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3 years ago

13

What is the value of

100 \times 4} " alt=" \sqrt{100 \times 4} " align="absmiddle" class="latex-formula">

2 answers:

Pavel [41]3 years ago

3 0

Answer: 20

explanation: simplify the radical by breaking the radicand up into a product of known factors, assuming positive real numbers

Amanda [17]3 years ago

3 0

Answer:

√{100×4)=√20²=±20 is your answer

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PLZ HELP ANYONE NOW THIS ILL GIVE 50 POINTS

Ahat [919]

Answer:

44

Step-by-step explanation:

<u>Plug in x = -4 and find the value:</u>

f(-4) = 3*(-4)² - 7(-4) - 32
f(-4) = 48 + 28 - 32
f(-4) = 44

8 0

2 years ago

When she subtracts 4 from both sides, Startfraction one-half EndFraction x equals negative StartFraction one-half EndFraction x.

timama [110]

The value of x in the equation 1/2x = -1/2x when 4 is subtracted from both sides is 0(not defined)

Subtraction of fraction

Startfraction one-half EndFraction x equals negative StartFraction one-half EndFraction

1/2x = -1/2x

Subtract 4 from both sides

1/2x - 4 = -1/2x - 4

(x-8) / 2 = (-x-8) / 2

(x - 8) / 2 = (-x-8) / 2

cross product

2(x - 8) = 2(-x - 8)

2x - 16 = -2x - 16

2x + 2x = -16 + 16

4x = 0

Learn more about fraction:

brainly.com/question/11562149

#SPJ1

7 0

2 years ago

Helppppppppp meee too simplify this <br><br><br>

Daniel [21]

Answer:

30,976

Step-by-step explanation:

=}.

$(11 {}^{3} \times 4 { }^{2} \binom{}{11 {}^{2} } ) {}^{2}$

=}.

$(11 \times 4 {}^{2} ) {}^{2}$

=}.

$121 \times (4 {}^{2} ) {}^{2}$

=}.

$121 \times 256$

=}. 30,976

3 0

2 years ago

Read 2 more answers

A tank has the shape of a surface generated by revolving the parabolic segment y = x2 for 0 ≤ x ≤ 3 about the y-axis (measuremen

Darina [25.2K]

Answer:

$100\pi\int\limits^9_0 {(\sqrt y)^2(14-y)} \, dy$ ft-lbs.

Step-by-step explanation:

Given:

The shape of the tank is obtained by revolving $y=x^2$ about y axis in the interval $0\leq x\leq 3$ .

Density of the fluid in the tank, $D=100\ lbs/ft^3$

Let the initial height of the fluid be 'y' feet from the bottom.

The bottom of the tank is, $y(0)=0^2=0$

Now, the height has to be raised to a height 5 feet above the top of the tank.

The height of top of the tank is obtained by plugging in $x=3$ in the parabolic equation . This gives,

$H=3^2=9\ ft$

So, the height of top of tank is, $y(3)=H=9\ ft$

Now, 5 ft above 'H' means $H+5=9+5=14$

Therefore, the increase in height of the top surface of the fluid in the tank is given as:

$\Delta y=(14-y)$ ft

Now, area of cross section of the tank is given as:

$A(y)=\pi r^2\\r\to radius\ of\ the\ cross\ section$

Radius is the distance of a point on the parabola from the y axis. This is nothing but the x-coordinate of the point.

We have, $y=x^2$

So, $x=\sqrt y$

Therefore, radius, $r=\sqrt y$

Now, area of cross section is, $A(y)=\pi (\sqrt y)^2$

Work done in pumping the contents to 5 feet above is given as:

$W=D\int\limits^{y(3)}_{y(0)} {A(y)(\Delta y)} \, dy$

Plug in all the values. This gives,

$W=100\int\limits^9_0 {\pi (\sqrt y)^2(14-y)} \, dy\\\\W=100\pi\int\limits^9_0 { (\sqrt y)^2(14-y)} \, dy\textrm{ ft-lbs}$

7 0

3 years ago

Picture of problem is above eeeeeeeeeeeeee

yan [13]

I believe it is -2 because it doubles and the pattern of negative then positive requires a negative b

5 0

3 years ago