When K is positive the parabola ____

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

learn more on equation here: https://brainly.com/question/17290525

Area of the yellow sector is 58.91 cm2

The spinner has a circular shape and would be solved as a circle.

Area of a circle = pi r^2

= 3.142 * 5^2

= 78.55 cm2

The circle is divided into 8

78.55 / 8 = 9.8188

Area of 6 parts = 6 * 9.8188
= 58.9128 cm2

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:

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]

Step-by-step explanation:

Given :-

To find:-

Solution :-

Formula for perimeter of rectangular swamp = 2(L+W)

124 ft. = 2( 5x -10 + x )

124 ft. = 2(6x - 10)

124 ft. = 12x - 20

124+20 = 12x

144/12 = x

12 = x

putting the values of x

Length = 5x-10

width = X

perimeter =2L+2W or 2(L+W)

124= 2(5W-10+W)

perimeter =2( 6W-10)

124 = 12W-20

124+20=(12W+20-20

144=12W

144÷12=12W÷12

12=W

Answer:

113.04cm^2

Step-by-step explanation:

The area of a circle is pi * r^2

pi=3.14

So 6^2 is 36

36*3.14=113.04

Hope it helps!

Answer:

Its 9.0 units

Step-by-step explanation:

For the side mostly to the left, its gonna be 3 units, since we can easlily count it

Now, for the diagonals, we have to use the formula for the area of a triangle, which is (l * w) / 2 = a.

For the bottom side, its 3 along the x axis and 1 along the y axis. 3 * 1 is 3, now divide that by 2 and you got 1.5

Now, for the right side, its 5 on the y, 1 on the x. Multiply 1 by 5, you got 5. Now divide that by 2 and you have 2.5

Finally, for the top side, its 4 on the x, and 1 on the y. 4 * 1 is 4, divided by 2 and its 2.

Alright, now we have to add 'em all together. 3 + 1.5 + 2.5 + 2 is 9 units

Theres your answer :)

Answer:

-60p -6

Step-by-step explanation:

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

Distribute

8 - 48p  -12p -14

Combine like terms

-60p -6

Answer:

it is very high because

Step-by-step explanation:

5-2+3 so maybe it a subtraction problem hope this works

Answer:

I can help!! answer in image

Step-by-step explanation:

Have a great day sir

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

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

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

Read more about quota at:

https://brainly.com/question/1499498

Answer:

$25.40

Step-by-step explanation:

Find the unit price. So what you do is just divide 304.80 by 12 to get the price of one which comes out to be $25.40

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:

x is equal to -34/16

Step-by-step explanation:

Hey there!

In order to solve for x, we need to first simplify everything

After simplifying, it will look like this:

[tex](4x^2+14x+16)-7x+14=4x^2-4-9x[/tex]

Now you can simplify by combining like terms

[tex]4x^2+7x+30=4x^2-9x-4[/tex]

You can cross out [tex]4x^2[/tex] from both sides and then simplify even more by again combining like terms

[tex]16x+34=0\\16x=-34\\x=-34/16[/tex]

So x is equal to -34/16

To solve the given equation, let's simplify and expand both sides:

[tex]\displaystyle\sf (2x+1)(2x+6)-7(x-2)=4(x+1)(x-1)-9x[/tex]

Expanding the brackets:

[tex]\displaystyle\sf (4x^{2}+13x+6)-7x+14=4(x^{2}-1)-9x[/tex]

Simplifying further:

[tex]\displaystyle\sf 4x^{2}+13x+6-7x+14=4x^{2}-4-9x[/tex]

Combining like terms:

[tex]\displaystyle\sf 4x^{2}+6x+20=4x^{2}-9x-4[/tex]

Subtracting [tex]\displaystyle\sf 4x^{2}[/tex] from both sides:

[tex]\displaystyle\sf 6x+20=-9x-4[/tex]

Bringing all the variables to one side and the constants to the other side:

[tex]\displaystyle\sf 6x+9x=-4-20[/tex]

[tex]\displaystyle\sf 15x=-24[/tex]

Dividing both sides by [tex]\displaystyle\sf 15[/tex] to solve for [tex]\displaystyle\sf x[/tex]:

[tex]\displaystyle\sf x=-\frac{24}{15}[/tex]

Simplifying the fraction:

[tex]\displaystyle\sf x=-\frac{8}{5}[/tex]

Therefore, the solution to the equation is [tex]\displaystyle\sf x=-\frac{8}{5}[/tex].

[tex]\huge{\mathfrak{\colorbox{black}{\textcolor{lime}{I\:hope\:this\:helps\:!\:\:}}}}[/tex]

♥️ [tex]\large{\underline{\textcolor{red}{\mathcal{SUMIT\:\:ROY\:\:(:\:\:}}}}[/tex]

Answer:

The answer is 340 ft²

To find the total area we make two squares and a big rectangle.

1st square

3x5=15

2nd square

5x5=25

Rectangle

20x15=300

Total Area

15+25+300 = 340ft²

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.

The area of the shaded part of the given figure is 60 square inches'

The given figure is made up of a rectangle and 2 similar triangles

Area of the shaded part = Area of rectangle + area of 2 triangles

From the figure

Area of rectangle = 12 * 4 = 48 square in

Area of one triangle = 0.5(6)(4) = 12 square in

Area  of the figure = 48 + 12

Area  of the figure = 60 square inches'

Hence the area of the shaded part is 60 square inches.

Learn more on area of composite figure here: https://brainly.com/question/15981553

Answer:

The distance is 20 units

Step-by-step explanation:

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: Its negetive. The answer is -52d^2

Hope this helped you!

Answer:

Step-by-step explanation:

[tex]-12\ d^{2}+-40\ d^{2} = -12\ d^{2}-\ 40\ d^{2} = -52\ d^{2}[/tex]

Since −52 is negative and d² is always positive for any number d

Then ,their product −52 d² is negative .

Answer:

3/4

Step-by-step explanation:

Give brainliest if it helped...(:

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. 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