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1 year ago

The number of hours worked, x, and the total dollars earned, y, have a strong positive relationship.

1 answer:

4 0

A strong positive relationship between number f hours worked, x, and the total dollars earned, y, means that the higher the number of hours worked, the higher the total dollars earned and vice versa.

Recall:

When two variables, x and y, have a strong positive relationship, it implies that if variable x moves higher, variable will also move in the same direction and with the same magnitude. If variable x moves lower, variable y will also do same with the same magnitude.

Therefore, a strong positive relationship between number f hours worked, x, and the total dollars earned, y, means that the higher the number of hours worked, the higher the total dollars earned and vice versa.

Learn more about strong positive relationship on:

brainly.com/question/15552860

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Lucy made a pumpkin pie to sell at a bake sale. She cut it into 6 slices and sold each slice for $1.25. How much money will she

Lelu [443]

she would make $7.50

7 0

2 years ago

Prove that sin20sin40sin60sin80=3/16

never [62]

We'll use the following properties of sine and cosine to prove this:

$sin(a)\cdot sin(b) = \frac12(cos(a-b) - cos(a+b))$

$cos(a)\cdot cos(b) = \frac12(cos(a+b) + cos(a-b))$

$cos(180+a) = cos(180-a)$

Then it's just a matter of filling it in...

sin20sin40 * sin60sin80 = 1/2(cos20 - cos60) * 1/2 (cos20 - cos140) =

1/8( cos40 + 1 - cos160 - cos120 - cos40 - cos80 + cos80 + cos200) =

1/8(1 - cos160 - cos120 + cos200) =

1/8(1 - cos160 - cos120 + cos160) =

1/8(1 - cos120 ) = 1/8( 1 + 1/2 ) = 3/16

5 0

2 years ago

Please help i am helping somebody and i think i know the answer but i dont want it to be wrong so please help here it is

Mumz [18]

Answer:

244

Step-by-step explanation:

6 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

2 years ago

Help I will give brainliest!!

Elena-2011 [213]

Answer:

I think it is A and D but i'm sorry if i'm wrong

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers