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artcher [175]
3 years ago
12

What is the answer to the question -2y-10+2x=0

Mathematics
2 answers:
alekssr [168]3 years ago
8 0

Answer:

if your solving for x: x = y + 5

If your solving for y: y = − 5 + x

Step-by-step explanation:

Pavlova-9 [17]3 years ago
3 0

Answer:

X-y=5

Step-by-step explanation:

-2y-10+2x=0

-2y+2x=10

-y+x=5

x-y=5

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What is the length of an arc with a central angle of 23π radians and a radius of 24 cm?
dalvyx [7]
We can solve for the arc length using the formula shown below:
Arc Lenght = 2pi*r(central angle/360)

We need to convert the central angle such as:
Central angle = 23pi rad * (180°/pi rad) = 23*180
Radius = 24cm 

Solving for arc length:
Arc length = 2*3.14 *24*(23*180/360)
Arc length = 1733.28 cm

7 0
3 years ago
A can of paint will cover 400 square feet. What is the length of the edge of the largest square that can be painted with this ca
Masteriza [31]
One gallon can of paint will cover up to 400 square feet, which is enough to cover a small room like a bathroom. Two gallon cans of paint cover up to 800 square feet, which is enough to cover an average size room. This is the most common amount needed, especially when considering second coat coverage.
7 0
3 years ago
2,17,82,257,626,1297 next one please ?​
In-s [12.5K]

The easy thing to do is notice that 1^4 = 1, 2^4 = 16, 3^4 = 81, and so on, so the sequence follows the rule n^4+1. The next number would then be fourth power of 7 plus 1, or 2402.

And the harder way: Denote the <em>n</em>-th term in this sequence by a_n, and denote the given sequence by \{a_n\}_{n\ge1}.

Let b_n denote the <em>n</em>-th term in the sequence of forward differences of \{a_n\}, defined by

b_n=a_{n+1}-a_n

for <em>n</em> ≥ 1. That is, \{b_n\} is the sequence with

b_1=a_2-a_1=17-2=15

b_2=a_3-a_2=82-17=65

b_3=a_4-a_3=175

b_4=a_5-a_4=369

b_5=a_6-a_5=671

and so on.

Next, let c_n denote the <em>n</em>-th term of the differences of \{b_n\}, i.e. for <em>n</em> ≥ 1,

c_n=b_{n+1}-b_n

so that

c_1=b_2-b_1=65-15=50

c_2=110

c_3=194

c_4=302

etc.

Again: let d_n denote the <em>n</em>-th difference of \{c_n\}:

d_n=c_{n+1}-c_n

d_1=c_2-c_1=60

d_2=84

d_3=108

etc.

One more time: let e_n denote the <em>n</em>-th difference of \{d_n\}:

e_n=d_{n+1}-d_n

e_1=d_2-d_1=24

e_2=24

etc.

The fact that these last differences are constant is a good sign that e_n=24 for all <em>n</em> ≥ 1. Assuming this, we would see that \{d_n\} is an arithmetic sequence given recursively by

\begin{cases}d_1=60\\d_{n+1}=d_n+24&\text{for }n>1\end{cases}

and we can easily find the explicit rule:

d_2=d_1+24

d_3=d_2+24=d_1+24\cdot2

d_4=d_3+24=d_1+24\cdot3

and so on, up to

d_n=d_1+24(n-1)

d_n=24n+36

Use the same strategy to find a closed form for \{c_n\}, then for \{b_n\}, and finally \{a_n\}.

\begin{cases}c_1=50\\c_{n+1}=c_n+24n+36&\text{for }n>1\end{cases}

c_2=c_1+24\cdot1+36

c_3=c_2+24\cdot2+36=c_1+24(1+2)+36\cdot2

c_4=c_3+24\cdot3+36=c_1+24(1+2+3)+36\cdot3

and so on, up to

c_n=c_1+24(1+2+3+\cdots+(n-1))+36(n-1)

Recall the formula for the sum of consecutive integers:

1+2+3+\cdots+n=\displaystyle\sum_{k=1}^nk=\frac{n(n+1)}2

\implies c_n=c_1+\dfrac{24(n-1)n}2+36(n-1)

