x-13=x+1

Please help mee

18 liters of 14% bleach contains 0.14×18 = 2.52 liters of bleach.

Adding [tex]x[/tex] liters of pure water to the solution increases the total volume to [tex]18+x[/tex] liters without changing the total amount of bleach.

To end up with a 10% bleach solution, we must add

[tex]\dfrac{2.52}{18+x} = 0.10 \implies 2.52 = 1.8 + 0.10x \implies 0.10x = 0.72 \implies x=\boxed{7.2}[/tex]

liters of water.

Answer:

b

Step-by-step explanation:

Step-by-step explanation:

3. B

because the figure to the left of the y-axis yields the one to the right when you reflect it across y = x.

4. C

Flip either of the figures vertically across x = -1 which is their intersection point and you'll get the other.

5. D

it can't be clockwise because the x value of the vertices would have been negative. so it is 270° counterclockwise resulting in ( -y , x )

6. A

Similar to #4 which except this time, you flip it horizontally like how you lay a book or pick it up.

Answer:

Step-by-step explanation:

The graph of the straight line [tex]y = -\frac{1}{3}x -2[/tex] is plotted. The graph is shown below.

A straight line is of the form y = mx + c, where m is the slope and c is the y-intercept.

To plot the given straight line [tex]y = -\frac{1}{3}x -2[/tex], follow the following steps:

Step 1: Substitute x = 0 in the given equation to obtain the point where the line intersects the y-axis.

y = 0 - 2

y = -2

The point at the y-axis is (0, -2).

Step 2: Substitute y = 0 in the given equation to obtain the point where the line intersects the x-axis.

[tex]0 = -\frac{1}{3}x -2\\x = -6[/tex]

The point at the x-axis is (-6, 0).

Step 3: Draw a straight line passing through both points.

Thus, the straight line [tex]y = -\frac{1}{3}x -2[/tex] is plotted.

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Answer:

x=-22

Step-by-step explanation:

Let's substitute the number you are thinking of with x

(x+4)*3=(x-5)*2

3x+12=2x-10

x+12=-10

x=-22

1) [tex]\triangle ABC[/tex] with [tex]\overline{AC} \cong \overline{BC}[/tex], [tex]\overline{AB} \parallel \vec{CE}[/tex] (given)

2) [tex]\angle A \cong \angle B[/tex] (base angles theorem)

3) [tex]\angle A \cong \angle 1[/tex] (corresponding angles theorem)

4) [tex]\angle B \cong \angle 2[/tex] (alternate interior angles theorem)

5) [tex]\angle 1 \cong \angle 2[/tex] (transitive property of congruence)

6) [tex]\vec{CE}[/tex] bisects [tex]\angle BCD[/tex] (if a ray splits an angle into two congruent parts, it is a bisector)

By evaluating the function, we conclude that f(10) = -7

Here we know that:

g(x) = 2x - 8

And f(x) = 5 - g(x).

Then we can write:

f(x) = 5 - (2x - 8) = 5 - 2x + 8 = -2x + 13

Now we want ot evaluate it in x = 10, this means replace the variable by the number 10.

f(10) = -2*10 + 13 = -20 + 13 = -7

Then, we conclude that f(10) = 7

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The solution of the differential equation is [tex]y = (C_{1} + C_{2} x)e^{-2x} +\frac{1}{2} [x^{3} -\frac{3}{2} x-6][/tex] .

According to the given question.

We have a differential equation

[tex]y^{"} + 4y^{'} + 4y = 2x^{3}[/tex]

The above differerntial equation acn be written as

[tex](D^{2} +4D+ 4)= 2x^{3}[/tex]

Now, the auxillary equation for the above differential equation is given by

[tex]m^{2} + 4m + 4 = 0[/tex]

[tex]\implies m^{2} + 2m + 2m + 4 = 0[/tex]

⇒ m (m + 2) + 2(m + 2) = 0

⇒ m(m + 2)(m + 2) = 0

Therefore,

[tex]C.F = (C_{1} + C_{2}x)e^{-2x}[/tex]

Now,

[tex]PI = \frac{1}{D^{2} +4D+4} 2x^{3}[/tex]

