-3(-5x-2u+1) use the distributive property to remove the parentheses

Answer:

Step-by-step explanation:

I answered this already a few minutes ago.

Answer:

yes you can

Step-by-step explanation:

you can write algebraic expressions and use variables for the unknown

Answer:

3 · |x+6|

Step-by-step explanation:

Write out what you see. "Three times" is 3 · something; "the absolute value of the sum of a number and 6" is |number + 6|. We'll use x for our number. Put it all together and you get 3 · |x+6|

The expression of the statement, Three times the absolute value of the sum of a number and 6 is  [tex]\[3\left| n+6 \right|\][/tex] .

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Answer:

[tex]\sin \left(\theta \right)-\frac{1}{2}\cos \left(2\theta \rightt)+C[/tex]

Step-by-step explanation:

We are given the graph of r = cos( θ ) + sin( 2θ ) so that we are being asked to determine the integral. Remember that [tex]\:r=cos\left(\theta \right)+sin\left(2\theta \right)[/tex] can also be rewritten as [tex]\int \cos \left(\theta \right)+\sin \left(2\theta \right)d\theta \right[/tex].

Let's apply the functional rule [tex]\int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx[/tex],

[tex]\int \cos \left(\theta \right)+\sin \left(2\theta \right)d\theta \right[/tex] = [tex]\int \cos \left(\theta \right)d\theta \right+\int \sin \left(2\theta \right)d\theta \right[/tex]

At the same time [tex]\int \cos \left(\theta \right)d\theta \right=\sin \left(\theta \right)[/tex] = [tex]sin( \theta \right ))[/tex], and [tex]\int \sin \left(2\theta \right)d\theta \right[/tex] = [tex]-\frac{1}{2}\cos \left(2\theta \right)[/tex]. Let's substitute,

[tex]\int \cos \left(\theta \right)d\theta \right+\int \sin \left(2\theta \right)d\theta \right[/tex] = [tex]\sin \left(\theta \right)-\frac{1}{2}\cos \left(2\theta \right)[/tex]

And adding a constant C, we receive our final solution.

[tex]\sin \left(\theta \right)-\frac{1}{2}\cos \left(2\theta \rightt)+C[/tex] - this is our integral

Answer:

1 2/3 ounces in each bowl

Step-by-step explanation:

We need to convert 5 cups to ounces

1 cup = 8 ounces

5 cups = 5*8 = 40 ounces

We divide the 40 ounces into 24 bowls

40 ounces / 24 bowl

5/3 ounces per bowl

1 2/3 ounces in each bowl

Answer:

each bowl can contain 5/3 oz. of soup.

Step-by-step explanation:

1 cup = 8 oz.

                   8 oz.

5 cups x --------------  =  40 oz.

                    1 cup

to get the measurement of each bowl,

40 oz. divided into 24 bowls.

therefore, each bowl can contain 5/3 oz. of soup.

Answer:

[tex]\displaystyle \iiint_E {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV = \frac{\pi (17e^5 - 2)}{2}[/tex]

General Formulas and Concepts:
Calculus

Integration

Integration Rule [Reverse Power Rule]:
[tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]

Integration Rule [Fundamental Theorem of Calculus 1]:
[tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]

Integration Property [Multiplied Constant]:
[tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]

Integration Property [Addition/Subtraction]:
[tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]

Integration Method [Integration by Parts]:
[tex]\displaystyle \int {u} \, dv = uv - \int {v} \, du[/tex]

Multivariable Calculus

Triple Integrals

Cylindrical Coordinate Conversions:

Spherical Coordinate Conversions:

Integral Conversion [Spherical Coordinates]:
[tex]\displaystyle \iiint_T {f( \rho, \phi, \theta )} \, dV = \iiint_T {\rho^2 \sin \phi} \, d\rho \, d\phi \, d\theta[/tex]

Step-by-step explanation:

*Note:

Recall that φ is bounded by 0 ≤ φ ≤ 0.5π from the z-axis to the x-axis.

