rewrite 4+2/3x=3/4 so it does not have fractions.

Answer: -1

Step-by-step explanation:

Answer:

A. Similar triangles

Step-by-step explanation:

Triangle ABC

35° + 90° + x = 180°

125° + x = 180°

x = 180° - 125°

x = 55°

They are similar due to the occurrence in angles

The radius of the silo is the distance from its center to the outside wall

The radius of the silo is 7 3/8" or 7 3/8 inches

From the question, we have the following parameters:

The radius is then calculated as:

Radius = 0.5 * Diameter

This gives

Radius = 0.5 * 14 3/4"

Evaluate the product

Radius = 7 3/8"

Hence, the radius of the silo is 7 3/8" or 7 3/8 inches

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1320

Step-by-step explanation:

the triangle,1/2×15×20×2=300 .rectangle,17×25=425,20×17=340,15×17=255. sum them up 300+255+425+340=1320

bought price= 2000

sold price= 2500

the percentage profit

[tex]profit\% = 100 \times \frac{(final - initial)}{initial} [/tex]

[tex]profit\% = 100 \times \frac{(2500 - 2000)}{2000} [/tex]

[tex] = \frac{(2500 - 2000)}{2000} [/tex]

[tex] = \frac{1}{4} [/tex]

[tex] = \frac{1}{4} \times 100\%[/tex]

[tex]profit\% = 25\%[/tex]

therefore, his percentage profit is 25.

Answer:

im thinking that is 3/74 because i count them

Step-by-step explanation:

3/74 rhe answer but 3/77 are wrong he just thinking

Answer:

3 over 77, 3/77

Step-by-step explanation:

You add up all of the marbles. (including the blue ones)

30+28+13+6 = 77.

So. You have a 3/77 chance of pulling a blue marble.

Answer:

Step-by-step explanation:

Opposite angles of a cyclic quadrilateral sum to 180 so

m < x = 180-105 = 75 degrees

m < y = 180 - 95 = 85 degrees.

Answer:

When you graph the equations, both equations represent the same line. If a system has no solution, it is said to be inconsistent . The graphs of the lines do not intersect, so the graphs are parallel and there is no solution.

Answer:

The distance is 20 units

Step-by-step explanation:

Answer:

3/4

Step-by-step explanation:

Give brainliest if it helped...(:

Base on the equation, the amount Laura pay for 2 tennis balls and 4 golf balls $8

Sam buys 4 tennis balls and 12 golf balls for a total of £19.

let

x = price of each tennis ball

y = price of each golf ball

Hence,

4x + 12y = 19

Luke buys 6 tennis ball and 10 golf balls for a total of £22.50. Hence,

6x + 10y = 22.50

Combine the equation

4x + 12y = 19

6x + 10y = 22.50

6x + 18y = 28.5

6x + 10y = 22.50

8y = 6

y = 6 / 8

y = $0.75

6x + 10(0.75) = 22.50

6x + 7.5 = 22.50

6x = 22.50 - 7.5

6x = 15

x = 15 / 6

x = $2.5

The cost when Laura pays for 2 tennis balls and 4 golf balls is as follows:

2(2.5) + 4(0.75) = 5 + 3 = $8

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Answer:

The answer is $1.85

Step-by-step explanation:

Now, we know that the cost of 5 chocolate bars are $5.75.

For 1 chocolate bar

5.75 ÷ 5 = $1.15

Here, we get the price of 1 chocolate bar

So, the cost of 2 chocolate bars and 3 packets of sweets are $7.85.

2 × chocolate bar = 2 × (1.15) = $2.3

Now, we want to find only the cost of one packet of sweets.