[tex]\implies PI = \frac{1}{4(1+\frac{D^{2}+4D }{4} )} 2x^{3}[/tex]

[tex]\implies PI = \frac{1}{4} [1+(\frac{D^{2}+4D }{4} )]^{-1} 2x^{3}[/tex]

[tex]\implies PI = \frac{1}{4} [ 1 - (\frac{D^{2} +4D}{4} )+(\frac{D^{2} +4D}{4} )^{2} -(\frac{D^{2}+4D }{4}) ^{3} ...]2x^{3}[/tex]

[tex]\implies PI =\frac{1}{2} [ x^{3} -\frac{1}{4} (6x)-3x^{2} +3x^{2} -6][/tex]

[tex]\implies PI = \frac{1}{2} [x^{3} -\frac{3}{2} x-6][/tex]

Therefore, the solution of the differential equation will be

y = CI + PI

[tex]y = (C_{1} + C_{2} x)e^{-2x} +\frac{1}{2} [x^{3} -\frac{3}{2} x-6][/tex]

Hence, the solution of the differential equation is [tex]y = (C_{1} + C_{2} x)e^{-2x} +\frac{1}{2} [x^{3} -\frac{3}{2} x-6][/tex] .

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Answer:

d. 9

Step-by-step explanation:

y= 0.22x^2 - 1.76x +4.75

x is the time in months so we should replace x by 10

y = 0.22 (10)^2 - 1.76(10) + 4.75

0.22 (100) - 17.6 + 4.75

22 - 17.6 + 4.75

9.15

So the number of laptops that will be sold during month 10 is 9. (you can't have 0.15 of a computer)

Answer:

d. 9

Step-by-step explanation:

The equation: y = 0.22x^2 - 1.76x + 4.75

x: Time in months

y: Number of laptops sold

Since we need to predict the number of laptops that will be sold during month 10, we can replace the x with 10.

The new equation will be: y = 22 - 17.6 + 4.75

Then y will be 9.15

Since there cannot be 0.15 of a laptop, we round the number 9.15 to a whole number, which is 9.

D) 9 is our final answer to this question.

Answer:

Jeffrey has 225 % chance to go to Cafe Shirley.

Step-by-step explanation:

given,

chance of going to Cafe

Chris = 3/4

Nina =3/4

iana = 3/4

To find the total probability you should add all of them:

Total probability= 3/4 + 3/4 + 3/4

= 9/4 = 2.25

to convert to percent you should times 100

2.25 x 100 = 225%//

Answer:

Step-by-step explanation:

A. a = $1350 b. b= 89%

B. 1350(.89)^t = 675

    t = 5.9 years

Answer:

36 units

Step-by-step explanation:

to find the distance take the absolute value of the difference, that is

| - 42 - (- 78) | = | - 42 + 78 | = | 36 | = 36

or

| - 78 - (- 42) | = | - 78 + 42 | = | - 36 | = 36

Answer:

-36

Step-by-step explanation:

-78 - (-42)

= -78 + 42

= -36

Answer:

D It repeats

Step-by-step explanation:

square root of 15 is 3.87298334621

11/25= 0.44

3/20=.6

D = 0.33333333333

The number of gigabytes of data that Christopher can use while staying within his budget exists 1.3 gigabytes.

The number of gigabytes of data that Christopher can use while staying within his budget is 1.8 gigabytes.

Based on the information, we get

59 + (3 × g) = 62.90

59 + 3g = 62.90

simplifying the equation, we get

3g = 62.90 - 59

3g = 3.9

Divide both sides by 3, and we get

3g/3 = 3.9/3

g = 1.3

The number of gigabytes of data that Christopher can use while staying within his budget exists 1.3 gigabytes.

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Using the binomial distribution, we have that you would expected to draw a face card 8 times.

It is the probability of exactly x successes on n repeated trials, with p probability of a success on each trial.

The expected value of the binomial distribution is:

E(X) = np

For this problem, the parameters are given as follows:

Then the expected value is found as follows:

E(X) = np = 35 x 0.2308 = 8

You would expected to draw a face card 8 times.