I will not show/explain any intermediate calculus steps as there isn't enough space.

Step 1: Define

Identify given.

[tex]\displaystyle \iiint_E {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV[/tex]

[tex]\displaystyle \text{Region E:} \ x^2 + y^2 + z^2 = 25 \ \text{bounded by first octant}[/tex]

Step 2: Integrate Pt. 1

Find ρ bounds.

Find θ bounds.

Find φ bounds.

Step 3: Integrate Pt. 2

We evaluate this spherical integral by using the integration rules, properties, and methods listed above:

[tex]\displaystyle \begin{aligned} \iiint_E {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV & = \int\limits^{\frac{\pi}{2}}_0 \int\limits^{\frac{\pi}{2}}_0 \int\limits^5_0 {e^{\rho} \rho^2 \sin \phi} \, d\rho \, d\phi \, d\theta \\ & = \int\limits^{\frac{\pi}{2}}_0 \int\limits^{\frac{\pi}{2}}_0 {\bigg[ (\rho^2 - 2 \rho + 2) e^{\rho} \sin \phi \bigg] \bigg| \limits^{\rho = 5}_{\rho = 0}} \, d\phi \, d\theta\end{aligned}[/tex]

[tex]\displaystyle \begin{aligned}\iiint_E {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV & = \int\limits^{\frac{\pi}{2}}_0 \int\limits^{\frac{\pi}{2}}_0 {(17e^5 - 2) \sin \phi} \, d\phi \, d\theta \\& = \int\limits^{\frac{\pi}{2}}_0 {\bigg[ -(17e^5 - 2) \cos \phi \bigg] \bigg| \limits^{\phi = \frac{\pi}{2}}_{\phi = 0}} \, d\theta \\& = \int\limits^{\frac{\pi}{2}}_0 {17e^5 - 2} \, d\theta \\& = (17e^5 - 2) \theta \bigg| \limits^{\theta = \frac{\pi}{2}}_{\theta = 0} \\& = \frac{\pi (17e^5 - 2)}{2}\end{aligned}[/tex]

∴ the given integral equals [tex]\displaystyle \bold{\frac{\pi (17e^5 - 2)}{2}}[/tex].

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Topic: Multivariable Calculus

Unit: Triple Integrals Applications

Answer:

B All real numbers

hope you wil understand

Answer:

[tex]\boxed{\sf B. \ All \ real \ numbers}[/tex]

Step-by-step explanation:

The domain is all possible values for x.

[tex]f(x)=(\frac{1}{4} )^x[/tex]

There are no restrictions on the value of x.

The domain is all real numbers.

Answer: Transitive property.

Step-by-step explanation:

First, for the equality we have:

Reflexive:

  For all real numbers x, x = x.

Symmetric:  

 For all real numbers x, y

 if x= y, then y = x.

Transitive:

 For reals x, y and z.

 if x = y, and y = z, then x = z.

Now, let's talk about inequalities.

first, the reflexive property will say that:

x > x.

This has no sense, so this property does not work for inequalities.

Now, the reflexive.

If x > y, then y > x.

Again, this has no sense, if x is larger than y, then we can never have that y is larger than x. This property does not work for inequalities.

Not, the transitive property.

if x > y, and y > z, then x > z.

This is true.

x is bigger than y, and y is bigger than z, then x should also be bigger than z.

x > y > z.

And this also works for the inverse case:

x < y and y < z, then x < z.

So the correct option is transitive property.

Answer:the one area < with line underneath then -4

St-by-step explanation: I’m pretty sure this is correct

Answer:

Step-by-step explanation:

[tex] \mathrm{16x - 7 \leqslant - 71}[/tex]

Move constant to Right hand side and change its sign

[tex] \mathrm{16x \leqslant - 71 + 7}[/tex]

Calculate

[tex] \mathrm{16x \leqslant - 64}[/tex]

Divide both sides of the equation by 16

[tex] \mathrm{ \frac{16x}{16} \leqslant \frac{ - 64}{16} }[/tex]

Calculate

[tex] \mathrm{x \leqslant - 4}[/tex]

Hope I helped!

