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OverLord2011 [107]

3 years ago

9

The total weight of 25 bags of rice, 23 bags of flour and 19 bags of sugar is 2,719 pounds. Each bag of rice weighs 35 pounds an

d each bag of flour weighs 43 pounds. What is the weight of a bag of sugar, assuming they all weigh the same?

1 answer:

Tems11 [23]3 years ago

4 0

Just do 25 23 then what

You might be interested in

Suppose there are n people on a bus. At the next stop, 9 more people get on the bus. No one gets of the bus. Draw a bar diagram

nordsb [41]

Answer:

Algebaric Expression & Bar Diagram graphic representation of number of people in bus.

Step-by-step explanation:

No. of people on bus = n

After 9 more boarding & no one deboarding the bus , no. of people = n + 9

Bar diagrams is graphical method for presentation of data. Presenting the given in bar diagram :

7 0

3 years ago

What are whole numbers

DedPeter [7]

A number without fractions; an integer.

6 0

3 years ago

Read 2 more answers

Least to greatest<br> 3/4,5/8,3/7,1/3,2/5

tangare [24]

Answer:

1/3 < 2/5 < 3/7 < 5/8 < 3/4

Step-by-step explanation:

Using the given inputs:

3/4 5/8 3/7 1/3 2/5

The least common denominator (LCD) is: 840.

Rewriting as equivalent fractions with the LCD:

3/4= 630/840

5/8= 525/840

3/7= 360/840

1/3= 280/840

2/5= 336/840

Sorting this by the numerators of the equivalent fractions in order from least to greatest:

1/3= 280/840= < 2/5= 336/840= < 3/7= 360/840= < 5/8= 525/840= < 3/4 = 630/840

Therefore, the sorted inputs in order from least to greatest is:

1/3 < 2/5 < 3/7 < 5/8 < 3/4

Hopes this helps!

Have a good evening! :)

5 0

4 years ago

The solution of the equation 3(p - 5) = 5p -27 is

inn [45]

Answer:

A) p = 6

Step-by-step explanation:

Given equation:

3(p - 5) = 5p -27

<u />

Apply the <u>Distributive Property of Multiplication over subtraction law</u>:

⇒ 3 · p - 3 · 5 = 5p - 27

⇒ 3p - 15 = 5p - 27

Add 27 to both sides:

⇒ 3p - 15 + 27 = 5p - 27 + 27

⇒ 3p + 12 = 5p

Subtract 3p from both sides:

⇒ 3p - 3p + 12 = 5p - 3p

⇒ 12 = 2p

⇒ 2p = 12

Divide both sides by 2:

⇒ 2p ÷ 2 = 12 ÷ 2

⇒ p = 6

3 0

2 years ago

Read 2 more answers

The polynomial P(x) = 2x^3 + mx^2-5 leaves the same remainder when divided by (x-1) or (2x + 3). Find the value of m and the rem

Zigmanuir [339]

Answer:

m=7

Remainder =4

If q=1 then r=3 or r=-1.

If q=2 then r=3.

They are probably looking for q=1 and r=3 because the other combinations were used earlier in the problem.

Step-by-step explanation:

Let's assume the remainders left when doing P divided by (x-1) and P divided by (2x+3) is R.

By remainder theorem we have that:

P(1)=R

P(-3/2)=R

$P(1)=2(1)^3+m(1)^2-5$

$=2+m-5=m-3$

$P(\frac{-3}{2})=2(\frac{-3}{2})^3+m(\frac{-3}{2})^2-5$

$=2(\frac{-27}{8})+m(\frac{9}{4})-5$

$=-\frac{27}{4}+\frac{9m}{4}-5$

$=\frac{-27+9m-20}{4}$

$=\frac{9m-47}{4}$

Both of these are equal to R.

$m-3=R$

$\frac{9m-47}{4}=R$

I'm going to substitute second R which is (9m-47)/4 in place of first R.

$m-3=\frac{9m-47}{4}$

Multiply both sides by 4:

$4(m-3)=9m-47$

Distribute:

$4m-12=9m-47$

Subtract 4m on both sides:

$-12=5m-47$

Add 47 on both sides:

$-12+47=5m$

Simplify left hand side:

$35=5m$

Divide both sides by 5:

$\frac{35}{5}=m$

$7=m$

So the value for m is 7.

$P(x)=2x^3+7x^2-5$

What is the remainder when dividing P by (x-1) or (2x+3)?

Well recall that we said m-3=R which means r=m-3=7-3=4.

So the remainder is 4 when dividing P by (x-1) or (2x+3).

Now P divided by (qx+r) will also give the same remainder R=4.

So by remainder theorem we have that P(-r/q)=4.

Let's plug this in:

$P(\frac{-r}{q})=2(\frac{-r}{q})^3+m(\frac{-r}{q})^2-5$

Let x=-r/q

This is equal to 4 so we have this equation:

$2u^3+7u^2-5=4$

Subtract 4 on both sides:

$2u^3+7u^2-9=0$

I see one obvious solution of 1.

I seen this because I see 2+7-9 is 0.

u=1 would do that.

Let's see if we can find any other real solutions.

Dividing:

1 | 2 7 0 -9

| 2 9 9

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

2 9 9 0

This gives us the quadratic equation to solve:

$2x^2+9x+9=0$

Compare this to $ax^2+bx+c=0$

$a=2$

$b=9$

$c=9$

Since the coefficient of $x^2$ is not 1, we have to find two numbers that multiply to be $ac$ and add up to be $b$ .

Those numbers are 6 and 3 because $6(3)=18=ac$ while $6+3=9=b$ .

So we are going to replace $bx$ or $9x$ with $6x+3x$ then factor by grouping:

$2x^2+6x+3x+9=0$

$(2x^2+6x)+(3x+9)=0$

$2x(x+3)+3(x+3)=0$

$(x+3)(2x+3)=0$

This means x+3=0 or 2x+3=0.

We need to solve both of these:

x+3=0

Subtract 3 on both sides:

x=-3

----

2x+3=0

Subtract 3 on both sides:

2x=-3

Divide both sides by 2:

x=-3/2

So the solutions to P(x)=4:

$x \in \{-3,\frac{-3}{2},1\}$

If x=-3 is a solution then (x+3) is a factor that you can divide P by to get remainder 4.

If x=-3/2 is a solution then (2x+3) is a factor that you can divide P by to get remainder 4.

If x=1 is a solution then (x-1) is a factor that you can divide P by to get remainder 4.

Compare (qx+r) to (x+3); we see one possibility for (q,r)=(1,3).

Compare (qx+r) to (2x+3); we see another possibility is (q,r)=(2,3).

Compare (qx+r) to (x-1); we see another possibility is (q,r)=(1,-1).

6 0

3 years ago