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The number of servings of pasta were in the dish Lois made is 31.67.

Number of servings of pasta were in the dish Lois made = Number of cups of pasta Lois made / A serving of the pasta

= 6 1/3 ÷ 0.2

= 19/3 ÷ 0.2

= 19/3 × 1/0.2

= (19×1) / (3 × 0.2)

= 19/0.6

= 31.6666666666666

Approximately,

Number of servings of pasta were in the dish Lois made is 31.67 servings

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The correct choices based on the integers are: 1a, 2b, 3d, and 4c.

In the question, we are asked to match the scenarios to their corresponding boundaries.

Thus, the correct choices based on the integers are: 1a, 2b, 3d, and 4c.

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Answer:

Step-by-step explanation:

4x - 10 + x + 7 = 4x

5x - 3 = 4x

-3 = -x

x = 3

4(3) - 10 = 12 - 10 = 2

3 + 7 = 10

10+2 = 12

4(3)= 12

Answer:

[tex]\dfrac{3}{2} \ln |x-4| - \dfrac{1}{2} \ln |x+2| + \text{C}[/tex]

Step-by-step explanation:

Fundamental Theorem of Calculus

[tex]\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))[/tex]

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

[tex]\displaystyle \int \dfrac{x+5}{(x-4)(x+2)}\:\:\text{d}x[/tex]

Take partial fractions of the given fraction by writing out the fraction as an identity:

[tex]\begin{aligned}\dfrac{x+5}{(x-4)(x+2)} & \equiv \dfrac{A}{x-4}+\dfrac{B}{x+2}\\\\\implies \dfrac{x+5}{(x-4)(x+2)} & \equiv \dfrac{A(x+2)}{(x-4)(x+2)}+\dfrac{B(x-4)}{(x-4)(x+2)}\\\\\implies x+5 & \equiv A(x+2)+B(x-4)\end{aligned}[/tex]

Calculate the values of A and B using substitution:

[tex]\textsf{when }x=4 \implies 9 = A(6)+B(0) \implies A=\dfrac{3}{2}[/tex]

[tex]\textsf{when }x=-2 \implies 3 = A(0)+B(-6) \implies B=-\dfrac{1}{2}[/tex]

Substitute the found values of A and B:

[tex]\displaystyle \int \dfrac{x+5}{(x-4)(x+2)}\:\:\text{d}x = \int \dfrac{3}{2(x-4)}-\dfrac{1}{2(x+2)}\:\:\text{d}x[/tex]

[tex]\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int ax^n\:\text{d}x=a \int x^n \:\text{d}x$\end{minipage}}[/tex]

If the terms are multiplied by constants, take them outside the integral:

[tex]\implies \displaystyle \dfrac{3}{2} \int \dfrac{1}{x-4}- \dfrac{1}{2} \int \dfrac{1}{x+2}\:\:\text{d}x[/tex]

[tex]\boxed{\begin{minipage}{5 cm}\underline{Integrating}\\\\$\displaystyle \int \dfrac{f'(x)}{f(x)}\:\text{d}x=\ln |f(x)| \:\:(+\text{C})$\end{minipage}}[/tex]

[tex]\implies \dfrac{3}{2} \ln |x-4| - \dfrac{1}{2} \ln |x+2| + \text{C}[/tex]

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For an alternative approach, expand and complete the square in the denominator to write

[tex](x-4)(x+2) = x^2 - 2x - 8 = (x - 1)^2 - 9[/tex]

In the integral, substitute [tex]x - 1 = 3 \sin(u)[/tex] and [tex]dx=3\cos(u)\,du[/tex] to transform it to

[tex]\displaystyle \int \frac{x+5}{(x - 1)^2 - 9} \, dx = \int \frac{3\sin(u) + 6}{9 \sin^2(u) - 9} 3\cos(u) \, du \\\\ ~~~~~~~~~~~~ = - \int \frac{\sin(u) + 2}{\cos(u)} \, du \\\\ ~~~~~~~~~~~~ = - \int (\tan(u) + 2 \sec(u)) \, du[/tex]

