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spayn [35]
3 years ago
9

Which shows a correct order to solve this story problem? Maxine had to pay $1.46 in sales tax on her purchases. She bought 3 bar

s of soap for $1.50 each, a bottle of shampoo for $5.75, and a bottle of conditioner for $5.95. How much did Maxine spend? A. Step 1: Add the price of a bar of soap to the price of the bottle of shampoo. Step 2: Multiply that total by 3. Step 3: Add the price of the bottle of conditioner to that total. Step 4: Add the tax to the final total for all the items. B. Step 1: Calculate the total price of the 3 bars of soap. Step 2: Calculate the total price of the shampoo and the conditioner. Step 3: Add the amount from Step 2 to the amount from Step 1. Step 4: Add the tax to the final total for all the items. C. Step 1: Add the price of a bar of soap and the bottle of shampoo. Step 2: Multiply that total by 3. Step 3: Add $5.95 to the total amount for soap and shampoo from Step 2. Step 4: Add the tax to the final total for all the items.
Mathematics
1 answer:
romanna [79]3 years ago
4 0
Choice B would be the correct answer. I hope that helps! :D
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What is the answer for this ?
alex41 [277]

Answer:

The slope of the line y = 3x - 1 is 3 , y intercept is -1

The slope of the line y = -6x - 10 is - 6 and y intercept is -10


Step-by-step explanation:

y = mx + b where m = slope and b = y intercept

So

The slope of the line y = 3x - 1 is 3 , y intercept is -1

The slope of the line y = -6x - 10 is - 6 and y intercept is -10

3 0
3 years ago
Please help me with this question thank you
gulaghasi [49]

Answer:

\mathrm{The\:solution\:is} :

x=\frac{-y^2-1}{2-y^2}\space\left\{y\ge \:0\right\}

Step-by-step explanation:

Given

y=\sqrt{\frac{2x+1}{x-1}}

Taking square of both sides

y^2=\left(\sqrt{\frac{2x+1}{x-1}}\right)\:^2

\mathrm{Subtract\:}\left(\sqrt{\frac{2x+1}{x-1}}\right)^2\mathrm{\:from\:both\:sides}

y^2-\left(\sqrt{\frac{2x+1}{x-1}}\right)^2=\left(\sqrt{\frac{2x+1}{x-1}}\right)^2-\left(\sqrt{\frac{2x+1}{x-1}}\right)^2

\mathrm{Simplify}

y^2-\left(\sqrt{\frac{2x+1}{x-1}}\right)^2=0

As we know that \mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)

\mathrm{Factor\:}y^2-\left(\sqrt{\frac{2x+1}{x-1}}\right)^2:\quad \left(y+\sqrt{\frac{2x+1}{x-1}}\right)\left(y-\sqrt{\frac{2x+1}{x-1}}\right)

so

\left(y+\sqrt{\frac{2x+1}{x-1}}\right)\left(y-\sqrt{\frac{2x+1}{x-1}}\right)=0        

\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

\mathrm{Solve\:}\:y+\sqrt{\frac{2x+1}{x-1}}=0

\mathrm{Subtract\:}y\mathrm{\:from\:both\:sides}

y+\sqrt{\frac{2x+1}{x-1}}-y=0-y

\sqrt{\frac{2x+1}{x-1}}=-y

\mathrm{Square\:both\:sides}

\left(\sqrt{\frac{2x+1}{x-1}}\right)^2=\left(-y\right)^2

\mathrm{Expand\:}\left(\sqrt{\frac{2x+1}{x-1}}\right)^2

\left(\sqrt{\frac{2x+1}{x-1}}\right)^2

\mathrm{Apply\:radical\:rule}:\quad \sqrt{a}=a^{\frac{1}{2}}

=\left(\left(\frac{2x+1}{x-1}\right)^{\frac{1}{2}}\right)^2

=\frac{2x+1}{x-1}

so equation  \left(\sqrt{\frac{2x+1}{x-1}}\right)^2=\left(-y\right)^2 becomes

\frac{2x+1}{x-1}=y^2

now

\mathrm{Solve\:}\:\frac{2x+1}{x-1}=y^2

\frac{2x+1}{x-1}=y^2

\mathrm{Multiply\:both\:sides\:by\:}x-1

\frac{2x+1}{x-1}\left(x-1\right)=y^2\left(x-1\right)

