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3 years ago

Please assist me with this problem

Mathematics

1 answer:

tatiyna3 years ago

3 0

Put yes
because i did this

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How do you do this question?

quester [9]

Step-by-step explanation:

F(x) = ∫ₐˣ t⁷ dt

F(x) is the area under f(t) between t=a and t=x. When x=a, the width of the interval is 0, so the area is zero.

F(6) = 0, so a = 6.

F(x) = ∫₆ˣ t⁷ dt

F(6) = ∫₆⁶ t⁷ dt

F(6) = 0

3 0

4 years ago

Arrange the following in descending order A] 9,7,4,1 B] 5,4,0,3

Masja [62]

A. correct as it is
B. 5, 4, 3, 0

5 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

4 years ago

Whats the distributive property of 85 divided by 5?

Ahat [919]

$85:5=17 5(17:1)$

6 0

4 years ago

Read 2 more answers

-5/6 and 8/3 on a number line

ki77a [65]

Answer:

I hope this helps. This is what a number line looks like and how you do it.

6 0

4 years ago