Using the known antiderivatives

[tex]\displaystyle \int \tan(x) \, dx = - \ln|\cos(x)| + C[/tex]

[tex]\displaystyle \int \sec(x) \, dx = \ln|\sec(x) + \tan(x)| + C[/tex]

we get

[tex]\displaystyle \int \frac{x+5}{(x - 1)^2 - 9} \, dx = \ln|\cos(u)| - 2 \ln|\sec(u) + \tan(u)| + C \\\\ ~~~~~~~~~~~~ = - \ln\left|\frac{(\sec(u) + \tan(u))^2}{\cos(u)}\right|[/tex]

Now, for [tex]n\in\Bbb Z[/tex],

[tex]\sin(u) = \dfrac{x-1}3 \implies u = \sin^{-1}\left(\dfrac{x-1}3\right) + 2n\pi[/tex]

so that

[tex]\cos(u) = \sqrt{1 - \dfrac{(x-1)^2}9} = \dfrac{\sqrt{-(x-4)(x+2)}}3 \implies \sec(u) = \dfrac3{\sqrt{-(x-4)(x+2)}}[/tex]

and

[tex]\tan(u) = \dfrac{\sin(u)}{\cos(u)} = -\dfrac{x-1}{\sqrt{-(x-4)(x+2)}}[/tex]

Then the antiderivative we found is equivalent to

[tex]\displaystyle - \int \frac{x+5}{(x - 1)^2 - 9} \, dx = - \ln\left|-\frac{3(x+2)}{(x-4) \sqrt{-(x-4)(x+2)}}\right| + C[/tex]

and can be expanded as

[tex]\displaystyle - \int \frac{x+5}{(x - 1)^2 - 9} \, dx = -\ln\left| \frac{3(x+2)^{1/2}}{(x-4)^{3/2}}\right| + C \\\\ ~~~~~~~~~~~~ = - \ln\left|(x+2)^{1/2}\right| + \ln\left|(x-4)^{3/2}\right| + C \\\\ ~~~~~~~~~~~~ = \boxed{\frac32 \ln|x-4| - \frac12 \ln|x+2| + C}[/tex]

Using it's concept, it is found that there is a 0.3 = 30% probability to select a pupil that uses the swimming pool but not the gym.

A probability is given by the number of desired outcomes divided by the number of total outcomes.

In this problem, we have that 50 - 7 = 43 pupils use at least one of the pool or the gym.

We use the following relation, considering the numbers of each:

Both = Pool + Gym - At least one

Hence:

Both = 31 + 28 - 43 = 16.

From this, we have that out of 50 pupils, there are 31 - 16 = 15 pupils who use the pool but not the gym, hence the probability is:

p = 15/50 = 0.3 = 30%.

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45 square feet is the area of the base of the pedestal.

what is rectangular prism?

we know that

The volume of the pedestal (rectangular prism) is given by the formula

   V =   B × h

where

B is the area of the rectangular base of pedestal

V = 405 ft³

h = 9 ft

put the given values in the formula and solve for B

405 = B × 9

B = 405/9

B = 45 ft²

Therefore, 45 square feet is the area of the base of the pedestal.

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Answer: c. a-15= 20

Step-by-step explanation:

The answer is c because a represents her age and 15 less than that would mean that you have to subtract 15 from her age which is represented by a, and then 15 less than a would equal 20, so a-15=20

Answer:

Step-by-step explanation:

let we suppose the ratios as x , 2x and 3x

we know that,

x + 2x + 3x = -----

6x= -------

x= -----/6

therefore x = ......

2x = 2 * x (value of x)

3x = 3 * x (value of x)

and the question is solved

Answer:

let the ratio be 1x,2x,3x

Now,

1x+2x+3x=180°(sum of angles of triangle are 180)

or,6x=180°

or,x=180/6

or,x=30°

Then,

1x=1*30

=30°

2x=2*30

=60°

3x=3*30

=90°

It will take the diver 1.5seconds for her to hit the water

Linear equations are equation that has a leading degree of 2. Given the expression that expresses the distance covered by the diver as a function of time as shown;

d(t)= -16(2t - 3)(t+1)

where;

d is the height of the diver above the water and;

t is the time in seconds.

