:

$a^{p+q} = a^p \times a^q$

Let us use it to solve the powers of 5 in the given expression.

$\therefore$ we can write:

$5^{n+2} = 5^{n+1}\times 5= 5^n\times 5^{2}$

$5^{n+1} = 5^n\times 5$

The given expression becomes:

$\dfrac{5^{n+1} \times 5-6\times 5^{n+1}}{13 \times5^{n}-2\times5^{n}\times 5}$

Taking common $5^{n+1}$ from the numerator and

Taking common $5^{n}$ from the denominator

$\Rightarrow \dfrac{5^{n+1} (5-6)} {5^{n}(13-2\times5)}\\\Rightarrow \dfrac{5^{n+1} (-1)} {5^{n}(13-10)}\\\Rightarrow -\dfrac{5^{n+1}} {5^{n}\times3}\\\Rightarrow -\dfrac{5^{n}\times 5} {5^{n}\times3}\\\Rightarrow -\dfrac{5}{3}$

$\therefore$ The answer is:

$\dfrac{5^{n+2}-6\times 5^{n+1}}{13 \times5^{n}-2\times5^{n+1}} = -\dfrac{5}{3}$

6 0

2 years ago

A coach is ordering shirts for a team

oee [108]

Answer:

Step-by-step explanation:

Yes its B C=8k+24

5 0

2 years ago

Read 2 more answers