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3 years ago

88th term of the sequence 4,-1,-6

Mathematics

1 answer:

MArishka [77]3 years ago

4 0

Answer:

-431

Step-by-step explanation:

4-5=-1 -1-5= -6 common difference is -5 and start at 4 to get 88th term which is -431

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Write the conversion factor for converting meters to centimeters

posledela

Since we are converting meters to centimeters we are converting large to small.

Expect the answer to be a larger number. Here are some example of large to small:
2 weeks (larger unit) = 14 days (smaller unit) 14 is larger than 2
4 days (larger unit ) = 96 hours (smaller unit) 96 is larger than 4
Since there are 100cm in a meter, when we change meters to cm, we need a bigger number. The conversion factor is 100 and should be used as a multiplier.
Example: 47m x 100 = 4700 cm

7 0

3 years ago

A bicycle wheel has a circumference of 62 inches. Find the diameter of the wheel. If necessary, round to the tenths place. A. 39

san4es73 [151]

Answer:

The answer to your question is diameter = 19.7 in

Step-by-step explanation:

Data

Circumference = 62 in

diameter = ?

Process

1.- Write the formula of Circumference, remember that circumference = perimeter.

Perimeter = 2πr

2.- Equal the equation to the value given

2πr = 62

-Solve it for r

r = 62/2(3.14)

-Result

r = 62/6.28

r = 9.87 in

3.- Find the diameter

diameter = 2r

diameter = 2(9.87)

diameter = 19.7 in

4 0

3 years ago

Read 2 more answers

Find the volume of the solid.

dmitriy555 [2]

In Cartesian coordinates, the region (call it $R$ ) is the set

$R = \left\{(x,y,z) ~:~ x\ge0 \text{ and } y\ge0 \text{ and } 2 \le z \le 4-x^2-y^2\right\}$

In the plane $z=2$ , we have

$2 = 4 - x^2 - y^2 \implies x^2 + y^2 = 2 = \left(\sqrt2\right)^2$

which is a circle with radius $\sqrt2$ . Then we can better describe the solid by

$R = \left\{(x,y,z) ~:~ 0 \le x \le \sqrt2 \text{ and } 0 \le y \le \sqrt{2 - x^2} \text{ and } 2 \le z \le 4 - x^2 - y^2 \right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\sqrt2} \int_0^{\sqrt{2-x^2}} \int_2^{4-x^2-y^2} dz \, dy \, dx$

While doable, it's easier to compute the volume in cylindrical coordinates.

$\begin{cases} x = r \cos(\theta) \\ y = r\sin(\theta) \\ z = \zeta \end{cases} \implies \begin{cases}x^2 + y^2 = r^2 \\ dV = r\,dr\,d\theta\,d\zeta\end{cases}$

Then we can describe $R$ in cylindrical coordinates by

$R = \left\{(r,\theta,\zeta) ~:~ 0 \le r \le \sqrt2 \text{ and } 0 \le \theta \le\dfrac\pi2 \text{ and } 2 \le \zeta \le 4 - r^2\right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\pi/2} \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta \, dr \, d\theta \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta\,dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} r((4 - r^2) - 2) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} (2r-r^3) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \left(\left(\sqrt2\right)^2 - \frac{\left(\sqrt2\right)^4}4\right) = \boxed{\frac\pi2}$

3 0

1 year ago

PLZ help Asap!!!!!!!!!!!!!!!!!!!!!!!

maria [59]

I'm may not be that good at math, but I'm going to try my best to help you, because I know how it feels to be in this situation!

The first one :
You might be right on D. But I'm thinking B also. The question asks "Which could be a cross section of a rectangular pyramid that has been intersected by a plane perpendicular". But a triangle is basically a flat pyramid.

(I'm typing really fast to help!!!!)

Second one: Its B!! Look at it really closely! You'll get it!!!

(Going really fast now!!!!)

Third one: Between B or D!!! But I think you're right on your choice!!!!

(FASTER FASTER!!!)

FINALLY: Copy and paste your answers into google. People have asked the same question!

I hope I helped. I don't want to make you fail, as I am in the SAME similar situation as you! I really REALLY REALLY HOPE YOU PASS!!!

3 0

3 years ago

Find m abc (2x) (5x+5)

Gemiola [76]

< ABD + > ABC = 180.....because these angles form a linear pair
2x + 5x + 5 = 180
7x + 5 = 180
7x = 180 - 5
7x = 175
x = 175/7
x = 25

< ABC = 5x + 5......5(25) + 5 = 130 <==

3 0

3 years ago

Read 2 more answers