Best regards!

Answer:

√(x)

Step-by-step explanation:

(1)/(x^-(1/2)) that's 3 goes into -3 leaving 1 and goes into 6 leaving 2

1/2 is same as 2^-1

so therefore we can simplify the above as

x^-(-1/2)

x^(1/2)

and 4^(1/2)

is same as √(4)

so we conclude as

√(x)

Answer:

5328 cups.

Step-by-step explanation:

Given that 333 gallons

We know that

1 gallons = 16 cups

1 cups = 0.0625 gallons

Therefore,from the above conversion we can say that

Now by putting the values in the above conversion

333 gallons = 16 x 333 cups

333 gallons = 5328 cups

So , we can say that 333 gallons is equal to 5328 cups.

Thus the answer will be 5328 cups.

Answer:

48 cups(BTW he meant 33 galons, IVE had this before). lol you need to put the double number line image. first u have to divide 64/4 to get 16, Then it says "How many cups are in 3 gallons". There fore, U multiply 16 to 3 to get ur answer "48".

Answer:

True

Step-by-step explanation:

We assume that is true for the nth case and prove it for the n+1 case

and show that it is true for the case when n=1

Answer:

a. 7k - 7

Step-by-step explanation:

Step 1: Write out expression

-k - (-8k + 7)

Step 2: Distribute negative

-k + 8k - 7

Step 3: Combine like terms

7k - 7

And we have our answer!

Answer:

Interquartile range is the distance between the first and third of a data.

Step-by-step explanation:

Hope it will help you :)

Answer:

Step-by-step explanation:

Hello, the slope of the tangent is the value of the derivative.

f'(x) = 2*0.2x + 5 = 0.4x + 5

So we are looking for

[tex]-1\leq f'(x) \leq 1 \\ \\<=> -1\leq 0.4x+5 \leq 1 \\ \\<=> -1-5=-6\leq 0.4x \leq 1-5=-4 \\ \\<=> \dfrac{-6}{0.4}\leq 0.4x \leq \dfrac{-4}{0.4} \\\\<=> \boxed{-15 \leq x\leq -10}[/tex]

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

Using derivatives, it is found that the x-values in which the slope belong to the interval (-1,1) are in the following interval is (-15,-10).

It is given by the derivative at x = x_0, that is:

m = f'(x_0)

In this problem, the function is:

f(x) = 0.2x^2 + 5x − 12

Hence the derivative is:

f'(x) = 0.4x + 5

For a slope of -1, we have that,

0.4x + 5 = -1

0.4x = -6

x = -15.

For a slope of 1, we have that,

0.4x + 5 = 1.

0.4x = -4

x = -10

Hence it is found that the x-values in which the slope belong to the interval (-1,1) are in the following interval is (-15,-10).

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Answer:

The balance in two separate bank accounts grows each month at different rates. the growth rates for both accounts are represented by the functions f(x) = 2x and g(x) = 4x 12. in what month is the f(x) balance greater than the g(x) balance?

Answer:

6 months

A function is a relationship between inputs where each input is related to exactly one output.

x = 5,

f(5) = [tex]2^5\\[/tex] = 32

g(5) = 4 x 5 + 12 = 20 + 12 = 32

x = 6,

f(6) = [tex]2^6[/tex] = 64

g(6) = 4 x 6 + 12 = 24 + 12 = 36

At month 6 the funds in the f(x) bank account exceed those in the g(x) bank account.

A function is a relationship between inputs where each input is related to exactly one output.