So,

7.85 – 2.3 = $5.55

3 packets of sweets cost $5.55

For 1 packets of sweets

5.55 ÷ 3 = $1.85

Thus, The cost of one packet of sweets = $1.85

Answer:

A. X < 5

Step-by-step explanation:

6(x -2) + 21 < 39

6x - 12 + 21 < 39

Group like terms

6x < 39 + 12 - 21

6x < 30

Divide through by 6

X < 30/6

X < 5

Step-by-step explanation:

A = 0.5(πr^2 )

A = 0.5(π × (20^2 ))

A = 628.32 units^2

Answer:

628.3 ft

Step-by-step explanation:

area of half circle = [tex]\frac{\pi r^{2}}{2}[/tex]

                             = [tex]\frac{\pi *20^{2} }{2}[/tex]

                             = 200[tex]\pi[/tex] = 628.3 ft

Answer:

n=23

Step-by-step explanation:

a right angle measures 90

67+n=90

subtract 67 on both sides

n=23

Use angle addition postulate:

Substitute values to get the required equation and solve it for n:

Answer:

r=14/2

r=7

[tex]circumference = 2\pi {r}\\ = 2 \times \frac{22}{7} \times 7 \\ = 44[/tex]

[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]

Let's find the circumference of given circle ~

Circumference :

[tex] \qquad \tt  \dashrightarrow \: \pi{d}^{} [/tex]

where, d = diameter

Now, let's calculate it ~

[tex] \qquad \tt  \dashrightarrow \: c = 14\pi{}^{} [/tex]

That's the answer I terms of π, but we can also plug in the approximate value of pi as 3.14

[tex] \qquad \tt  \dashrightarrow \: c = 14 \times 3.14{}^{} [/tex]

[tex] \qquad \tt  \dashrightarrow \: c = 43.96 \: \: m[/tex]

[tex] \qquad \tt  \dashrightarrow \: c \approx 44 \: \: m[/tex]

Answer:

I think so none

answer to the following steps below:-

I hope my answer helps.

5. Let [tex]x = \sin(\theta)[/tex]. Note that we want this variable change to be reversible, so we tacitly assume 0 ≤ θ ≤ π/2. Then

[tex]\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - x^2}[/tex]

and [tex]dx = \cos(\theta) \, d\theta[/tex]. So the integral transforms to

[tex]\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = \int \frac{\sin^3(\theta)}{\cos(\theta)} \cos(\theta) \, d\theta = \int \sin^3(\theta) \, d\theta[/tex]

Reduce the power by writing

[tex]\sin^3(\theta) = \sin(\theta) \sin^2(\theta) = \sin(\theta) (1 - \cos^2(\theta))[/tex]

Now let [tex]y = \cos(\theta)[/tex], so that [tex]dy = -\sin(\theta) \, d\theta[/tex]. Then

[tex]\displaystyle \int \sin(\theta) (1-\cos^2(\theta)) \, d\theta = - \int (1-y^2) \, dy = -y + \frac13 y^3 + C[/tex]

Replace the variable to get the antiderivative back in terms of x and we have

[tex]\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\cos(\theta) + \frac13 \cos^3(\theta) + C[/tex]

[tex]\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\sqrt{1-x^2} + \frac13 \left(\sqrt{1-x^2}\right)^3 + C[/tex]

[tex]\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\frac13 \sqrt{1-x^2} \left(3 - \left(\sqrt{1-x^2}\right)^2\right) + C[/tex]

[tex]\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = \boxed{-\frac13 \sqrt{1-x^2} (2+x^2) + C}[/tex]

6. Let [tex]x = 3\tan(\theta)[/tex] and [tex]dx=3\sec^2(\theta)\,d\theta[/tex]. It follows that

[tex]\cos(\theta) = \dfrac1{\sec(\theta)} = \dfrac1{\sqrt{1+\tan^2(\theta)}} = \dfrac3{\sqrt{9+x^2}}[/tex]

since, like in the previous integral, under this reversible variable change we assume -π/2 < θ < π/2. Over this interval, sec(θ) is positive.

Now,

[tex]\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = \int \frac{27\tan^3(\theta)}{\sqrt{9+9\tan^2(\theta)}} 3\sec^2(\theta) \, d\theta = 27 \int \frac{\tan^3(\theta) \sec^2(\theta)}{\sqrt{1+\tan^2(\theta)}} \, d\theta[/tex]

The denominator reduces to

[tex]\sqrt{1+\tan^2(\theta)} = \sqrt{\sec^2(\theta)} = |\sec(\theta)| = \sec(\theta)[/tex]

and so

[tex]\displaystyle 27 \int \tan^3(\theta) \sec(\theta) \, d\theta = 27 \int \frac{\sin^3(\theta)}{\cos^4(\theta)} \, d\theta[/tex]

