Ask question

Ask question

All categories

2 years ago

8

A grocery store sells a bag of 8 oranges for $4.40. How much would it cost for 9 oranges?

2 answers:

Nuetrik [128]2 years ago

7 0

Answer: $4.95

Step-by-step explanation:

divide 4.40 by 8 to get unit price:

4.40/8 = 0.55

Add 0.55 to 4.40 to get the cost of 9 oranges

0.55 + 4.40 = $4.95

kolbaska11 [484]2 years ago

3 0

To solve this exercise, we apply the rule of three:

Oranges......................Price

8.................................$4.40

9.......................................x

We solve as follows:

$\boldsymbol{\sf{x=\dfrac{9\times4.40}{8}=\dfrac{39.6 }{8}=4.9 }}$

Therefore, 9 oranges would cost $4.9.

<h2>{ Pisces04 }</h2>

You might be interested in

800+900×25-65(89+5)-12÷67-7

vlada-n [284]

Answer:

As there is Bracket lats solve it first but as there is '-' sign before the bracket sign in the bracket will change. so we will have,

'800+900×25-65(89-5)-12÷67-7'

Solving the bracket gives us:

'800+900×25-65(84)-12÷67-7'

Now avoiding the Bracket:

'800+900×25-5460-12÷67-7'

Now first we have to do MUltiplication, so we will get,

'800+22500-5460-12÷67-7'

Now let's do the division, so we will get,

'800+22500-5460-0.17-7'

Now lets do the additions and subtractions we will get,

<h2><u>17,832.83</u></h2>

4 0

3 years ago

In the ordered pair -3/2,-7 is called the

trapecia [35]

Answer:

Coordinates.

3 0

3 years ago

How does the value of b affect the graph of y= mx+b?

Virty [35]

Answer:

I think it moves it up on the y axis

8 0

4 years ago

Read 2 more answers

For the function y=3x2: (a) Find the average rate of change of y with respect to x over the interval [3,6]. (b) Find the instant

nirvana33 [79]

Answer:

The instantaneous rate of change of $y$ with respect to $x$ at the value $x = 3$ is 18.

Step-by-step explanation:

a) Geometrically speaking, the average rate of change of $y$ with respect to $x$ over the interval by definition of secant line:

$r = \frac{y(b) -y(a)}{b-a}$ (1)

Where:

$a$ , $b$ - Lower and upper bounds of the interval.

$y(a)$ , $y(b)$ - Function exaluated at lower and upper bounds of the interval.

If we know that $y = 3\cdot x^{2}$ , $a = 3$ and $b = 6$ , then the average rate of change of $y$ with respect to $x$ over the interval is:

$r = \frac{3\cdot (6)^{2}-3\cdot (3)^{2}}{6-3}$

$r = 27$

The average rate of change of $y$ with respect to $x$ over the interval $[3,6]$ is 27.

b) The instantaneous rate of change can be determined by the following definition:

$y' = \lim_{h \to 0}\frac{y(x+h)-y(x)}{h}$ (2)

Where:

$h$ - Change rate.

$y(x)$ , $y(x+h)$ - Function evaluated at $x$ and $x+h$ .

If we know that $x = 3$ and $y = 3\cdot x^{2}$ , then the instantaneous rate of change of $y$ with respect to $x$ is:

$y' = \lim_{h \to 0} \frac{3\cdot (x+h)^{2}-3\cdot x^{2}}{h}$

$y' = 3\cdot \lim_{h \to 0} \frac{(x+h)^{2}-x^{2}}{h}$

$y' = 3\cdot \lim_{h \to 0} \frac{2\cdot h\cdot x +h^{2}}{h}$

$y' = 6\cdot \lim_{h \to 0} x +3\cdot \lim_{h \to 0} h$

$y' = 6\cdot x$

$y' = 6\cdot (3)$

$y' = 18$

The instantaneous rate of change of $y$ with respect to $x$ at the value $x = 3$ is 18.

5 0

3 years ago

the half-life of radium-226 is about 1,590 years. How much of a 100mg sample will be left in 500 years

Vedmedyk [2.9K]

Aa Bb Cc Dd Ee Ff Gg Hh Ii Jj Kk Ll Mm Nn Oo Pp Qq Rr Ss Tt Uu Vv Ww Xx Yy Zz

I know the best answer ever (*says sarcastically*)

6 0

3 years ago