Example:

f(x) = 2x + 1

f(1) = 2 + 1 = 3

f(2) = 2 x 2 + 1 = 4 + 1 = 5

The outputs of the functions are 3 and 5

The inputs of the function are 1 and 2.

We have,

f(x) = [tex]2^{x}[/tex]

g(x) = 4x + 12

x = number of months

Now,

x = 3,

f(3) = 2³ = 8

g(3) = 4 x 3 + 12 = 12 + 12 = 24

x = 4,

f(4) = [tex]2^4[/tex] = 16

g(4) = 4 x 4 + 12 = 16 + 12 = 28

x = 5,

f(5) = [tex]2^5\\[/tex] = 32

g(5) = 4 x 5 + 12 = 20 + 12 = 32

x = 6,

f(6) = [tex]2^6[/tex] = 64

g(6) = 4 x 6 + 12 = 24 + 12 = 36

We see that,

At x = 6,

f(5) = 64

g(5) = 36

Thus,

At month 6 the funds in the f(x) bank account exceed those in the g(x) bank account.

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Answer:

C≈31.42

Step-by-step explanation:

C=2πr

C=2xπx5

C≈31.42

pls mark as brainliest

Answer:

x = 200.674

Step-by-step explanation:

tan∅ = opposite/adjacent

Step 1: Find length of z

tan70° = 119/z

ztan70° = 119

z = 119/tan70°

z = 43.3125

Step 2: Find length z + x (denoted as y)

tan26° = 119/y

ytan26° = 119

y = 119/tan26°

y = 243.986

Step 3: Find x

y - z = x

243.986 - 43.3125 = x

x = 200.674

Answer:

Part a ) The degrees of freedom for the given two sample non-pooled t test is 24

Part b ) The degrees of freedom for the given two sample non-pooled t test is 30

Part c ) The degrees of freedom for the given two sample non-pooled t test is 30

Part d ) The degrees of freedom for the given two sample non-pooled t test is 25

Step-by-step explanation:

Degrees of freedom for a non-pooled two sample t-test is given by;

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Now given the information;

a) :- m = 12, n = 15, s₁ = 4.0, s₂ = 6.0

we substitute

Δf =  {[ 4²/12 + 6²/15 ]²} / {[( 4²/12)²/12-1] + [(6²/15)²/15-1]}

Δf  = 30184 / 1241

Δf  = 24.3223 ≈ 24 (down to the nearest whole number)

b) :- m = 12, n = 21, s₁ = 4.0, s₂ = 6.0

we substitute using same formula

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Δf = {[ 4²/12 + 6²/21 ]²} / {[( 4²/12)²/12-1] + [(6²/21)²/21-1]}

Δf = 56320 / 1871

Δf = 30.1015 ≈ 30 (down to the nearest whole number)

c) :- m = 12, n = 21, s₁ = 3.0, s₂ = 6.0

we substitute using same formula

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Δf = {[ 3²/12 + 6²/21 ]²} / {[( 3²/12)²/12-1] + [(6²/21)²/21-1]}

Δf = 29095 / 949

Δf = 30.6585 ≈ 30 (down to the nearest whole number)

d) :- m = 10, n = 24, s₁ = 4.0, s₂ = 6.0

we substitute using same formula

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Δf = {[ 4²/10 + 6²/24 ]²} / {[( 4²/10)²/10-1] + [(6²/24)²/24-1]}

Δf = 1044 / 41  

Δf = 25.4634 ≈ 25 (down to the nearest whole number).

Answer:

Step-by-step explanation:

[tex]distance = 15 miles\\time = 0.1 hours\\\\Speed = \frac{Distance}{time}\\ Speed = \frac{15}{0.1}\\ Speed =150[/tex]

Answer:

[tex]150mph[/tex]

Step-by-step explanation:

Given:

s=15miles

t=0.1hours

Required:

v=?

