Ask question

Ask question

All categories

2 years ago

14

Translate each algebraic expression into a verbal expression. Then compare and contrast the two verbal expressions. What do you

notice?
\sqrt{x}-4
\sqrt{x-4}

1 answer:

Lostsunrise [7]2 years ago

3 0

Answer:

Step-by-step explanation:y=Mx+b

You might be interested in

Express 3cm: 3m in the ratio of its simplest form

Dmitry_Shevchenko [17]

Answer:

1cm:100cm

Step-by-step explanation:

You have to convert the ratio into the same unit (I chose cm). If there are 100cm in a metre, there will be 300cm in 3m.

That leaves you with the ratio 3cm:300cm. To simplify, divide both sides by three leaving you with 1cm:100cm

3 0

3 years ago

Read 2 more answers

Question 2..plz help meee

Art [367]

refer to the attachment

7 0

2 years ago

Haley's age is 6 years less than half her mother's

Mrrafil [7]

Answer:
X-6= X-16
Hope this helps

4 0

3 years ago

5. The temperature was

Vesna [10]

Answer:

6.2 is the difference

Step-by-step explanation:

4 0

3 years ago

Find the length of the curve y = 3/5x^5/3 - 3/4x^1/3 + 6 for 1 < = x < = 8. The length of the curve is . (Type an exact an

Mashutka [201]

Answer:

$\sqrt\frac{387}{20}$

Step-by-step explanation:

$Arc Length =\int\limits^a_b {\sqrt{1+(\frac{dy}{dx})^2 } } \, dx$

$y=\dfrac{3}{5}x^{\frac{5}{3}}- \dfrac{3}{4}x^{\frac{1}{3}}+6$

$\frac{dy}{dx} =x^{\frac{2}{3}}-\dfrac{1}{4}x^{-\frac{2}{3}}$

$1+(\frac{dy}{dx})^2 }=1+(x^{\frac{2}{3}}-\dfrac{1}{4}x^{-\frac{2}{3}})^2\\=1+(x^{\frac{4}{3}}-\dfrac{1}{2}+ \dfrac{1}{16}x^{-\frac{4}{3}})$

$=\dfrac{1}{2}+x^{\frac{4}{3}}+ \dfrac{1}{16}x^{-\frac{4}{3}}$

For the Interval $1\leq x\leq 8$

Length of the Curve $=\int\limits^8_1 {\sqrt{\dfrac{1}{2}+x^{\frac{4}{3}}+ \dfrac{1}{16}x^{-\frac{4}{3}} } } \, dx\\$

Using T1-Calculator

$=\sqrt\frac{387}{20}$

3 0

3 years ago