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3 years ago

8.97 in word decimal form

Mathematics

2 answers:

kobusy [5.1K]3 years ago

8 0

Answer:

eight point ninety-seven

Step-by-step explanation:

hope this helped<3

Mamont248 [21]3 years ago

3 0

Answer: Eight point ninety-seven or Eight and ninety-seven

Step-by-step explanation:

Eight (8)

point (decimal point)

ninety-seven (97)

<u>Some teachers also like to say it like this:</u>

Eight (8)

and (decimal point)

ninety-seven (97)

Hope this helps!

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Evaluate the integral Integral from 0 to 1 Integral from 0 to 3 Integral from 3 y to 9 StartFraction 6 cosine (x squared )Over 5

Natalka [10]

Answer:

$\int^1_0\int^3_0\int^9_{3y}\frac{6 cos x^2}{5\sqrt z}dxdydz$ $=\frac{18}{5}(1+\frac{sin2}{2})$

Step-by-step explanation:

cosine x²= cos x²

Rule

$\int x^ndx= \frac{x^{n+1}}{n+1}+c$

$\int cos \ mx \ dx = \frac{sin \ mx}{m}+c$
$\int \frac{1}{\sqrt x}dx = \frac{\sqrt x}{\frac{1}{2}} +c= 2\sqrt x+c$

Given that,

$\int^1_0\int^3_0\int^9_{3y}\frac{6 cos x^2}{5\sqrt z}dxdydz$

$=\int ^1_0[\int^3_0(\int^9_{3y} \frac{6cos x^2}{5\sqrt z}dz)dy]dz$

$=\int^1_0[\int^3_0([\frac{6cos x^2 \times \sqrt z}{5\times \frac{1}{2}}]^9_{3y})dy]dx$

$=\int^1_0[\int^3_0([\frac{12cos x^2 \times( \sqrt 9-\sqrt{3y})}{5}])dy]dx$

$=\int^1_0[\int^3_0([\frac{12cos x^2 \times( 3-\sqrt{3y})}{5}])dy]dx$

$=\int^1_0[\frac{12cos x^2 \times( 3y-\frac{\sqrt{3}y^\frac{3}{2}}{\frac{3}{2}})}{5}]^3_0dx$

$=\int^1_0[\frac{12cos x^2 \times( 3.3-\frac{2\sqrt{3}.3^\frac{3}{2}}{3})}{5}]^3_0dx$

$=\int^1_0[\frac{12cos x^2 \times( 9-6)}{5}]dx$

$=\frac{18}{5}\int^1_02cos x^2dx$

$=\frac{18}{5}\int^1_0(1+cos2x)dx$

$=\frac{18}{5}[(x+\frac{sin2x}{2})]^1_0$

$=\frac{18}{5}(1+\frac{sin2}{2})$

6 0

3 years ago

Please answer it now in two minutes

Gemiola [76]

Answer:

8.4 in

Step-by-step explanation:

Solution:-

- We consider the large right angle triangle namely, " XVW "

- We will recall all the trigonometric ratios that are applicable to all right angled triangles.

- While we are dealing with trigonometric ratios we have the following terms that needs to be correlated with the given specific problem:

Hypotenuse ( H ): Side opposite to 90 degrees angle

Base (B): The side adjacent to the available angle ( θ )

Perpendicular (P): The side opposite to the available angle ( θ )

- We will go ahead and mark our respective sides as follows:

Angle ( θ ) : 34°

Hypotenuse ( H ) : XW = 15 in

Base ( B ) : VW

Perpendicular ( P ) : VX

- Now recall all the trigonometric ratios studied:

sin ( θ ) = P / H = VX / XW

cos ( θ ) = B / H = VW / XW

tan ( θ ) = P / B = VX / VW

- Now choose the appropriate trigonometric ratio with two values given and one ( VX ) that needs to be determined as follows:

sin ( θ ) = P / H = VX / XW

sin ( 34° ) = VX / 15

VX = 15*sin ( 34° )

VX = 8.387 .. ( 8.4 ) in

4 0

3 years ago

Using the numbers -2,-3,5,2 create one expression that equal 10

Nostrana [21]

One equation that you could make is 5+2-(-3)

3 0

3 years ago

100 Points PLEASE ANSWER ASP! Will Give BRAINLIST!

VikaD [51]

Let's work to solve this system of equations:

y = 2x ~~~~~~~~\gray{\text{Equation 1}}y=2x Equation 1

x + y = 24 ~~~~~~~~\gray{\text{Equation 2}}x+y=24 Equation 2

The tricky thing is that there are two variables, xx and yy. If only we could get rid of one of the variables...

Here's an idea! Equation 11 tells us that \goldD{2x}2x and \goldD yy are equal. So let's plug in \goldD{2x}2x for \goldD yy in Equation 22 to get rid of the yy variable in that equation:

\begin{aligned} x + \goldD y &= 24 &\gray{\text{Equation 2}} \\\\ x + \goldD{2x} &= 24 &\gray{\text{Substitute 2x for y}}\end{aligned}

x+y

x+2x

=24

Equation 2

Substitute 2x for y

Brilliant! Now we have an equation with just the xx variable that we know how to solve:

x+2x3x 3x3x=24=24=243=8Divide each side by 3

Nice! So we know that xx equals 88. But remember that we are looking for an ordered pair. We need a yy value as well. Let's use the first equation to find yy when xx equals 88:

\begin{aligned} y &= 2\blueD x &\gray{\text{Equation 1}} \\\\ y &= 2(\blueD8) &\gray{\text{Substitute 8 for x}}\\\\ \greenD y &\greenD= \greenD{16}\end{aligned}

=2x

=2(8)

=16

Equation 1

Substitute 8 for x

Sweet! So the solution to the system of equations is (\blueD8, \greenD{16})(8,16). It's always a good idea to check the solution back in the original equations just to be sure.

Let's check the first equation:

\begin{aligned} y &= 2x \\\\ \greenD{16} &\stackrel?= 2(\blueD{8}) &\gray{\text{Plug in x = 8 and y = 16}}\\\\ 16 &= 16 &\gray{\text{Yes!}}\end{aligned}

=2x

2(8)

=16

Plug in x = 8 and y = 16

Yes!

Let's check the second equation:

\begin{aligned} x +y &= 24 \\\\ \blueD{8} + \greenD{16} &\stackrel?= 24 &\gray{\text{Plug in x = 8 and y = 16}}\\\\ 24 &= 24 &\gray{\text{Yes!}}\end{aligned}

x+y

8+16

=24

Plug in x = 8 and y = 16

Yes!

Great! (\blueD8, \greenD{16})(8,16) is indeed a solution. We must not have made any mistakes.

Your turn to solve a system of equations using substitution.

Use substitution to solve the following system of equations.

4x + y = 284x+y=28

y = 3xy=3x

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3 years ago

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