2x+1=y^2\left(x-1\right)

2x+1=xy^2-y^2         ∵  y^2\left(x-1\right):\quad xy^2-y^2

2x=xy^2-y^2-1

2x-xy^2=-y^2-1

x\left(2-y^2\right)=-y^2-1         ∵ \mathrm{Factor}\:2x-xy^2:\quad x\left(2-y^2\right)

\mathrm{Divide\:both\:sides\:by\:}2-y^2

\frac{x\left(2-y^2\right)}{2-y^2}=-\frac{y^2}{2-y^2}-\frac{1}{2-y^2}

x=\frac{-y^2-1}{2-y^2}

so

y+\sqrt{\frac{2x+1}{x-1}}=0:\quad x=\frac{-y^2-1}{2-y^2}\space\left\{y\le \:0\right\}

similarly

y-\sqrt{\frac{2x+1}{x-1}}=0:\quad x=\frac{-y^2-1}{2-y^2}\space\left\{y\ge \:0\right\}

\mathrm{Verify\:Solutions}:\quad x=\frac{-y^2-1}{2-y^2}

\mathrm{Check\:the\:solutions\:by\:plugging\:them\:into\:}y^2=\left(\sqrt{\frac{2x+1}{x-1}}\right)^2

\mathrm{Remove\:the\:ones\:that\:don't\:agree\:with\:the\:equation.}

\mathrm{Plug}\quad x=\frac{-y^2-1}{2-y^2}

y^2=\left(\sqrt{\frac{2\left(\frac{-y^2-1}{2-y^2}\right)+1}{\left(\frac{-y^2-1}{2-y^2}\right)-1}}\right)^2

\mathrm{Subtract\:}\left(\sqrt{\frac{2\frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2\mathrm{\:from\:both\:sides}

y^2-\left(\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2=\left(\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2-\left(\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2

y^2-\left(\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2=0

\mathrm{Factor\:}y^2-\left(\sqrt{\frac{2\frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2:\quad \left(y+\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)\left(y-\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)

so

\left(y+\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)\left(y-\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)=0

\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

\mathrm{Solve\:}\:y+\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}=0:\quad y\le \:0

\mathrm{Solve\:}\:y-\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}=0:\quad y\ge \:0

\mathrm{True\:for\:all}\:y

Therefore,  \mathrm{The\:solution\:is} :

x=\frac{-y^2-1}{2-y^2}\space\left\{y\ge \:0\right\}

6 0
3 years ago
Which one of the following numbers is an irrational number? A. 3.741 B. 7.676767… C. 8.12131415… D. 134⁄675
Hunter-Best [27]
I think it’s C cause it’s never ending but not repeating at the same time
3 0
3 years ago
Lines
kirill [66]

Answer:

line r: y=-2x-2

line s: -2x+1

lime t: -2x+3

Step-by-step explanation:

every slope must be -2 since that's stated in the question

to find b in y=mx+b, you substitute 0 for x and the y coordinate when x is 0 into the equation (for example for slope r, when x is 0 then y is -2, so -2=-2(0)+b) then solve for b (-2=-2(0)+b is -2=b)

basically, b is the y component when x is 0

8 0
4 years ago
Read 2 more answers
I need answers fast, I have other questions on my account btw. Thanks
rjkz [21]

Answer:

X=4

Step-by-step explanation:

It is vertical so it is X=__.  It is on x=4 so that is the answer

3 0
4 years ago
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