Given the following

d = 0 (the distance on the ground)

Substitute into the formula below;

-16(2t - 3)(t+1)= 0

Divide through by -16

(2t - 3)(t+1) = 0

Determine the time

2t - 3 = 0

t = 3/2

Similarly;

t +1 = 0
t = -1

Hence it will take the diver 1.5seconds for her to hit the water

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Answer:

Answer:

  48 minutes

Step-by-step explanation:

The rate of filling from both pipes is the sum of the rates in units of tanks per hour. We want the time for 2/3 of a tank.

The sum of the filling rates is ...

  rate A + rate B = (1/3) tank/hour + (1/2) tank/hour = 5/6 tank/hour

The time to fill a tank or part thereof will be the number of tanks divided by the rate in tanks per hour:

  time = (2/3 tank)/(5/6 tank/hour) = ((4/6)/(5/6)) hour = 4/5 hour

There are 60 minutes in an hour, so the fill time is ...

  (4/5 hour)×(60 min/hour) = 48 min

It will take 48 minutes for both pipes to fill 2/3 of the tank.

Answer: 144

Step-by-step explanation:

The length of DF is 16.

The horizontal distance from DF to E is 18.

So, the area is [tex]\frac{1}{2}(16)(18)=144[/tex]

The area of the triangle DEF is approximately equal to 144.014 square units.

Triangles can be generated on a Cartesian plane by marking three non-colinear points on there. Heron's formula offers the possibility of calculating the area of a triangle by only using the lengths of its three sides, whose formula is now introduced:

A = √ [s · (s - DE) · (s - EF) · (s - DF)]     (1)

s = (DE + EF + DF) / 2     (2)

Where s is the semiperimeter of the triangle.

First, we determine the lengths of the sides DE, EF and DF by Pythagorean theorem:

Side DE

DE = √ [[10 - (- 8)]² + (- 2 - 8)²]

DE ≈ 20.591

Side EF

EF = √ [(- 8 - 10)² + [- 8 - (- 2)]²]

EF ≈ 18.974

Side DF

DF = √[[- 8 - (- 8)]² + (- 8 - 8)²]

DF = 16

Then, the area of the triangle DEF is by Heron's formula:

s = (16 + 18.974 + 20.591) / 2

s = 27.783

A = √[27.783 · (27.783 - 20.591) · (27.783 - 18.974) · (27.783 - 16)]

A ≈ 144.014

The area of the triangle DEF is approximately equal to 144.014 square units.

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Answer:

40

Step-by-step explanation:

7 - 3 + 9 x 8 / 2

Multiplication and division left - right.

7 - 3 + 72 / 2

7 - 3 + 36

Add and subtract left to right.

4 + 36 = 40

The height of the tree and the building are 7.6 m and 28.6 m respectively.

An equation is an expression that shows the relationship between two or more variables and numbers.

Two triangles are similar if they have the same shape and the ratio of their corresponding sides are in the same proportion.

Triangle BGD, CHD and AFD are similar triangles.

CH = 1.6, DH = 8, GH = 30,  GF = 105

FH = GF + GH = 105 + 30 = 135

FD = FH + DH = 135 + 8 = 143; GD = 30 + 8 = 38

Hence, using similar triangles:

BG/CH = GD / DH

BG / 1.6 = 38/8

BG = 7.6 m

Also:

building/CH = DF / DH

building / 1.6 = 143/8

building = 28.6 m

The height of the tree and the building are 7.6 m and 28.6 m respectively.

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The number of contemporary titles and classic titles in Jarred DVDs collection is 2,921 and 579 respectively.

Simultaneous equation is an equation which involves the solving for two unknown values at the same time.

x + y = 3,500

x – y = 2,342

Add both be equation

x + x = 3,500 + 2,342

2x = 5,842

x = 5,842 ÷ 2

x = 2,921

Substitute x = 2,921 into

x – y = 2,342

2,921 - y = 2, 342

-y = 2,342 - 2,921

-y = -579

y = 579

Therefore, the number of contemporary titles and classic titles in Jarred DVDs collection is 2,921 and 579 respectively.

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