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3 years ago

Answer the sum correctly

Mathematics

1 answer:

Katyanochek1 [597]3 years ago

7 0

it would be 7 friends because it says at least.

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Katie bought 4 sweaters that each cost the same amount and 1 skirt that cost $20 .the items she bought cost a total of $160 befo

lorasvet [3.4K]

In order to solve this you first need to take $20 subtracted from 160 that leaves you with 140 next because she bought 4 sweaters you need to divide your remaining cash spent by 4 to get a total of $35 a sweater.

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3 years ago

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Consider the equation 5-3(2x - 7) = 12-5(x - 2). Is x = 2 a solution to this equation? Is x = 4 a solution to this equation? Exp

mario62 [17]

Answer:

x=4

Step-by-step explanation:

5-3(2x-7)=5-6x+21=-6x+26

--------------------------------------

12-5(x-2)=12-5x+10=-5x+22

--------------------------------------

-6x+26=-5x+22

-6x-(-5x)=22-26

-6x+5x=-4

-x=-4

x=4

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3 years ago

What is the original price of the item? Explain. Help

kicyunya [14]

The original price is $68.26 I think lmk if I’m wrong:)

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3 years ago

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Find the value of the expression 2x squared + 3y squared - 17 when x = 3 and y = 4

ser-zykov [4K]

Answer:

the value would be 49

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4 years ago

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2. The functions ???? and ???? represent the population of two different kinds of bacteria, where x is the time (in hours) and ?

klio [65]

Answer:

(a) Bacterial 1 had a faster rate of growth.

(b) The population of f(x) always exceed the population of g(x). In other words, population of g(x) cannot exceed the population of f(x).

Step-by-step explanation:

Consider the given functions are

$f(x)=2x^2+7$

$g(x)=2x$

where, x is the time (in hours) and f(x) and g(x) are the number of bacteria (in thousands).

(a)

The rate of change of a function f(x) on [a,b] is

$m=\frac{f(b)-f(a)}{b-a}$

Rate of change between third and sixth hour of first function is

$m_1=\frac{f(6)-f(3)}{6-3}$

$m_1=\frac{(2(6)^2+7)-(2(3)^2+7)}{6-3}$

$m_1=\frac{79-25}{3}$

$m_1=\frac{54}{3}$

$m_1=18$

Rate of change between third and sixth hour of second function is

$m_2=\frac{g(6)-g(3)}{6-3}$

$m_2=\frac{2(6)-2(3)}{6-3}$

$m_2=\frac{12-6}{3}$

$m_2=\frac{6}{3}$

$m_2=2$

Since $m_1>m_2$ , therefore bacterial 1 had a faster rate of growth.

(b)

The initial population of f(x) is 7 and it increases exponentially.

The initial population of g(x) is 0 and it increases linearly.

It means population of f(x) always exceed the population of g(x).

In other words, population of g(x) cannot exceed the population of f(x).

8 0

3 years ago