Formula:

[tex]v = \frac{s}{t} [/tex]

Solution:

[tex]v = \frac{s}{t} = \frac{15m}{0.1h} = \frac{150m}{1h} = 150mph[/tex]

Hope this helps ;) ❤❤❤

Answer:

Step-by-step explanation:

The height on the side of the deck (h) can be found using the Pythagorean theorem. It tells you ...

  6^2 + h^2 = 10^2

  h = √(10^2 -6^2) = √64 = 8

The ladder touches the deck 8 feet above the ground. Tom is not afraid of that height.

Answer:

The Width = 28 inches

The Height = 21 inches

Step-by-step explanation:

We are told in the question that:

The width and height of an older 35-inch television whose screen has an aspect ratio of 4:3

Using Pythagoras Theorem

Width² + Height² = Diagonal²

Since we known that the size of a television is the length of the diagonal of its screen in inches.

Hence, for this new TV

Width² + Height² = 35²

We are given ratio: 4:3 as aspect ratio

Width = 4x

Height = 3x

(4x)² +(3x)² = 35²

= 16x² + 9x² = 35²

25x² = 1225

x² = 1225/25

x² = 49

x = √49

x = 7

Hence, for the 35 inch tv set

The Width = 4x

= 4 × 7

= 28 inches.

The Height = 3x

= 3 × 7

= 21 inches

Answer: S₁₉ = 855

Step-by-step explanation:

T₄ = a + ( n - 1 )d  = 5 , from the statement above , but n = 4

       a + 3d  = 5 -------------------------1

S₆ = ⁿ/₂[(2a + ( n - 1 )d]  =  10, where n = 6

    = ⁶/₂( 2a + 5d )         = 10

    = 3( 2a + 5d ) = 10

    = 6a + 15d      = 10 -----------------2

Now solve the two equation together simultaneously to get the values of a and d

   a + 3d     = 5

   6a + 15d = 10

from 1,

a = 5 - 3d -------------------------------3

Now put (3) in equation 2 and open the brackets

6( 5 - 3d )  + 15d = 10

30 - 18d + 15d      = 10

30 - 3d                 = 10

            3d            = 30 - 10

             3d           = 20

                         d = ²⁰/₃.

Now substitute for d to get a in equation 3

           a = 5 - 3( ²⁰/₃)

           a = 5 - 3 ₓ ²⁰/₃

              = 5 - 20

          a  = -15.

Now to find the sum of the first 19 terms,

we use the formula

S₁₉ = ⁿ/₂( 2a + ( n - 1 )d )

     = ¹⁹/₂( 2 x -15 + 18 x ²⁰/₃ )

     = ¹⁹/₂( -30 + 6 x 20 )

     = ¹⁹/₂( -30 + 120 )

     = ¹⁹/₂( 90 )

     = ¹⁹/₂ x 90

     = 19 x 45

     = 855

Therefore,

S₁₉ = 855

Answer:

a.  k = -0.01014 s⁻¹

b.  [tex]\mathbf{\dfrac{dy}{dt} = - \dfrac{In(\dfrac{3}{2})}{40}(y-60)}[/tex]

c.  [tex]\mathbf{y(t) = 60+ \dfrac{60 \sqrt{3}}{\sqrt{2}} \ e^{\dfrac{-In(\dfrac{3}{2})\ t}{40}}}[/tex]

d.  y(t) = 130.485°F

Step-by-step explanation:

A hot metal bar is submerged in a large reservoir of water whose temperature is 60°F. The temperature of the bar 20 s after submersion is 120°F. After 1 min submerged, the temperature has cooled to 100°F.

(Let y be measured in degrees Fahrenheit, and t be measured in seconds.)

We are to determine :

a.  Determine the cooling constant k. k = s−1

By applying the new law of cooling

[tex]\dfrac{dT}{dt} = k \Delta T[/tex]

[tex]\dfrac{dT}{dt} = k(T_1-T_2)[/tex]

[tex]\dfrac{dT}{dt} = k (T - 60)[/tex]

Taking the integral.

