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3 years ago

Angle 1 is 50 degrees. Which angle is also 50 degrees?

Mathematics

1 answer:

solong [7]3 years ago

4 0

Answer:

Step-by-step explanation:

angle 1= 50 degrees

x angle=50 degrees

cross multiply and you get and you get 1 as your answer

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Explane why kequally 10 is the consentof proporitomally

Oxana [17]

I don't understand the question, could you please re-word it and if I know it I will give u the answer

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3 years ago

ALGEBRA 1 HELP URGENT!!

butalik [34]

Here, Sequence : -2, -6, -18, -54
Common Ratio = a₂/a₁ = -6/-2 = 3
As the expression is in negative sign, it's sequence will be -3ˣ
Where x = number of terms in the sequence [ By this rule, you can calculate the unknown nth number]

In short, Option C is your answer!!!!

Let me know, if you have some doubt.

5 0

2 years ago

A toy manufacturer makes lightweight balls for indoor play. The large basketball uses 4 ounces of foam and 20 minutes of labor a

trapecia [35]

Answer:

(24, 224 ) i.e. 24 basketball and 224 footballs

Step-by-step explanation:

Basketball :

uses 4 ounces of foam

20 minutes of labor

profit = $2.50

Football :

uses 3 ounces of foam

30 minutes of labor

profit = $2

Manufacturer has :

768 ounces of foam , 120 labor hours per week,

using the simplex method to determine the optimal production schedule so as to maximize profits

lets assume

x = number of basketball

y = number of footballs

maximize = 2.5 x + 2y from a linear programming problem

attached below is the detailed solution

6 0

2 years ago

What is the 5th term of this geometric sequence? 100, 20, 4, 0.8, …

Alex777 [14]

First we have to find out the common number of them.
For that purpose, the first number should be divided by next number.
100/20 = 5
20/4 = 5
4/0.8 = 5

we have found that 5 is the common number.
0.8/5 = 0.16

therefore the answer is c

6 0

2 years ago

How do I find the integral ∫10(x−1)(x2+9)dxint10/((x-1)(x^2+9))dx ?

defon

$\int\frac{10}{(x-1)(x^2+9)}\ dx=(*)\\\\\frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}=\frac{A(x^2+9)+(Bx+C)(x-1)}{(x-1)(x^2+9)}\\\\=\frac{Ax^2+9A+Bx^2-Bx+Cx-C}{(x-1)(x^2+9)}=\frac{(A+B)x^2+(-B+C)x+(9A-C)}{(x-1)(x^2+9)}\\\Updownarrow\\10=(A+B)x^2+(-B+C)x+(9A-C)\\\Updownarrow\\A+B=0\ and\ -B+C=0\ and\ 9A-C=10\\A=-B\ and\ C=B\to9(-B)-B=10\to-10B=10\to B=-1\\A=-(-1)=1\ and\ C=-1$

$(*)=\int\left(\frac{1}{x-1}+\frac{-x-1}{x^2+9}\right)\ dx=\int\left(\frac{1}{x-1}-\frac{x+1}{x^2+9}\right)\ dx\\\\=\int\frac{1}{x-1}\ dx-\int\frac{x+1}{x^2+9}\ dx=\int\frac{1}{x-1}-\int\frac{x}{x^2+9}\ dx-\int\frac{1}{x^2+9}\ dx=(**)\\\\\#1\ \int\frac{1}{x-1}\ dx\Rightarrow\left|\begin{array}{ccc}x-1=t\\dx=dt\end{array}\right|\Rightarrow\int\frac{1}{t}\ dt=lnt+C_1=ln(x-1)+C_1$

$\#2\ \int\frac{x}{x^2+9}\ dx\Rightarrow \left|\begin{array}{ccc}x^2+9=u\\2x\ dx=du\\x\ dx=\frac{1}{2}\ du\end{array}\right|\Rightarrow\int\left(\frac{1}{2}\cdot\frac{1}{u}\right)\ du=\frac{1}{2}\int\frac{1}{u}\ du\\\\\\=\frac{1}{2}ln(u)+C_2=\frac{1}{2}ln(x^2+9)+C_2$

$\#3\ \int\frac{1}{x^2+9}\ dx=\int\frac{1}{x^2+3^2}\ dx=\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3\\\\therefore:\\\\\#1;\ \#2;\ \#3\Rightarrow(**)=ln(x-1)+C_1-\frac{1}{2}ln(x^2+9)+C_2-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3$

$\boxed{=ln(x-1)-\frac{1}{2}ln(x^2+9)-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C}$

4 0

3 years ago