Rewrite sin³(θ) just like before,

[tex]\displaystyle 27 \int \frac{\sin(\theta) (1-\cos^2(\theta))}{\cos^4(\theta)} \, d\theta[/tex]

and substitute [tex]y=\cos(\theta)[/tex] again to get

[tex]\displaystyle -27 \int \frac{1-y^2}{y^4} \, dy = 27 \int \left(\frac1{y^2} - \frac1{y^4}\right) \, dy = 27 \left(\frac1{3y^3} - \frac1y\right) + C[/tex]

Put everything back in terms of x :

[tex]\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = 9 \left(\frac1{\cos^3(\theta)} - \frac3{\cos(\theta)}\right) + C[/tex]

[tex]\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = 9 \left(\frac{\left(\sqrt{9+x^2}\right)^3}{27} - \sqrt{9+x^2}\right) + C[/tex]

[tex]\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = \boxed{\frac13 \sqrt{9+x^2} (x^2 - 18) + C}[/tex]

2(b). For some constants a, b, c, and d, we have

[tex]\dfrac1{x^2+x^4} = \dfrac1{x^2(1+x^2)} = \boxed{\dfrac ax + \dfrac b{x^2} + \dfrac{cx+d}{x^2+1}}[/tex]

3(a). For some constants a, b, and c,

[tex]\dfrac{x^2+4}{x^3-3x^2+2x} = \dfrac{x^2+4}{x(x-1)(x-2)} = \boxed{\dfrac ax + \dfrac b{x-1} + \dfrac c{x-2}}[/tex]

5(a). For some constants a-f,

[tex]\dfrac{x^5+1}{(x^2-x)(x^4+2x^2+1)} = \dfrac{x^5+1}{x(x-1)(x+1)(x^2+1)^2} \\\\ = \dfrac{x^4 - x^3 + x^2 - x + 1}{x(x-1)(x^2+1)^2} = \boxed{\dfrac ax + \dfrac b{x-1} + \dfrac{cx+d}{x^2+1} + \dfrac{ex+f}{(x^2+1)^2}}[/tex]

where we use the sum-of-5th-powers identity,

[tex]a^5 + b^5 = (a+b) (a^4-a^3b+a^2b^2-ab^3+b^4)[/tex]

Answer:

-60p -6

Step-by-step explanation:

8(1 - 6p) - 2(6p + 7)

Distribute

8 - 48p  -12p -14

Combine like terms

-60p -6

Answer:

I can help!! answer in image

Step-by-step explanation:

Have a great day sir

Answer:

45

Step-by-step explanation:

Calculate the least common multiple LCM(9; 5)

LCM(9; 5) = 45

Find additional factors for each fraction, for this divide LCM by the denominators of each fraction:

45 : 9 = 5 (factor of 1 fraction)

45 : 5 = 9 (factor of 2 fraction)

We multiply the numerator and denominator of each fraction by an additional factor of this fraction:

6/9=6/9 ⋅ 5/5=30/45

1/5=1/9 ⋅ 9/9=9/45

Result

6/9=30/45

1/5=9/45

Hope this Helps!

Answer:

$20

Step-by-step explanation:

take 8 and multiply it by 2 and you get 16

then take half of 8 which is 4 and add it on

16+4=20

Answer:

5/4

Step-by-step explanation:

csc theta is essentially 1/sintheta

so you can say that csc theta is 5/4.

another way to do this is to draw the right triangle and find the values of all the sides. this way you'll be able to find all the trig values of the triangle.

post a comment if you want any further guidance or help.

Answer:

7k + 6

Steps:

Subtract the numbers 20+6k-14k+k

6+6k+k

Combine the terms:

6+6k+k

6+7k

Rearange terms:

6+7k

7k+6

The apportion of Bronx quota is the alloted profit to Bronx

The apportion for Bronx quota is 20

The dataset is given as:

From the dataset, we have:

Bronx: 20.73

Remove the numbers after the decimal point

Bronx =  20

Hence, the apportion for Bronx quota is 20

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Answer:

a. 6 kg.

b. 110 m long.

c. 60 ml of juice.