[tex]\int \dfrac{dT}{T-60} = \int kdt[/tex]

㏑ (T -60) = kt + C

T - 60 = [tex]e^{kt+C}[/tex]

[tex]T = 60+ C_1 e^{kt} ---- (1)[/tex]

After 20 seconds, the temperature of the bar submersion is 120°F

T(20) = 120

From equation (1) ,replace t = 20s and T = 120

[tex]120 = 60 + C_1 e^{20 \ k}[/tex]

[tex]120 - 60 = C_1 e^{20 \ k}[/tex]

[tex]60 = C_1 e^{20 \ k} --- (2)[/tex]

After 1 min i.e 60 sec , the temperature  = 100

T(60) = 100

From equation (1) ; replace t = 60 s and T = 100

[tex]100 = 60 + c_1 e^{60 \ t}[/tex]

[tex]100 - 60 =c_1 e^{60 \ t}[/tex]

[tex]40 =c_1 e^{60 \ t} --- (3)[/tex]

Dividing equation (2) by (3) , we have:

[tex]\dfrac{60}{40} = \dfrac{C_1e^{20 \ k } }{C_1 e^{60 \ k}}[/tex]

[tex]\dfrac{3}{2} = e^{-40 \ k}[/tex]

[tex]-40 \ k = In (\dfrac{3}{2})[/tex]

- 40 k = 0.4054651

[tex]k = - \dfrac{0.4054651}{ 40}[/tex]

k = -0.01014 s⁻¹

b. What is the differential equation satisfied by the temperature y(t)?

Recall that :

[tex]\dfrac{dT}{dt} = k \Delta T[/tex]

[tex]\dfrac{dT}{dt} = \dfrac{- In (\dfrac{3}{2})}{40}(T-60)[/tex]

Since y is the temperature of the body , then :

[tex]\mathbf{\dfrac{dy}{dt} = - \dfrac{In(\dfrac{3}{2})}{40}(y-60)}[/tex]

(c) What is the formula for y(t)?

From equation (1) ;

where;

[tex]T = 60+ C_1 e^{kt} ---- (1)[/tex]

Let y be measured in degrees Fahrenheit

[tex]y(t) = 60 + C_1 e^{-\dfrac{In (\dfrac{3}{2})}{40}t}[/tex]

From equation (2)

[tex]C_1 = \dfrac{60}{e^{20 \times \dfrac{-In(\dfrac{3}{2})}{40}}}[/tex]

[tex]C_1 = \dfrac{60}{e^{-\dfrac{1}{2} {In(\dfrac{3}{2})}}}[/tex]

[tex]C_1 = \dfrac{60}{e^ {In(\dfrac{3}{2})^{-1/2}}}}[/tex]

[tex]C_1 = \dfrac{60}{\sqrt{\dfrac{2}{3}}}[/tex]

[tex]C_1 = \dfrac{60 \times \sqrt{3}}{\sqrt{2}}}[/tex]

[tex]\mathbf{y(t) = 60+ \dfrac{60 \sqrt{3}}{\sqrt{2}} \ e^{\dfrac{-In(\dfrac{3}{2})\ t}{40}}}[/tex]

(d) Determine the temperature of the bar at the moment it is submerged.

At the moment it is submerged t = 0

[tex]\mathbf{y(0) = 60+ \dfrac{60 \sqrt{3}}{\sqrt{2}} \ e^{\dfrac{-In(\dfrac{3}{2})\ 0}{40}}}[/tex]

[tex]\mathbf{y(t) = 60+ \dfrac{60 \sqrt{3}}{\sqrt{2}} }[/tex]

y(t) = 60 + 70.485

y(t) = 130.485°F

Answer:

[tex] \frac{2}{3} [/tex]

Step-by-step explanation:

Area of Octagon A = 4 m²

Side length of Octagon A = a

Area of Octagon B = 9 m²

Side length of Octagon B = b

The scale factor of their side lengths = [tex] \frac{a}{b} [/tex]

