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3 years ago

PLEASE!!!!!!!!!!!!!!!!!!!!THIS IS DUE TODAY

Mathematics

2 answers:

rjkz [21]3 years ago

6 0

3 + 1.75m

This represents 1.75 per mile plus a one time 3 charge.

Another way this can be written is

(1.75m + 6) - 3

This is a less efficient way, but represents the same. It is 1.75 per mile, this is plus 6, therefore we add the -3 outside the parenthesis so it evens out to +3 as the base charge.

Example: Lets say its 3 miles.

3 + 1.75 (3) = 8.25

(1.75 (3) + 6) - 3 = 8.25

serg [7]3 years ago

3 0

Answer:

When m= 0,25

(8m-m ) + 3

3+ 8m-1/4m

With the first one you can add any number of miles infront of the brackets; ie) 4 miles

4(8m-m) +3 = 0.25t

32m-4m /28 (+3) = 0.25t x 28

28m +3 = 0.25t x 28

28m + 3 = 7t + 3t

31m = 10t

Step-by-step explanation:

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3 years ago

Read 2 more answers

One fourth of marked price of a watch is given as discount. What is the discount percent

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3 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

Write a rule for a geometric sequence that has -6 as its first term and a common ratio of 2. What is the tenth term of this sequ

slega [8]

The general form of the geometric sequence is $a_{n} = a_{1} (r) ^{n-1}$ , where a sub n is the number term you're looking for (we're looking for the tenth term). a sub 1 is the first term in the sequence (ours is -6), r is the common ratio, and n-1 is the numbered term you're looking for minus 1. Our formula then looks like this: $a_{10} =-6(2) ^{10-1}$ . Simplify it to $a_{10} =-6(2)^9$ . Take 2 to the 9th power then multiply it by -6 to get -3072. C is your answer.

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3 years ago