All categories

3 years ago

Which expression shows the amount would you pay if your lunch bill is X and you leave(pay) a 15% tip?

Mathematics

1 answer:

ddd [48]3 years ago

8 0

Answer:

Step-by-step explanation:

(b) is correct. x represents the bill and 0.15 represents the 15% tip on that bill.

You might be interested in

23 degrees below 0, what is the answer please help

IrinaVladis [17]

Answer:

-23

Step-by-step explanation:

23 degrees below 0 is 0-23, or -23

Hope this helps plz mark brainliest :D

6 0

3 years ago

Read 2 more answers

How do you re-write this using only positive exponents? <br><br> 3x^-4/3 (1+2x^5/3)

Kay [80]

$\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}
\\\\\\
and\qquad \quad a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
-------------------------------\\\\
$

$\bf \cfrac{3x^{-\frac{4}{3}}}{3(1+2x^{\frac{5}{3}})}\implies \cfrac{3}{3}\cdot \cfrac{1}{x^{\frac{4}{3}}}\cdot \cfrac{1}{(1+2x^{\frac{5}{3}})}\implies \cfrac{1}{x^{\frac{4}{3}}(1+2x^{\frac{5}{3}})}$

$\bf \cfrac{1}{x^{\frac{4}{3}}+2x^{\frac{5}{3}}x^{\frac{4}{3}}}\implies \cfrac{1}{x^{\frac{4}{3}}+2x^{\frac{5}{3}+\frac{4}{3}}}\implies \cfrac{1}{x^{\frac{4}{3}}2x^{\frac{9}{3}}}\implies \cfrac{1}{x^{\frac{4}{3}}+2x^3}
\\\\\\
\cfrac{1}{\sqrt[3]{x^4}+2x^3}\implies \cfrac{1}{x\sqrt[3]{x}+2x^3}\implies \cfrac{1}{x(\sqrt[3]{x}+2x^2)}$

8 0

4 years ago

Use the GFC to factor 45x+75

Rom4ik [11]

Answer:

I believe it's 15 (3x+5)

Step-by-step explanation:

15 times 3 is 45 and 15 times 5 is 75

3 0

3 years ago

Read 2 more answers

Find the measures of the three angles, in radians, of the triangle with the given vertices: d(1,1,1), e(1,−5,2), and f(−2,2,7).

Oduvanchick [21]

Consider triangle DEF with vertices D(1,1,1), E(1,-5,2) and F(-2,2,7).

1. Find

$\overrightarrow{DE}=(1-1,-5-1,2-1)=(0,-6,1),\\ \\\overrightarrow{DF}=(-2-1,2-1,7-1)=(-3,1,6).$

Then

$\cos \angle D=\dfrac{0\cdot (-3)+(-6)\cdot 1+1\cdot 6}{\sqrt{0^2+(-6)^2+1^2}\cdot \sqrt{(-3)^2+1^2+6^2}}=\dfrac{0}{\sqrt{37} \cdot \sqrt{46} }=0.$

2. Find

$\overrightarrow{ED}=(1-1,1-(-5),1-2)=(0,6,-1),\\ \\\overrightarrow{EF}=(-2-1,2-(-5),7-2)=(-3,7,5).$

Then

$\cos \angle E=\dfrac{0\cdot (-2)+6\cdot 7+(-1)\cdot 5}{\sqrt{0^2+6^2+(-1)^2}\cdot \sqrt{(-3)^2+7^2+5^2}}=\dfrac{37}{\sqrt{37} \cdot \sqrt{83} }=\sqrt{\dfrac{37}{83}}.$

3. Find

$\overrightarrow{FE}=(1-(-2),-5-2,2-7)=(3,-7,-5),\\ \\\overrightarrow{FD}=(1-(-2),1-2,1-7)=(3,-1,-6).$

Then

$\cos \angle F=\dfrac{3\cdot 3+(-7)\cdot (-1)+(-5)\cdot (-6)}{\sqrt{3^2+(-7)^2+(-5)^2}\cdot \sqrt{3^2+(-1)^2+(-6)^2}}=\dfrac{46}{\sqrt{83} \cdot \sqrt{46} }=\sqrt{\dfrac{46}{83}}.$

$\angle E=\arccos0=\dfrac{\pi}{2},\\ \\
\angle D=\arccos\letf(\sqrt{\dfrac{37}{83}}\right)\approx 0.27\pi,\\ \\
\angle F=\arccos\letf(\sqrt{\dfrac{46}{83}}\right)\approx 0.23\pi.$

3 0

3 years ago

Which statement is true about a number and its multiplicative inverse? A. Their product is always zero. B. Their product is alwa

stiv31 [10]

Answer: their product is always one

Step-by-step explanation:

6 0

4 years ago

Read 2 more answers