3⁄4 of 2⁄3 =? givin brainliest

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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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: -1

Step-by-step explanation:

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:

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

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:

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:

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

Answer:

I can help!! answer in image

Step-by-step explanation:

Have a great day sir

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:

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

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:

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:

I think so none

answer to the following steps below:-

I hope my answer helps.

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:

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:

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.

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:

3/4

Step-by-step explanation:

Give brainliest if it helped...(:

Answer:

-60p -6

Step-by-step explanation:

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

Distribute

8 - 48p  -12p -14

Combine like terms

-60p -6

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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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:

a. 6 kg.

b. 110 m long.

c. 60 ml of juice.

Answer:

The distance is 20 units

Step-by-step explanation:

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