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3 years ago

What’s the answer ??

Mathematics

1 answer:

scoundrel [369]3 years ago

6 0

So to find the answer, you'll have to substitute b for 6 and a for 1/4 as shown. Then, you will get 5(6)/32(1/4)^2.
Now, all you have to do is just calculate it, which you will get 15 as your final answer.
I hope this helps :)

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Find the distance between the two points rounding to the nearest tenth (if

lana [24]

Answer:

5.7

Step-by-step explanation:

Creating the point into a triangle, both sides a 4.

As the Pythagorean theorem goes, a^2+b^2=c^2

4^2+4^2=

16+16=

The square root of 32 is about 5.7

The distance between the points is about 5.7 units

6 0

3 years ago

Let f(x) = 7x − 13. Find f−1(x)

motikmotik

$f(x)=7x-13\\\\y=7x-13\ \ \ \ |add\ 13\ to\ both\ sides\\\\y+13=7x\ \ \ \ \ |divide\ both\ sides\ by\ 7\\\\x=\dfrac{y+13}{7}\\\\Answer:f^{-1}(x)=\dfrac{y+13}{7}$

6 0

3 years ago

I need to know the answer to those two equations

sladkih [1.3K]

6(3x-5)-7x=25
18x-30-7x=25
11x=55
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3 0

3 years ago

What is the midpoint of coordinates (6, 1) and (9, -3)?

ZanzabumX [31]

Answer:

(7.5, -1)

Step-by-step explanation:

hope this helps you!

4 0

3 years ago

Determine the formula for the nth term of the sequence:<br>-2,1,7,25,79,...

rodikova [14]

A plausible guess might be that the sequence is formed by a degree-4* polynomial,

$x_n = a n^4 + b n^3 + c n^2 + d n + e$

From the given known values of the sequence, we have

$\begin{cases}a+b+c+d+e = -2 \\ 16 a + 8 b + 4 c + 2 d + e = 1 \\ 81 a + 27 b + 9 c + 3 d + e = 7 \\ 256 a + 64 b + 16 c + 4 d + e = 25 \\ 625 a + 125 b + 25 c + 5 d + e = 79\end{cases}$

Solving the system yields coefficients

$a=\dfrac58, b=-\dfrac{19}4, c=\dfrac{115}8, d = -\dfrac{65}4, e=4$

so that the n-th term in the sequence might be

$\displaystyle x_n = \boxed{\frac{5 n^4}{8}-\frac{19 n^3}{4}+\frac{115 n^2}{8}-\frac{65 n}{4}+4}$

Then the next few terms in the sequence could very well be

$\{-2, 1, 7, 25, 79, 208, 466, 922, 1660, 2779, \ldots\}$

It would be much easier to confirm this had the given sequence provided just one more term...

* Why degree-4? This rests on the assumption that the higher-order forward differences of $\{x_n\}$ eventually form a constant sequence. But we only have enough information to find one term in the sequence of 4th-order differences. Denote the k-th-order forward differences of $\{x_n\}$ by $\Delta^{k}\{x_n\}$ . Then

• 1st-order differences:

$\Delta\{x_n\} = \{1-(-2), 7-1, 25-7, 79-25,\ldots\} = \{3,6,18,54,\ldots\}$

• 2nd-order differences:

$\Delta^2\{x_n\} = \{6-3,18-6,54-18,\ldots\} = \{3,12,36,\ldots\}$

• 3rd-order differences:

$\Delta^3\{x_n\} = \{12-3, 36-12,\ldots\} = \{9,24,\ldots\}$

• 4th-order differences:

$\Delta^4\{x_n\} = \{24-9,\ldots\} = \{15,\ldots\}$

From here I made the assumption that $\Delta^4\{x_n\}$ is the constant sequence {15, 15, 15, …}. This implies $\Delta^3\{x_n\}$ forms an arithmetic/linear sequence, which implies $\Delta^2\{x_n\}$ forms a quadratic sequence, and so on up $\{x_n\}$ forming a quartic sequence. Then we can use the method of undetermined coefficients to find it.

5 0

2 years ago