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3 years ago

Let f(x) = 4x – 5 and g(x) = 3x + 7. find f(x) + g(x) and state its domain.

Mathematics

1 answer:

Tema [17]3 years ago

5 0

We are given the functions:

f(x) = 4 x – 5 ---> 1

g(x) = 3 x + 7 ---> 2

To find for the value of f(x) + g(x), all we have to do is to add equations 1 and 2:

f(x) + g(x) = 4 x – 5 + 3 x + 7

f(x) + g(x) = 7 x + 2 = y

In this case, for any real number value assign to x, we get a real number value of y. This is because the function is linear.

Therefore the domain of the function is all real numbers.

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Which of the following is equivalent to –2i(6 – 7i)? (0 – 2i)(6 – 7i) (0 + 2i)(6 – 7i) (–2 + i)(6 – 7i) (6 – 2i) – (7i – 2i)

Mariulka [41]

Answer:

The expression which is equivalent to –2i(6 – 7i) is; (0 – 2i)(6 – 7i)

4 0

2 years ago

Find an equation of the tangent line to the curve at the given point. y = x + tan(x) at (pi, pi)

S_A_V [24]

Differentiating :

y' = 1 + sec^2(x),

cosπ=−1

plug in pi = -1 ,for x:

1 + sec^2(pi) = 1 + (-1)^2
= 2

8 0

3 years ago

If a ladder is 5.5 m long and it is placed 3 m from the wall, solve and show the work when you use the Pythagorean Theorem to fi

natali 33 [55]

Answer:

4.609 meters

Step-by-step explanation:

The hypotenuse is the ladder because it is the slanted side. 3m is one of the other side lengths

A^2 + B^2 = C^2

3^2 + B^2 = 5.5^2

9 + B^2 = 30.25

30.25 - 9= B^2

21.25= B^2

4.609=B

4 0

3 years ago

Help please! (no one was answering it in physics and i really need help)

aivan3 [116]

Let $v_0$ be the cat's speed just as it leaves the edge of the table. Then taking the point 1.3 m below the edge of the table to be the origin, the cat's horizontal position at time $t$ is given by

$x(t)=v_0t$

and its height is

$y(t)=1.3\,\mathrm m-gt^2$

where $g$ is 9.8 m/s^2, the magnitude of the acceleration due to gravity.

The time it takes for the cat to hit the ground is $t$ with

$0=1.3\,\mathrm m-gt^2\implies t=\sqrt{\dfrac{1.3\,\rm m}g}\approx0.36\,\mathrm s$

(Unfortunately, this doesn't match any of the given options...)

The cat lands 0.75 m away (horizontally) from the edge of the table, so that its speed $v_0$ was

$0.75\,\mathrm m=v_0(0.36\,\mathrm s)\implies v_0\approx2.08\dfrac{\rm m}{\rm s}$

(Again, not one of the answer choices...)

I'm guessing there's either a typo in the question or answers.

6 0

3 years ago

Several students from boxerville middle stood in a circle, evenly spaced, to play a game. If the 6th student was directly across

myrzilka [38]

Answer:

There were 14 students in the circle

Step-by-step explanation:

* Lets explain how to solve the problem

- The students stood in a circle

∴ The line joining each two opposite students is the diameter

of the circle

- The 6th student was directly across from the 13th student

∴ The 6th student is opposite to the 13th student

- Lets count how many students between 6th student and

13th student

∵ 7th , 8th , 9th , 10th , 11th , 12th students between the 6th

student and 13th student

∴ There are 6 students between them

- Lets count how many students between 13th student and

6th student

∵ 14th , 1st , 2nd , 3rd , 4th , 5th students between the 13th

student and 6th student

∴ There are 6 students between them

∵ There are 6 students between 6th and 13th

∵ There are 6 students between 13th and 6th

∵ There are 2 students in 6th and 13th

∴ The number of the students = 6 + 6 + 2 = 14

∴ There were 14 students in the circle

8 0

3 years ago