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777dan777 [17]
3 years ago
13

The total price of a bag of peaches varies directly with the cost per pound. If 3 pounds of peaches cost $3.60, how much would 5

.5 pounds cost?
A: $1.20
B: $6.60
C: $6.00
D: $1.96
Mathematics
1 answer:
Levart [38]3 years ago
8 0
<h3>Therefore the cost of 5.5 pound of peaches is $ 6.60.</h3>

Step-by-step explanation:

Given , the total price of a bag of peaches varies directly with the cost per pound. If 3 pound of peaches cost $3.60.

Therefore,

{\textrm{beg of peaches}}\propto{\textrm{cost}}

\Leftrightarrow {\textrm{beg of peaches}}= {\textrm{k  cost}}.......(1)      [ k is a constant]

cost = $3.60 when beg of peaches = 3 pounds

3 = 3.60 k

\Leftrightarrow k = \frac{3}{3.60}

\Leftrightarrow k = \frac{1}{1.20}

Therefore the equation (1) becomes

{\textrm{beg of peaches}}=\frac {cost}{1.20}

\Leftrightarrow {\textrm{beg of peaches}}\times 1.20={cost}

When beg of peaches = 5.5 pound

cost = 5.5 \times 1.20

\Leftrightarrow cost = 6.60

Therefore the cost of 5.5 pound of peaches is $ 6.60.

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What are the common factors of 20 and 24
Mandarinka [93]

Answer:

4

Step-by-step explanation:

4 x 5= 20

4x6 = 24

So 4 can go in 20 and 24 making it the common factor

7 0
3 years ago
A right circular cone is undergoing a transformation in such a way that the radius of the cone is increasing at a rate of 1/2 in
Ivenika [448]

Answer:

The volume is decreasing at the rate of 1.396 cubic inches per minute

Step-by-step explanation:

Given

Shape: Cone

\frac{dr}{dt} =\frac{1}{2} --- rate of the radius

\frac{dh}{dt} =-\frac{1}{3} --- rate of the height

r = 2

h = \frac{1}{3}

Required

Determine the rate of change of the cone volume

The volume of a cone is:

V = \frac{\pi}{3}r^2h

Differentiate with respect to time (t)

\frac{dV}{dt} = \frac{\pi}{3}(2rh \frac{dr}{dt} + r^2 \frac{dh}{dt})

Substitute values for the known variables

\frac{dV}{dt} = \frac{\pi}{3}(2*2*\frac{1}{3}* \frac{1}{2} - 2^2 *\frac{1}{3})

\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}* \frac{1}{2} - \frac{4}{3})

\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}(\frac{1}{2} - 1))

\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}*- 1)

\frac{dV}{dt} = -\frac{\pi}{3}*\frac{4}{3}

\frac{dV}{dt} = -\frac{22}{7*3}*\frac{4}{3}

\frac{dV}{dt} = -\frac{22}{21}*\frac{4}{3}

\frac{dV}{dt} = -\frac{88}{63}

\frac{dV}{dt} =-1.396in^3/min

The volume is decreasing at the rate of 1.396 cubic inches per minute

3 0
3 years ago
According to the Fundamental Theorem of Algebra, the graph of f(x) = x2 - 4x + 3, has roots. From the graph we can see that it h
andreyandreev [35.5K]

Answer:

The graph f(x)=x^2-4x+3  has two zeros namely 3 and 1.

Step-by-step explanation:

Consider the given equation of graph f(x)=x^2-4x+3

According to the Fundamental Theorem of Algebra

For a given polynomial of degree n can have a maximum of n roots.

Thus, for the given equation f(x)=x^2-4x+3  the degree of polynomial is 2 , thus the function can have maximum of 2 roots.

We know at roots the value of function is 0 that is f(x) = 0,

Substitute f(x) = 0 , we get, f(x)=x^2-4x+3=0

This is a quadratic equation, x^2-4x+3=0

We first solve it manually and then check by plotting graph.

Quadratic equation can be solved using middle term splitting method,

here, -4x can be written as -x-3x,

x^2-4x+3=0 \Rightarrow x^2-x-3x+3=0

\Rightarrow x(x-1)-3(x-1)=0

\Rightarrow (x-3)(x-1)=0

Using zero product property, a\cdot b=0 \Rightarrow a=0\ or \ b=0

\Rightarrow (x-3)=0 or \Rightarrow (x-1)=0

\Rightarrow x=3 or \Rightarrow x=1

Thus, the two zero of f(x) are 3 and 1.

We can also see on graph attached below that the graph f(x)=x^2-4x+3  has two zeros  namely 3 and 1.  

3 0
3 years ago
Help <br> Fjdjdjdjdjdjdjdjdjdjdjd
Zepler [3.9K]

Answer:

i belive its 6 please tell me if im wrong

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Could someone please help me asap? Thanks in advance! :)
Elden [556K]

Answer:

A

Step-by-step explanation:

the standard form of a quadratic equation is

ax² + bx + c = 0 : a ≠ 0

obtain (5 + x)(5 - x) = 7 in this form by expanding the factors

25 - x ² = 7 ( subtract 7 from both sides )

18 - x² = 0 ( multiply through by - 1 )

x² - 18 = 0 ← in standard form

with a = 1, b= 0 and c = - 18 → A


4 0
3 years ago
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