\implies c_n=12n^2+24n+14

\begin{cases}b_1=15\\b_{n+1}=b_n+12n^2+24n+14&\text{for }n>1\end{cases}

b_2=b_1+12\cdot1^2+24\cdot1+14

b_3=b_2+12\cdot2^2+24\cdot2+14=b_1+12(1^2+2^2)+24(1+2)+14\cdot2

b_4=b_3+12\cdot3^2+24\cdot3+14=b_1+12(1^2+2^2+3^2)+24(1+2+3)+14\cdot3

and so on, up to

b_n=b_1+12(1^2+2^2+3^2+\cdots+(n-1)^2)+24(1+2+3+\cdots+(n-1))+14(n-1)

Recall the formula for the sum of squares of consecutive integers:

1^2+2^2+3^2+\cdots+n^2=\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6

\implies b_n=15+\dfrac{12(n-1)n(2(n-1)+1)}6+\dfrac{24(n-1)n}2+14(n-1)

\implies b_n=4n^3+6n^2+4n+1

\begin{cases}a_1=2\\a_{n+1}=a_n+4n^3+6n^2+4n+1&\text{for }n>1\end{cases}

a_2=a_1+4\cdot1^3+6\cdot1^2+4\cdot1+1

a_3=a_2+4(1^3+2^3)+6(1^2+2^2)+4(1+2)+1\cdot2

a_4=a_3+4(1^3+2^3+3^3)+6(1^2+2^2+3^2)+4(1+2+3)+1\cdot3

\implies a_n=a_1+4\displaystyle\sum_{k=1}^3k^3+6\sum_{k=1}^3k^2+4\sum_{k=1}^3k+\sum_{k=1}^{n-1}1

\displaystyle\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}4

\implies a_n=2+\dfrac{4(n-1)^2n^2}4+\dfrac{6(n-1)n(2n)}6+\dfrac{4(n-1)n}2+(n-1)

\implies a_n=n^4+1

4 0
3 years ago
Historical data indicates that only 20% of cable customers are willing to switch companies. If a binomial process is assumed, th
Natasha_Volkova [10]

Answer:

a. The probability is 0.735

b. The probability is 0.6296

c. The probability is 0

Step-by-step explanation:

If we assume a binomial process, the probability that x customer are willing to switch companies is:

P(x)=nCx*p^{x}*(1-p)^{n-x}

nCx is calculated as:

nCx=\frac{n!}{x!(n-x)!}

Where n are the 20 cable customers and p is the probability 0.2 that the cable customer are willing to switch companies.

Then, P(x) is:

P(x)=20Cx*0.2^{x}*(1-0.2)^{20-x}

The probability that between 2 and 5 (inclusive) customers are willing to switch companies is:

P(2≤x≤5) = P(2) + P(3) + P(4) + P(5)

Where P(2), P(3), P(4) and P(5) are equal to:

P(2)=20C2*0.2^{2}*(1-0.2)^{20-2}=0.1369

P(3)=20C3*0.2^{3}*(1-0.2)^{20-3}=0.2054

P(4)=20C4*0.2^{4}*(1-0.2)^{20-4}=0.2182

P(5)=20C5*0.2^{5}*(1-0.2)^{20-5}=0.1745

So, P(2≤x≤5) is:

P(2≤x≤5) = 0.1369 + 0.2054 + 0.2182 + 0.1745 = 0.735

At the same way, the probability that less than 5 customers are willing to switch is:

P(x<5)=P(0)+P(1)+P(2)+P(3)+P(4)

P(x<5)=0.6296

Finally, the probability that more than 16 customers are willing to switch is:

P(x>16)=P(17)+P(18)+P(19)+P(20)

P(x>16)=0

7 0
4 years ago
The sum of three numbers is 131. The second of three numbers is seven more than twice the first. The third number is 12 less tha
Norma-Jean [14]
First number is (x).
Second number is (2x+7).
Third number is (x-12).
Their sum is  (x+2x+7+x-12).
x+2x+7+x-12 =131

You can simplify this equation
4x-5=131,

4x=136
x=136/4=34 (first number), 
(2x+7)=34*2+7=75 (second number)
34-x=34-12 =22(third number)

Check: 34+75+22=131
6 0
3 years ago
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