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Diano4ka-milaya [45]
3 years ago
11

PLSSS HELP IF YOU TURLY KNOW THISS

Mathematics
1 answer:
ExtremeBDS [4]3 years ago
7 0

Answer:

0.81

Step-by-step explanation:

how this helps!

btw: you spelled truly wrong

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1. What is the approximate volume of the cone? Use 3.14 for π.
iren2701 [21]
The volume of a cone is V = (1/3)*pi*h*r^2
V = (1/3)*3.14*8*12^2 = 1206 cu.cm
4 0
4 years ago
Y=1/20x^2<br> The directrix of the parabola is:<br> x=-5<br> x=5<br> y=-5
rusak2 [61]

Answer:

it's y=5

Step-by-step explanation:

5 0
3 years ago
Oliver has created a plot of his backyard on the given coordinate plane, so that each unit length is one yard. He wants to desig
Karolina [17]

Answer:

The possible locations are;

(3, 4), (3, 12), (7, 4), and (7, 12)

Please find attached the plot of the points created with Microsoft Excel      

Step-by-step explanation:

The given parameters of the plot of Oliver's backyard are;

The area of the right triangular plant bed Oliver wants to design, A = 8 yd²

The length of a/each unit of the plot = 1 yard

The location of one of the vertex of the right triangular area = (3, 8)

The location of the other vertex of the right triangular area = (7, 8)

We note that the area of a triangle, A = 1/2 × Base length, B × Height, h

Let the distance between the two plotted point represent the base length, 'B', therefore;

B = 7 - 3 = 4 (We subtract only the x-values (7 and 3) given that the y-values (8 and 8) are the same;

Therefore, the base length, B = 4 yards

A = 1/2 × B × h

∴ A = 8 yd.² = 1/2 × 4 yd. × h

h = 8 yd.²/ 2 yd. = 4 yd.

The height of the right triangle, h = 4 yd.

Therefore the possible locations of the last vertex the right triangle on the coordinate plane, given that the first two points are parallel with the x-axis, are points 4 units above or the y-axis value of the plotted points with the x-value of the last vertex being either 3, or 7 as follows;

(3, 8 ± 4) or (7, 8 ± 4) which give points, (3, 4), (3, 12), (7, 4), and (7, 12).

8 0
3 years ago
30 pts+
Veseljchak [2.6K]

Answer:

I dont know the answer I'm just scrolling down this app because I like going through my brothers phone exceptionally mommy and daddy wrestling

7 0
2 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
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