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4 years ago

Please help me with math i need it soon

Mathematics

2 answers:

Bas_tet [7]4 years ago

8 0

Divide each price by the number of cubic yards it buys to find the cost per cubic yard.

A. $165/(3 yd³) = $55/yd³

B. $240/(4 yd³) = $60/yd³

C. $300/(6 yd³) = $50/yd³

D. $360/(8 yd³) = $45/yd³ . . . . . . . this is the best price per cubic yard.

AnnyKZ [126]4 years ago

6 0

Lets get started :)

We are told that Mrs. Thymes is buying mulch for her gardens.

Let us check the price for each option:

$\boxed {Option \ A}$
3 yd³ for $165
Divide 165 by 3, to find the price per cubic yard:
$\frac{165}{3}$ = $55 per cubic yard

$\boxed {Option \ B }$
4 yd³ for $240
Do the same:
$\frac{240}{4}$ = $60 per cubic yard

$\boxed {Option \ C }$
6 yd³ for $300
$\frac{300}{6}$ = $50 per cubic yard

$\boxed {Option \ D }$
8 yd³ for $360
$\frac{360}{8}$ = $45 per cubic yard

Therefore, we can see that 8 yd³ for $360 is the best price

$\boxed {Your \ Answer \ will \ be \ Option \ D }$

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german

Answer:

$\dfrac{x^4y^4}{81}$

Step-by-step explanation:

$(\dfrac{3}{xy})^{-4} =$

$= (\dfrac{xy}{3})^{4}$

$= \dfrac{(xy)^4}{3^4}$

$= \dfrac{x^4y^4}{81}$

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3 years ago

An item sells for $30 and is on sale for 15% off and the sales tax is 8.2%. What is the final cost to the nearest cent? (Calcula

mestny [16]

Answer:

$27.59

Step-by-step explanation:

You multiply 30 dollars by 0.85 (1-0.15, with the 0.15 being for the 15 percent off). Which gets you to $25.50, and you then multiply it by 1.082 (the 1.082 being the tax, and the .082 part equating to the 8.2% tax percentage).

8 0

3 years ago

Solve 2cos ²y -siny -1=0 for 0° ≤y≤360°

natima [27]

$\displaystyle\bf\\2cos^2y -sin\,y -1=0~~~for~~0^o\leq y \leq360^o\\\\cos^2y=1-sin^2y\\\\2(1-sin^2y) -sin\,y -1=0\\\\2-2sin^2y-sin\,y-1=0\\\\-2sin^2y-sin\,y+2-1=0\\\\-2sin^2y-sin\,y+1=0~~~\Big|\times(-1)\\\\2sin^2y+sin\,y-1=0$

$\displaystyle\bf\\\boxed{\bf sin\,y=x}\\\\2x^2+x-1=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-1\pm\sqrt{1-4\cdot2\cdot(-1)}}{2\cdot2}=\\\\=\frac{-1\pm\sqrt{1+8}}{4}=\frac{-1\pm\sqrt{9}}{4}=\frac{-1\pm3}{4}\\\\x_1=\frac{-1+3}{4}=\frac{2}{4}=\boxed{\bf\frac{1}{2}}\\\\x_2=\frac{-1-3}{4}=\frac{-4}{4}=\boxed{\bf-1}\\\\sin\,y=\frac{1}{2}\\\\\boxed{\bf y_1=\frac{\pi}{6}~~or~~(30^o)}\\\\\boxed{\bf y_2=\frac{5\pi}{6}~~or~~(150^o)}\\\\sin\,y=-1\\\\\boxed{\bf y_3=\frac{3\pi}{2}~~or~~(270^o)}$

5 0

3 years ago

. Use the quadratic formula to solve each quadratic real equation. Round

Liono4ka [1.6K]

Answer:

A. No real solution

B. 5 and -1.5

C. 5.5

Step-by-step explanation:

The quadratic formula is:

$\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}$ , with a being the x² term, b being the x term, and c being the constant.

Let's solve for a.

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {5^2 - 4\cdot1\cdot11} }}{{2\cdot1}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {25 - 44} }}{{2}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {-19} }}{{2}}} \end{array}$

We can't take the square root of a negative number, so A has no real solution.

Let's do B now.

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {7^2 - 4\cdot-2\cdot15} }}{{2\cdot-2}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {49 + 120} }}{{-4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {169} }}{{-4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm 13 }}{{-4}}} \end{array}$

$\frac{7+13}{4} = 5\\\frac{7-13}{4}=-1.5$

So B has two solutions of 5 and -1.5.

Now to C!

$\begin{array}{*{20}c} {\frac{{ -(-44) \pm \sqrt {-44^2 - 4\cdot4\cdot121} }}{{2\cdot4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 44 \pm \sqrt {1936 - 1936} }}{{8}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 44 \pm 0}}{{8}}} \end{array}$

$\frac{44}{8} = 5.5$

So c has one solution: 5.5

Hope this helped (and I'm sorry I'm late!)

4 0

3 years ago

PLEASE LOOK AT PICURE THEN ANWSER, WHOEVER IS CORRECT I WILL MARK BRAINIEST :)

irga5000 [103]

Answer:

b.g(x)= $x^{2}$ +3

Step-by-step explanation:

If two graphs have the same shape and you only want to change one's y intercept, you just have to add or subtract the number at which you want the graph to intercept the y axis from the equation

8 0

3 years ago