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Lady_Fox [76]
3 years ago
5

5 pounds of apples cost $8.20. How much is it for 1 pound?

Mathematics
2 answers:
Anarel [89]3 years ago
6 0
Just divide 8.2 by 5 and the answer is 1.64 
so if u were going to write it down on a piece of paper it would look like this... 
8.2/5=1.64 
so for one pound it would be... 1.64
Julli [10]3 years ago
3 0
$8.20÷5=$1.64 so that will be your answer. $1.64 per pound
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Len [333]

Answer:

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7 0
3 years ago
10. Write an equation in point-slope form and slope-intercept form for the line.
olga2289 [7]

We can use the points (2, -2) and (4, -1) to solve.

Slope formula: y2-y1/x2-x1

= -1-2/4-(-2)

= -3/6

= -1/2

Point slope form: y - y1 = m(x - x1)

y - 2 = -1/2(x + 2)

Solve for y-intercept.

-2 = -1/2(2)  + b

-2 = -1 + b

-2 + 1 = -1 + 1 + b

-1 = b

Slope Intercept Form: y = mx + b

y = -1/2x - 1

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5 0
2 years ago
To put a narrow border around a square photo, Alicia has 32 inches of trim. The area of the photo is 60 square inches. Will she
olga nikolaevna [1]

Answer: Yes, she will have enough trim for all four sides of the square, because the perimeter of the square photo (30.98 inches) is less than  32 inches of trim she has.

Step-by-step explanation:

The formula for calculate the area of a square is:

A=s^2

Where "s" is the lenght of any side of the square.

The formula for calculate the perimeter of a square is:

P=4s

Where "s" is the lenght of any side of the square.

We know that that the area of the photo is 60 square inches, therefore, we can solve for "s" from the formula  A=s^2 and find its value:

A=s^2\\\\s=\sqrt{A}[\\\\s=\sqrt{60in^2}\\\\s=2\sqrt{15}in

Substituting the value of "s" into the formula P=4s, we get that the perimeter of the photo is:

P=4(2\sqrt{15}in)=30.98in

Therefore, since Alicia has 32 inches of trim and 30.98in, we can conclude that she will have enough trim for all four sides of the square.

8 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
¿give me the answer please?
nevsk [136]

Answer:

the volume is 60 cm cubed

Step-by-step explanation:

(5 x 8) / 2 x 3

5 x 8 = 40

40 / 2 = 20

20 x  3 = 60

6 0
2 years ago
Read 2 more answers
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