According to the area of similar polygons theorem, [tex] \frac{4}{9} = (\frac{a}{b})^2 [/tex]

Thus,

[tex] \sqrt{\frac{4}{9}} = \frac{a}{b} [/tex]

[tex] \frac{\sqrt{4}}{\sqrt{9}} = \frac{a}{b} [/tex]

[tex] \frac{2}{3} = \frac{a}{b} [/tex]

Scale factor of their sides = [tex] \frac{2}{3} [/tex]

Answer:

3:5

Step-by-step explanation:

square root of 9 is 3.

square root if 25 is 5.

therefore, 3:5.

Step-by-step explanation:

83P (E)=17-17P (E),

P (E)=17/100=0.17

Answer:

Solution : [tex]-\frac{3}{4}-\frac{3}{4}i[/tex]

Step-by-step explanation:

[tex]-3\left[\cos \left(\frac{-\pi }{4}\right)+i\sin \left(\frac{-\pi \:}{4}\right)\right]\:\div \:2\sqrt{2}\left[\cos \left(\frac{-\pi \:\:}{2}\right)+i\sin \left(\frac{-\pi \:\:\:}{2}\right)\right][/tex]

Let's apply trivial identities here. We know that cos(-π / 4) = √2 / 2, sin(-π / 4) = - √2 / 2, cos(-π / 2) = 0, sin(-π / 2) = - 1. Let's substitute those values,

[tex]\frac{-3\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)}{2\sqrt{2}\left(0-1\right)i}[/tex]

=[tex]-3\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)[/tex] ÷ [tex]2\sqrt{2}\left(0-1\right)i[/tex]

= [tex]3\left(-\frac{\sqrt{2}i}{2}+\frac{\sqrt{2}}{2}\right)[/tex] ÷ [tex]-2\sqrt{2}i[/tex]

= [tex]\frac{3\left(1-i\right)}{\sqrt{2}}[/tex]÷ [tex]2\sqrt{2}i[/tex] = [tex]-3-3i[/tex] ÷ [tex]4[/tex] = [tex]-\frac{3}{4}-\frac{3}{4}i[/tex]

As you can see your solution is the last option.

Answer:

90

Step-by-step explanation:

Answer:

90

Step-by-step explanation:

Here is the equation

[tex]20+3\times(7+4)+5+2\times(7+9)[/tex]

In the order of operations parentheses go first so we get

[tex]20+3\times11+5+2\times16[/tex]

Next we do the multiplication

[tex]20+33+5+32\\[/tex]

And finally we add them all up

[tex]20+33+5+32=90\\[/tex]

Thus, 90 is the answer of [tex]20+3\times(7+4)+5+2\times(7+9)[/tex] or [tex]20+3(7+4)+5+2(7+9)[/tex]

Answer:

a). BLUE color

b). 20%

Step-by-step explanation:

a). "Which color was chosen by approximately one fourth of the students?"

  Since one fourth of the students will be represented by one fourth area of the circle given.

That means color of choice represented by the quarter of the circle will be the color liked by one fourth students.

In the figure attached, BLUE color is the choice of one fourth students in the class.

b). Area represented by purple, green and other colors is a quarter of the circle.

If we divide this quarter into five equal sections, then the total of purple and green will be  [tex]4\times \frac{1}{5}[/tex] of the the quarter of the circle.

Measure of the angle defined by purple or green sections = [tex]\frac{4}{5}\times 90[/tex]

                                                                                                     = 72°

Percentage of the students who preferred purple or green = [tex]\frac{72}{360}\times 100[/tex]

                                                                                                     = 20%

Answer:

blue

20%

Step-by-step explanation:

Answer:

B. the product of 7 and a number x

Step-by-step explanation:

7x is 7 multiplied by x.

Answer:

b is the product

Step-by-step explanation: