20 liters of mixture contain milk nad water in the ratio 5:3 of 4 liters of the mixture are replaced by 4 liters of milk find the new ratio of milk to water

Answer:

A. 55, 89

Step-by-step explanation:

In a Fibonacci sequence, you start with 2 given numbers. Then each subsequent number is the sum of the last two numbers.

12, 21, 34

12 + 21 = 34

34 + 21 = 55

55 + 34 = 89

Answer: 55, 89

Aswer:I am sure of the answer it is 6 and 42

Step-by-step explanation:

Answer:

8

Step-by-step explanation:

Ham with or without cheese-2 choices

Bologna with or without cheese-2 choices

Bologna with cheese with water or juice-2 choices

Bologna without cheese with juice or water-2 choices

Ham with cheese with juice or water -2 choices

Ham without cheese with juice or water -2 choices

2+2+2+2=8

Kile has 8 choices for lunch

Answer:

B. The graph of function f is stretched horizontally by a scale factor of 3 to create the graph of function g.

Step-by-step explanation:

The rules for linear transformations are that

 g(x) = a·f(b·(x-c)) +d

stretches the graph vertically by a factor of "a" (before the shift)

compresses the graph horizontally by a factor of "b" (before the shift)

shifts it to the right by amount "c"

shifts it up by amount "d".

Your equation has b=1/3, so the graph is compressed by a factor of 1/3, which is equivalent to a stretch by a factor of 3.

The appropriate choice of description is ...

 b) the graph of g(x) is horizontally stretched by a factor of 3

Answer:

B

Step-by-step explanation:

Correct on Plato

Answer:

Step-by-step explanation:

Using FV = PV(1 + r)^n where FV = future value, PV = present value, r = interest rate per period, and n = # of periods

1/PV (FV) = (PV(1 + r^n)1/PV divide by PV

ln(FV/PV) = ln(1 + r^n) convert to natural log function

ln(FV/PV) = n[ln(1 + r)] by simplifying

n = ln(FV/PV) / ln(1 + r) solve for n

n = ln(2/1) / ln(1 + .08) solve for n, letting FV + 2, PV = 1 and rate = 8% or .08 compound annually

n = 9

n = ln(2/1) / ln(1 + .08/12) solve for n, letting FV + 2, PV = 1 and rate = .08/12 compound monthly

n = 104 months or 8.69 years

n = ln(2/1) / ln(1 + .08/365) solve for n, letting FV + 2, PV = 1 & rate = .08/365 compound daily

n = 3163 days or 8.67 years

Alternatively

A = P e ^(rt)

Given that r = 8%

= 8/100

= 0.08

2 = e^(0.08t)

ln(2)/0.08 = t

0.6931/0.08 = t

t= 8.664yrs

t = 8.67yrs

Which ever approach you choose to use,you will still arrive at the same answer.

Answer:

Problem 1)       [tex] m = \dfrac{1}{4} [/tex]     [tex] slope_{perpendicular} = -4 [/tex]

Problem 2)      [tex] m = \dfrac{1}{3} [/tex]     [tex] slope_{perpendicular} = -3 [/tex]

Step-by-step explanation:

[tex] slope = m = \dfrac{y_2 - y_1}{x_2 - x_1} [/tex]

[tex] slope_{perpendicular} = \dfrac{-1}{m} [/tex]

Problem 1) M(9,6), N(1,4)

[tex] slope = m = \dfrac{6 - 4}{9 - 1} = \dfrac{2}{8} = \dfrac{1}{4} [/tex]

[tex] slope_{perpendicular} = \dfrac{-1}{\frac{1}{4}} = -4 [/tex]

Problem 2) M(-2,2), N(4,-4)

[tex] slope = m = \dfrac{4 - 2}{4 - (-2)} = \dfrac{2}{6} = \dfrac{1}{3} [/tex]

[tex] slope_{perpendicular} = \dfrac{-1}{\frac{1}{3}} = -3 [/tex]

Answer:

81.432 minutes

Step-by-step explanation:

Given the following :

Video Games (Mins) - - - Time with Family(Mins)

40 - - - - - - - - - - - - - - - - - - - 80

55 - - - - - - - - - - - - - - - - - - - 75

70 - - - - - - - - - - - - - - - - - - - 69

85 - - - - - - - - - - - - - - - - - - - 64

Best fit line:

ý = -0.363x +94.5

For someone who spent 36 minutes playing video games, the predicted number of minutes spent with family according to the best fit line will be:

Here number of minutes playing video games '36' is the independent variable

ý is the dependent or predicted variable ;

94.5 is the intercept

ý = -0.363(36) +94.5

ý = −13.068 + 94.5

ý = 81.432 minutes

Which is about 81 minutes to the nearest whole number.

Answer:

137, 280 feet

Step-by-step explanation:

There are 5,280 feet in a mile.

26 * 5,280 = 137,280

There are 137, 280 feet in 26 miles.

There are 137,280 feet in 26 miles.

The unitary method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value.

We know that there are 5,280 feet in a mile.

So, the solution would be;

26 x 5,280 = 137,280

Thus, There are 137,280 feet in 26 miles.

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Answer:

Hey there!

The slope is -1/3, because the rise over run is -1/3.

Let me know if this helps :)

Answer:

E(w) = 1600000

v(w) = 240000

Step-by-step explanation:

given data

sequence = 1 million iid  (+1 and +2)

probability of transmitting a +1 =  0.4

solution

sequence will be here as

P{Xi = k } = 0.4              for k = +1

                  0.6              for k = +2

and define is

x1  + x2 + ................ + X1000000

so for expected value for W

E(w) = E( x1  + x2 + ................ +  X1000000 )   ......................1

as per the linear probability of expectation

E(w) = 1000000 ( 0.4 × 1 + 0.6 × 2)

E(w) = 1600000

and

for variance of W

v(w) = V ( x1  + x2 + ................ + X1000000 )    ..........................2

v(w) = V x1  + V x2 + ................  + V  X1000000

here also same as that xi are i.e d so cov(xi, xj ) = 0 and i ≠ j

so

v(w) = 1000000 ( v(x) )

v(w) = 1000000 ( 0.24)

v(w) = 240000

Answer:

2 sqrt(41) = x

Step-by-step explanation:

This is a right triangle so we can use the Pythagorean theorem

a^2 + b^2 = c^2

8^2 + 10 ^2 = x^2

64+ 100 = x^2

164 = x^2

Take the square root of each side

sqrt(164) = sqrt(x^2)

sqrt(4) sqrt(41) = x

2 sqrt(41) = x

Answer:

- At any time t, the population is:

P = 375t² + 3000t + 1000

- At time t = 3 days, the population is:

P = 13,375

Step-by-step explanation:

Given the rate of change of the population of bacteria as:

dP/dt = 3000/(1 + 0.25t)

we need to rewrite the given differential equation, and solve.

Rewriting, we have:

dP/3000 = (1 + 0.25t)dt

Integrating both sides, we have

P/3000 = t + (0.25/2)t² + C

P/3000 = t + 0.125t² + C

When t = 0, P = 1000

So,

1000/3000 = C

C = 1/3

Therefore, at any time t, the population is:

P/3000 = 0.125t² + t + 1/3

P = 375t² + 3000t + 1000

At time t = 3 days, the population is :

P = 375(3²) + 3000(3) + 1000

= 3375 + 9000 + 1000

P = 13,375

Answer:

Note that orthogonal to the plane means perpendicular to the plane.

Step-by-step explanation:

-1x+3y-3z=1 can also be written as -1x+3y-3z=0

The direction vector of the plane -1x+3y-3z-1=0 is (-1,3,-3).

Let us find a point on this  line for which the vector from this point to (0,0,5) is perpendicular to the given line. The point is x-0,y-0 and z-0 respectively

Therefore, the vector equation is given as:

-1(x-0) + 3(y-0) + -3(z-5) = 0

-x + 3y + (-3z+15) = 0

-x + 3y -3z + 15 = 0

Multiply through by - to get a positive x coordinate to give

x - 3y + 3z - 15 = 0

Answer:

If a polygon has 3 sides, then the polygon is a triangle.

Step-by-step explanation:

Bold = hypothesis

Italic = conclusion

Statement:

If p, then q.

Converse: If q, then p.

To find the converse, switch the hypothesis and conclusion.

Statement:

If a polygon is a triangle, then it  has 3 sides.

Now we switch the hypothesis and the conclusion to write the converse of the statement.

If it  has 3 sides, then a polygon is a triangle.

We fix a little the wording:

If a polygon has 3 sides, then it is a triangle.

Answer: If a polygon has 3 sides, then the polygon is a triangle.

The converse statement will be;

⇒ If a polygon has 3 sides, then the polygon is a triangle.

A triangle is a three sided polygon, which has three vertices and three angles which has the sum 180 degrees.

Given that;

The statement is,

''If a polygon is a triangle, then it has 3 sides. ''

Now,

Since, The statement is,

''If a polygon is a triangle, then it has 3 sides. ''

We know that;

The converse of statement for p → q will be q → p.

Thus, The converse statement is find as;

⇒ If a polygon has 3 sides, then the polygon is a triangle.

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Answer:

[tex] \sqrt{4 {}^{2} + ( - 4) {}^{2} } [/tex]

[tex] \sqrt{32} [/tex]

and the angle

[tex] \tan( \alpha ) = - 4 \div 4 = - 1[/tex]

and since the sin component is -ve, we have our angle on 4th quadrant, which equals 315 degrees

Options:

Determine two pairs of polar coordinates for the point (-4, 4) with 0° ≤ θ < 360°. (5 points)

Group of answer choices

(4  , 135°), (-4  , 315°)

(4  , 45°), (-4  , 225°)

(4  , 315°), (-4  , 135°)

(4  , 225°), (-4  , 45°)

Step-by-step explanation:

The guy asking forgot to provide the options you can comment the awnswe in the comments just do it before brainly turns off comments to try and prevent people from learning

Answer:

B) x=80°

Step-by-step explanation:

This is a hexagon, so it has interior angles equaling 720°.  (N-2)*180

So the equation would be

78+134+136+132+2x+x=720

480+3x=720

3x=720-480

3x=240

x=80°

Answer:

ig

Step-by-step explanation:

[tex](9-\sqrt{-8} )- (5 + \sqrt{-32} ) \\(9-5) + (-\sqrt{-8}- \sqrt{-32} )\\4 - \sqrt{-8} -\sqrt{-32} \\4-2i\sqrt{2} -4i\sqrt{2} \\4-6i\sqrt{2}[/tex]

Answer:

A

Step-by-step explanation:

The height is always perpinducular to the base. The height here is perpendicular to line segment A.

Answer:

A

Step-by-step explanation:

Assuming the cup is a right circular cylinder, it's volume is [tex]V=\pi r^2 h[/tex]

$h=10$, $r=\frac 42$

So the volume is $\pi\cdot(2)^2\cdot10=125.66$

hence you can fill up to 125.66 cubic Inches of milkshake

.

Answer:

The answer is:

C. [tex]\bold{x = -1.25h+220}[/tex]

Step-by-step explanation:

Given:

[tex]h=0.8( 220-x )[/tex]

Where [tex]h[/tex] is the heartbeats per minute and

[tex]x[/tex] is the age of person

To find:

Age of person in terms of heartbeats per minute = ?

To choose form the options:

[tex]A.\ x=176-h\\B.\ x=176-0.8h\\C.\ x=-1.25h+220\\D.\ x=h-0.8220[/tex]

Solution:

First of all, let us have a look at the given equation:

[tex]h=0.8( 220-x )[/tex]

It is value of [tex]h[/tex] in terms of [tex]x[/tex].

We have to find the value of [tex]x[/tex] in terms of [tex]h[/tex].

Let us divide the equation by 0.8 on both sides:

[tex]\dfrac{h}{0.8}=\dfrac{0.8( 220-x )}{0.8}\\\Rightarrow \dfrac{1}{0.8}h=220-x\\\Rightarrow 1.25h=220-x[/tex]

Now, subtracting 220 from both sides:

[tex]\Rightarrow 1.25h-220=220-x-220\\\Rightarrow 1.25h-220=-x[/tex]

Now, multiplying with -1 on both sides:

[tex]-1.25h+220=x\\OR\\\bold{x = -1.25h+220}[/tex]

So, the answer is:

C. [tex]\bold{x = -1.25h+220}[/tex]

Answer: x < 28

Step-by-step explanation:

[tex] 24 = 3 \cdot 2^3 [/tex]

[tex]96=3\cdot 2^5 [/tex]

[tex] 384=3\cdot2^7[/tex]

hence it is a geometric progression, with a multiplied constant [tex]3[/tex]

Sum of G.P. of [tex]n[/tex] terms [tex] S_n = a\dfrac{r^n-1}{r-1}\quad \text{where } r \text{ is the common ratio and } a \text{ is the first term} [/tex]

and [tex] r=-2^2=-4[/tex]

Note that the constant should be separated, so

[tex] a= -8 [\tex]

after plugging the values, you'll get the answer

[tex] -26216 \times 3 [/tex]

which option C

Answer:

C

Step-by-step explanation:

-24+96-384+...

a=-24

r=96/(-24)=-4

[tex]s_{7}=a\frac{1-r^7}{1-r} \\=-24\frac{1-(-4)^7}{1-(-4)}\\=-24\frac{1+4^7}{1+4} \\=-24\frac{1+16384}{5} \\=-24\frac{16385}{5} \\=-24 \times 3277\\=-78648[/tex]

Answer:

Step-by-step explanation:

Hello, first, let's use the product rule.

Derivative of uv is u'v + u v', so it gives:

[tex]f(x)=(x^3-2x+1)(x-3)=u(x) \cdot v(x)\\\\f'(x)=u'(x)v(x)+u(x)v'(x)\\\\ \text{ **** } u(x)=x^3-2x+1 \ \ \ so \ \ \ u'(x)=3x^2-2\\\\\text{ **** } v(x)=x-3 \ \ \ so \ \ \ v'(x)=1\\\\f'(x)=(3x^2-2)(x-3)+(x^3-2x+1)(1)\\\\f'(x)=3x^3-9x^2-2x+6 + x^3-2x+1\\\\\boxed{f'(x)=4x^3-9x^2-4x+7}[/tex]

Now, we distribute the expression of f(x) and find the derivative afterwards.

[tex]f(x)=(x^3-2x+1)(x-3)\\\\=x^4-2x^2+x-3x^3+6x-4\\\\=x^4-3x^3-2x^2+7x-4 \ \ \ so\\ \\\boxed{f'(x)=4x^3-9x^2-4x+7}[/tex]

Hope this helps.

Do not hesitate if you need further explanation.

Thank 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:

choice 1,2,4,5 from top to bottom

Step-by-step explanation:

1:the points given are in the line where both planes intersect

2:point H is not on any plane

3:in the diagram point F is on plane R so false

4:if you connect the points given they will intersect so not collinear

5:the points F and G are on the plane R

6:so F is on plane R but H is not on any do false

Answer:

Step-by-step explanation:

Hello, please consider the following.

[tex]x=x(t)=2cos(t)\\\\dx=\dfrac{dx}{dt}dt=x'(t)dt=-2sin(t)dt[/tex]

and

[tex]y=y(t)=2sin(t)\\\\dy=\dfrac{dy}{dt}dt=y'(t)dt=2cos(t)dt[/tex]

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

The values of dx and dy are give as -2sin(t)dt and 2cos(t)dt respectively. The answer to the given problem can be stated as,

dy = 2cos(t)dt

And,  dx = -2sin(t)dt.

The integration can be defined as the inverse operation of differentiation. If a function is the integration of some function f(x) , then differentiation of that function is f(x).

The given integral over C is ∫ (x − y) dx + (x + y) dy.

And, the parameters for C are as follows,

x = 2cos(t)

y = 2sin(t)

0 ≤ t ≤ 2π

Now, on the basis of these parameters dx and dy can be found as follows,

x =  2cos(t)

Differentiate both sides with respect to t as follows,

dx/dt = 2d(cos(t))/dt

=> dx/dt = -2sin(t)

=> dx =  -2sin(t)dt

And, y = 2sin(t)

Differentiate both sides with respect to t as follows,

dy/dt = 2d(sin(t))/dt

=> dy/dt = 2cos(t)

=> dy = 2cos(t)dt

Hence, the value of dx and dy as per the given parameters is -2sin(t)dt and 2cos(t)dt respectively.

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Answer:

the relation is not one-to-one.

Step-by-step explanation:

it can't because every number is in the 4th quadrant.  

Answer:

Linear, Quadratic, Radical, Absolute Value, Cubic, Cube Root

Step-by-step explanation:

To find:

Which functions have an intercept at (0, 0).

That means, when we put a value [tex]x=0[/tex] in the [tex]y =f(x)[/tex], value of [tex]y=0[/tex].

Let us discuss each parent function one by one:

1. Linear:

[tex]y = x[/tex]

When we put x = 0, y = 0

Therefore, it has intercept at (0, 0).

2. Quadratic:

[tex]y = x^2[/tex]

When we put x = 0, y = 0

Therefore, it has intercept at (0, 0).

3. Radical:

[tex]y = \sqrt x[/tex]

When we put x = 0, y = 0

Therefore, it has intercept at (0, 0).

4. Absolute Value:

[tex]y = |x|[/tex]

When we put x = 0, y = 0

Therefore, it has intercept at (0, 0).

5. Rational:

[tex]y = \dfrac{1}{x}[/tex]

When we put [tex]x = 0\Rightarrow y \rightarrow \infty[/tex]

Therefore, it does not have intercept at (0, 0).

6. Exponential:

[tex]y = b^x[/tex]

b is any base

When we put [tex]x = 0\Rightarrow y =1[/tex]

Therefore, it does not have intercept at (0, 0).

7. Logarithmic:

[tex]y = logx[/tex]

When we put [tex]x = 0 \Rightarrow y\rightarrow[/tex] Not defined

Therefore, it does not have intercept at (0, 0).

8. Cubic:

[tex]y = x^3[/tex]

When we put [tex]x = 0\Rightarrow y =0[/tex]

Therefore, it has intercept at (0, 0).

9. Cube Root:

[tex]y = \sqrt[3]x[/tex]

When we put [tex]x = 0\Rightarrow y =0[/tex]

Therefore, it has intercept at (0, 0).

This sequence converges to 0.

Proof: Recall that

[tex]\displaystyle\lim_{n\to\infty}\frac1{\sqrt n}=0[/tex]

is to say that for any given [tex]\varepsilon>0[/tex], there is some [tex]N[/tex] for which [tex]\left|\frac1{\sqrt n}-0\right|=\frac1{\sqrt n}<\varepsilon[/tex] for all [tex]n>N[/tex].

Let [tex]N=\left\lceil\frac1{\varepsilon^2}\right\rceil[/tex]. Then

[tex]n>\left\lceil\dfrac1{\varepsilon^2}\right\rceil\ge\dfrac1{\varepsilon^2}[/tex]

[tex]\implies\dfrac1n<\varepsilon^2[/tex]

[tex]\implies\dfrac1{\sqrt n}<\varepsilon[/tex]

as required.