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trasher [3.6K]
3 years ago
14

2 1/4 pounds of chocolate cost $9.00. What is the price per pound?

Mathematics
2 answers:
garik1379 [7]3 years ago
8 0
9/2.25= $4 per pound is the answer to the question
melisa1 [442]3 years ago
7 0
4 dollars per pound.

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-2(x + y)2 for for x =-3, y = 2 with steps
svlad2 [7]

Step-by-step explanation:

-2(-3+2)2

(-2×-3)+(-2×2)2

(6-4)2

2(6-4)

(2×6)+(2×-4)

12-8=4

6 0
3 years ago
A regular pentagon has side lengths of 10 cm and an
Fittoniya [83]

Answer: 172 square centimeters.

Step-by-step explanation:

Hi, to answer this question we have to apply the next formula:

Area of a pentagon (A) = 1/2× 5s × a

Where:

s : side of the pentagon

a: apothem

Replacing with the values given and solving for A (area):

A =1/2 x (5 x10) x 6.88

A = 172 square centimeters.

Feel free to ask for more if needed or if you did not understand something.

7 0
2 years ago
Remember to show work and explain. Use the math font.
MrMuchimi

Answer:

\large\boxed{1.\ f^{-1}(x)=4\log(x\sqrt[4]2)}\\\\\boxed{2.\ f^{-1}(x)=\log(x^5+5)}\\\\\boxed{3.\ f^{-1}(x)=\sqrt{4^{x-1}}}

Step-by-step explanation:

\log_ab=c\iff a^c=b\\\\n\log_ab=\log_ab^n\\\\a^{\log_ab}=b\\\\\log_aa^n=n\\\\\log_{10}a=\log a\\=============================

1.\\y=\left(\dfrac{5^x}{2}\right)^\frac{1}{4}\\\\\text{Exchange x and y. Solve for y:}\\\\\left(\dfrac{5^y}{2}\right)^\frac{1}{4}=x\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\dfrac{(5^y)^\frac{1}{4}}{2^\frac{1}{4}}=x\qquad\text{multiply both sides by }\ 2^\frac{1}{4}\\\\\left(5^y\right)^\frac{1}{4}=2^\frac{1}{4}x\qquad\text{use}\ (a^n)^m=a^{nm}\\\\5^{\frac{1}{4}y}=2^\frac{1}{4}x\qquad\log_5\ \text{of both sides}

\log_55^{\frac{1}{4}y}=\log_5\left(2^\frac{1}{4}x\right)\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\dfrac{1}{4}y=\log(x\sqrt[4]2)\qquad\text{multiply both sides by 4}\\\\y=4\log(x\sqrt[4]2)

--------------------------\\2.\\y=(10^x-5)^\frac{1}{5}\\\\\text{Exchange x and y. Solve for y:}\\\\(10^y-5)^\frac{1}{5}=x\qquad\text{5 power of both sides}\\\\\bigg[(10^y-5)^\frac{1}{5}\bigg]^5=x^5\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(10^y-5)^{\frac{1}{5}\cdot5}=x^5\\\\10^y-5=x^5\qquad\text{add 5 to both sides}\\\\10^y=x^5+5\qquad\log\ \text{of both sides}\\\\\log10^y=\log(x^5+5)\Rightarrow y=\log(x^5+5)

--------------------------\\3.\\y=\log_4(4x^2)\\\\\text{Exchange x and y. Solve for y:}\\\\\log_4(4y^2)=x\Rightarrow4^{\log_4(4y^2)}=4^x\\\\4y^2=4^x\qquad\text{divide both sides by 4}\\\\y^2=\dfrac{4^x}{4}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\y^2=4^{x-1}\Rightarrow y=\sqrt{4^{x-1}}

6 0
3 years ago
Help with this find the image of (1 ,2) after a reflection about y=x followed by a reflection about y=-x
natali 33 [55]

Answer:

(0,-3)

Step-by-step explanation:

sold by graphing

6 0
3 years ago
Jade and Juliette are riding their bikes across the country to promote autism awareness. They rode their bikes 45.4 miles on the
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They need to ride 7 more days
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