Having trouble here

Consider the function, f (x) = 4x – 5

Anastaziya [24]

I don't see a table lol, link it and I'll solve it

4 0

3 years ago

The length of the hypotenuse of a 30-60 -90 triangle is 14 units. what is the length of the short leg

kifflom [539]

Recall that the short leg is the side opposite the 30 degrees angle. And the ratio of the short leg to the hypotenuse is 1:2.

Let the length of the short leg be x, then,

$\cfrac{x}{14}=\cfrac{1}{2}\\ \\\Rightarrow2x=14\\ \\\Rightarrow x=\cfrac{14}{2}=7$

6 0

4 years ago

Seth’s father is thinking of buying his son a six-month movie pass for $140. With the pass,matinees cost 1.00.If nastiness are n

Gala2k [10]

Answer:

40 shows

Step-by-step explanation:

Seth’s father has spent $140.

At the regular price, the number (n) of shows Seth could attend is

n = 140/3.50

n = 40

If Seth attends 40 shows at that price, he will have spent $140.

Every show after that will cost only $1 instead of $3.50.

Seth must attend 40 shows for it to benefit his father.

5 0

3 years ago

Does 2x+18=25 equal X=3.5

alina1380 [7]

Answer:

Yes

Step-by-step explanation:

2x+18=25

2x=25-18

2x=7

x=7÷2

x=3.5

8 0

3 years ago

Read 2 more answers

Let R be the region bounded by

loris [4]

a. The area of $R$ is given by the integral

$\displaystyle \int_1^2 (x + 6) - 7\sin\left(\dfrac{\pi x}2\right) \, dx + \int_2^{22/7} (x+6) - 7(x-2)^2 \, dx \approx 9.36$

b. Use the shell method. Revolving $R$ about the $x$ -axis generates shells with height $h=x+6-7\sin\left(\frac{\pi x}2\right)$ when $1\le x\le 2$ , and $h=x+6-7(x-2)^2$ when $2\le x\le\frac{22}7$ . With radius $r=x$ , each shell of thickness $\Delta x$ contributes a volume of $2\pi r h \Delta x$ , so that as the number of shells gets larger and their thickness gets smaller, the total sum of their volumes converges to the definite integral

$\displaystyle 2\pi \int_1^2 x \left((x + 6) - 7\sin\left(\dfrac{\pi x}2\right)\right) \, dx + 2\pi \int_2^{22/7} x\left((x+6) - 7(x-2)^2\right) \, dx \approx 129.56$

c. Use the washer method. Revolving $R$ about the $y$ -axis generates washers with outer radius $r_{\rm out} = x+6$ , and inner radius $r_{\rm in}=7\sin\left(\frac{\pi x}2\right)$ if $1\le x\le2$ or $r_{\rm in} = 7(x-2)^2$ if $2\le x\le\frac{22}7$ . With thickness $\Delta x$ , each washer has volume $\pi (r_{\rm out}^2 - r_{\rm in}^2) \Delta x$ . As more and thinner washers get involved, the total volume converges to

$\displaystyle \pi \int_1^2 (x+6)^2 - \left(7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \pi \int_2^{22/7} (x+6)^2 - \left(7(x-2)^2\right)^2 \, dx \approx 304.16$ <em />

d. The side length of each square cross section is $s=x+6 - 7\sin\left(\frac{\pi x}2\right)$ when $1\le x\le2$ , and $s=x+6-7(x-2)^2$ when $2\le x\le\frac{22}7$ . With thickness $\Delta x$ , each cross section contributes a volume of $s^2 \Delta x$ . More and thinner sections lead to a total volume of

$\displaystyle \int_1^2 \left(x+6-7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \int_2^{22/7} \left(x+6-7(x-2)^2\right) ^2\, dx \approx 56.70$

7 0

2 years ago