Ask question

Ask question

All categories

3 years ago

14

At Shop and Save Grocery Store, apples cost $3.59 a pound. If you buy 2.5 pounds, how much will it cost? Round your answer to th

e nearest cent.

2 answers:

Alex73 [517]3 years ago

8 0

The answer would be $8.98. If each pound of apples costs $3.59 cents and you buy 2.5 pounds, your equation would be 2.5(3.59)=8.98.

12345 [234]3 years ago

3 0

Answer:

$8.98

Step-by-step explanation:

3.59+3.59+1.80

You might be interested in

A certain brand of golf balls comes in packs of 24. Each pack has three yellow golf balls in it. Willam orders 360 golf balls on

iogann1982 [59]

Start by taking the total number of golf balls, 360, and dividing it by how many golf balls come in each pack, to find the number of packs.

360 ÷ 24 = 15

So there are 15 packs, each with 3 yellow golf balls.

15 · 3 = 45

Meaning 45 of the golf balls would be yellow

3 0

4 years ago

Read 2 more answers

A person x inches tall has a pulse rate of y beats per minute, as given approximately by yequals600 x Superscript negative 1 di

tankabanditka [31]

Answer:

a) 5.13beats/min

b) 2.82 beats/min

Step-by-step explanation:

Given the pulse rate of a person modelled by the equation y = 600x^-1/3 for 30≤x≤75

If the height is 39inches, the instantaneous rate of change of pulse rate for the heights will be expressed as;

y = 600(39)^-1/3

y = {600(1/39)}/3

y = 600/39×3

y = 600/117

y ≈ 5.13beats/min

The instantaneous rate for a 39 inches tall person is 5.13 beats per min

b) For a 71inches tall person, the beat rate will be expressed as;

y = 600(71)^-1/3

y = {600(1/71)}/3

y = 600/71×3

y = 600/213

y ≈ 2.82 beats per minute

The instantaneous rate for a 71 inches tall person is 2.82 beats per min

7 0

4 years ago

Omg help with my homework PLEASE ASAP. what donu graph and how do i suow my workkkkk whats the answer

Gnoma [55]

Answer:

you have to start with you b which in your problem is 5 so put a point at 5 on the y-axis on your graph. then with the M (the fraction numbers) you have to move either up or down depending if there is a negative number. so yours is 1/2 so then you would start at the 5 point and go up 1 and right 2 then you just keep repeating till you get a few points and can make a line then you just do the same thing for the other equation. start at 3 on the y-axis and then go up 3 and right 4 then just keep repeating till you get a few point and can create a line at the end the lines should cross each other like perpendicular lines

Step-by-step explanation:

hope this helps:)

7 0

2 years ago

Not sure what to do here can someone please help

kumpel [21]

Answer:

x = 2

Step-by-step explanation:

These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.

__

<h3>Squaring</h3>

The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.

$\sqrt{x+2}+1=\sqrt{3x+3}\qquad\text{given}\\\\(x+2)+2\sqrt{x+2}+1=3x+3\qquad\text{square both sides}\\\\2\sqrt{x+2}=(3x+3)-(x+3)=2x\qquad\text{isolate the root term}\\\\x+2=x^2\qquad\text{divide by 2, square both sides}\\\\x^2-x-2=0\qquad\text{write in standard form}\\\\(x-2)(x+1)=0\qquad\text{factor}$

The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.

x = -1: √(-1+2) +1 = √(3(-1)+3) ⇒ 1+1 = 0 . . . . not true

x = 2: √(2+2) +1 = √(3(2) +3) ⇒ 2 +1 = 3 . . . . true . . . x = 2 is the solution

__

<h3>Substitution</h3>

Another way to solve this is using substitution for one of the radicals. We choose ...

$u=\sqrt{x+2}\qquad\text{requires $u\ge0$}\\\\u^2-2=x\qquad\text{solve for x}\\\\u+1=\sqrt{3(u^2-2)+3}\qquad\text{substitute for x in the original equation}\\\\(u+1)^2=3u^2-3\qquad\text{square both sides, simplify a little}\\\\2u^2-2u-4=0\qquad\text{subtract $(u+1)^2$}\\\\2(u-2)(u+1)=0\qquad\text{factor}$

Solutions to this equation are ...

u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution

The value of x is ...

x = u² -2 = 2² -2

x = 2 . . . . the solution to the equation

_____

<em>Additional comment</em>

Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.

6 0

2 years ago

Solve the matrix equation for a, b, c, and d. [1 2] [a b] [6 5][3 4] [c d]= [19 8]

Anit [1.1K]

Answer:

The answer is " $\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}$ ".

Step-by-step explanation:

$\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}$

Solve the L.H.S part:

$\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]$

After calculating the L.H.S part compare the value with R.H.S:

$\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\$

$\to a+2c =6....(i)\\\\\to b+2d =5....(ii)\\\\\to 3a+4c =19....(iii)\\\\\to 3b+4d = 8 ....(iv)\\\\$

In equation (i) multiply by 3 and subtract by equation (iii):

$\to 3a+6c=18\\\to 3a+4c=19\\\\\text{subtract}... \\\\\to 2c = -1\\\\\to c= - \frac{1}{2}$

put the value of c in equation (i):

$\to a+ 2 (- \frac{1}{2})=6\\\\\to a- 2 \times \frac{1}{2}=6\\\\\to a- 1=6\\\\\to a =6 +1\\\\\to a = 7\\$

In equation (ii) multiply by 3 then subtract by equation (iv):

$\to 3b+6d=15\\\to 3b+4d=8\\\\\text{subtract...}\\\\\to 2d = 7\\\\\to d= \frac{7}{2}\\$

put the value of d in equation (iv):

$\to 3b+4 (\frac{7}{2})=8\\\\\to 3b+4 \times \frac{7}{2}=8\\\\\to 3b+14=8\\\\\to 3b =8-14\\\\\to 3b = -6\\\\\to b= \frac{-6}{3}\\\\\to b= -2$

The final answer is " $\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}$ ".

4 0

4 years ago