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3 years ago

Y – 9.9 = 6.4 What are the steps to solve this and the answer

Mathematics

2 answers:

Olin [163]3 years ago

6 0

to be honest I'm looking for the delay but if they are not going through it and it will work and I am going through it.

monitta3 years ago

6 0

Hope the answer I gave you helps!!!
:)

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One of the solutions to the equation 72 - 57– 24 – is 8. What is the other solution?

gulaghasi [49]

Answer:

72-57-24=-9

Step-by-step explanation:

simple subtraction

8 0

3 years ago

Question 4, Pre calc, leave answer in bold, bad WiFi, if I get disconnected please continue.

Alenkasestr [34]

It is given that

A stone is thrown into a pond, A circular ripple spreads over the pond in such a way that the radius is increasing at a rate of 2.5 feet per second.

In one second the increment in radius is 2.5 feet

After t seconds, the increment in radius is given by

$r(t)=2.5t$

The area of the circular ripple is

$A(r)=\pi r^2$

$(A\circ r)(t)=A\lbrack r(t)\rbrack$

Substitute r(t)=2.5t, we get

$(A\circ r)(t)=A\lbrack2.5t\rbrack$

$UseA(r)=\pi r^2,\text{ we get}$

$(A\circ r)(t)=\pi(2.5t)^2$

The required function is

$(A\circ r)(t)=6.25\pi t^2$

Hence the fourth option is correct.

3 0

1 year ago

Multiply (1.78 x 106) by (3.4 x 102). Express your answer in scientific notation. A) .6052 x 1012 B) 6.052 x 108 C) 6.052 x 1012

AVprozaik [17]

Answer:

Step-by-step explanation:

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3 years ago

Evaluate the integral of 17/(x^(3)-125)

kvv77 [185]

$x^3-125=(x-5)(x^2+5x+25)$

$\dfrac{17}{(x-5)(x^2+5x+25)}=\dfrac a{x-5}+\dfrac{bx+c}{x^2+5x+25}$
$\dfrac{17}{(x-5)(x^2+5x+25)}=\dfrac{a(x^2+5x+25)+(bx+c)(x-5)}{(x-5)(x^2+5x+25)}$
$\implies17=(a+b)x^2+(5a-5b+c)x+(25a-5c)$
$\implies\begin{cases}a+b=0\\5a-5b+c=0\\25a-5c=17\end{cases}\implies a=\dfrac{17}{75},b=-\dfrac{17}{75},c=-\dfrac{34}{15}$

$\displaystyle\int\dfrac{17}{x^3-125}\,\mathrm dx=\dfrac{17}{75}\int\dfrac{\mathrm dx}{x-5}-\dfrac{17}{75}\int\frac x{x^2+5x+25}\,\mathrm dx-\dfrac{34}{15}\int\dfrac{\mathrm dx}{x^2+5x+25}$
$\displaystyle=\dfrac{17}{75}\int\dfrac{\mathrm dx}{x-5}-\dfrac{17}{150}\int\frac{2x+5}{x^2+5x+25}\,\mathrm dx-\dfrac{17}{10}\int\dfrac{\mathrm dx}{x^2+5x+25}$
$\displaystyle=\dfrac{17}{75}\int\dfrac{\mathrm dx}{x-5}-\dfrac{17}{150}\int\frac{2x+5}{x^2+5x+25}\,\mathrm dx-\dfrac{17}{10}\int\dfrac{\mathrm dx}{\left(x+\frac52\right)^2+\frac{75}4}$

The first integral is trivial. For the second, replace $y=x^2+5x+25$ so that $\mathrm dy=(2x+5)\,\mathrm dx$ . For the third, take $x+\dfrac52=\dfrac{5\sqrt3}2\tan z$ so that $\mathrm dx=\dfrac{5\sqrt3}2\sec^2z\,\mathrm dz$ .

$\displaystyle=\dfrac{17}{75}\ln|x-5|-\dfrac{17}{150}\int\frac{\mathrm dy}y-\dfrac{17}{10}\int\dfrac{\frac{5\sqrt3}2\sec^2z}{\left(\frac{5\sqrt3}2\tan z\right)^2+\frac{475}4}\,\mathrm dz$
$\displaystyle=\dfrac{17}{75}\ln|x-5|-\dfrac{17}{150}\ln|y|-\dfrac{17}{25\sqrt3}\int\dfrac{\sec^2z}{\tan^2z+1}\,\mathrm dz$
$\displaystyle=\dfrac{17}{75}\ln|x-5|-\dfrac{17}{150}\ln(x^2+5x+25)-\dfrac{17}{25\sqrt3}\int\mathrm dz$
$\displaystyle=\dfrac{17}{75}\ln|x-5|-\dfrac{17}{150}\ln(x^2+5x+25)-\dfrac{17}{25\sqrt3}z+C$
$\displaystyle=\dfrac{17}{75}\ln|x-5|-\dfrac{17}{150}\ln(x^2+5x+25)-\dfrac{17}{25\sqrt3}\arctan\dfrac{2x+5}{5\sqrt3}+C$

8 0

3 years ago

Which of the following are exterior angles? Check all that apply. 3 2 A. 26 [ B. 24 C. 75

ASHA 777 [7]

Answer:

E and A

Step-by-step explanation:

7 0

3 years ago