LAST ATTEMPT MARKING AS BRAINLIEST!! ( write a rule to describe each transformation)

Answer:

1

Step-by-step explanation:

(3,3) has a positive x value and a positive y value which means it is in the first quadrant

Answer:

4 fluid ounces

Step-by-step explanation:

A waitress is filling coffee mugs at a diner. She pours 12 fluid ounces into each of 5 mugs from a full pot of coffee. If her coffee pot holds 8 cups, how many fluid ounces does she have left?

Remember

8 fluid ounces = 1 cup.

First you need to do find out how many fluid ounces she poured in total.

So do,

5 mugs × 12 fluid ounces.

which is 60 fluid ounces.

Now that you now that you need to convert 8 cups into fluid ounces.

Like I said at the top 8 fluid ounces = 1 cup.

So you need to do,

8 cups × 8 fluid ounces.

Which is 64 fluid ounces she can hold in her coffee pot.

Now you subtract,

64 fluid ounces - 60 fluid ounces.

Which is 4 fluid ounces left in her coffee pot.

The number of fluid ounces she has left is 4 fluid ounces if the waitress is filling coffee mugs at a diner. She pours 12 fluid ounces into each of the 5 mugs from a full pot of coffee.

It is defined as the conversion from one quantity unit to another quantity unit followed by the process of division, and multiplication by a conversion factor.

It is given that:

A waitress is filling coffee mugs at a diner. She pours 12 fluid ounces into each of 5 mugs from a full pot of coffee.

Let x be the number of fluid ounces she has left.

4 fluid ounces

As we know,

8 fluid ounces = 1 cup

The total amount of coffee = 5×12

The total amount of coffee = 60

Total amount = 8 cups × 8 fluid ounces.

x = 64 fluid ounces - 60 fluid ounces.

x = 4 fluid ounces

Thus, the number of fluid ounces she has left is 4 fluid ounces if the waitress is filling coffee mugs at a diner. She pours 12 fluid ounces into each of the 5 mugs from a full pot of coffee.

Learn more about the unit conversion here:

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Factorization involves breaking down an expression.

The sum of cubes is given as:

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

(a) Verify the formula

Expand the expression on the right-hand side

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

Collect like terms

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

[tex](a^3 + b^3) = a^3 + b^3[/tex]

The formula has been verified

(b) Factorized 8x^3+ 27

We have:

[tex]8x^3 + 27[/tex]

Express 27 as 3^3

[tex]8x^3 + 27 =8x^3 + 3^2[/tex]

Express 8 as 2^3

[tex]8x^3 + 27 =2^3x^3 + 3^3[/tex]

Rewrite as:

[tex]8x^3 + 27 =(2x)^3 + 3^3[/tex]

Given that:

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

The expression becomes

[tex]8x^3 + 27 =(2x + 3)((2x)^2 - (2x)3 +3^2)[/tex]

[tex]8x^3 + 27 =(2x + 3)(4x^2 -8x3 +9)[/tex]

(c)The other factor of 2^3 - b^3

By difference of cubes, we have:

[tex]x^3 - y^3=(x-y)(x^2+xy+y^2)[/tex]

So, the equation becomes

[tex]2^3 - b^3=(2-b)(2^2+2b+b^2)[/tex]

This gives

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

Hence, the other factor of [tex]2^3 - b^3[/tex] is [tex]4+2b+b^2[/tex]

(c) Factor x^3 - 1

We have:

[tex]x^3 - y^3=(x-y)(x^2+xy+y^2)[/tex]

Express 1 as 1^3 in x^3 - 1

[tex]x^3 - 1 =x^3 - 1^3[/tex]

[tex]x^3 - y^3=(x-y)(x^2+xy+y^2)[/tex] becomes

[tex]x^3 - 1^3=(x-1)(x^2+x+1^2)[/tex]

[tex]x^3 - 1^3=(x-1)(x^2+x+1)[/tex]

Read more about factorization at:

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Answer:

631 m²

Step-by-step explanation:

Outer length of park = Total area - Area of pond

Outer length of park = 1000 - 369

Outer length of park = 631 m²

Answer:

Hope it will help you a lot.

Answer:

A - 144

Step-by-step explanation:

everything else would be too far. thx

Answer:

2 feet: $18

4 feet: $36

5 feet: $45

$81 = 4.5 feet

10 feet = $90

Step-by-step explanation:

2 feet: $18

4 feet: $36

18 + 18

5 feet: $45

18 + 18 + 9

$81 = 4.5 feet

81/18

10 feet = $90

18 + 18 + 18 + 18 + 18

The answer it 9300 so 9296 round to is 9300

Answer:

[tex]Understand \: how \: to \: work \: with \: negative \: bases \\ \: and \: negative \: exponents. \\

\bold{answer - } \\ {5}^{2} = 5 \times 5 = 25 \\ {5}^{ - 2} = \frac{1}{ {5}^{2} } = \frac{1}{25} = 0.04 \\ {( - 5)}^{2} = ( - 5) \times ( - 5) = 25 \\ - {5}^{2} = ( - 5) \times ( - 5) = 25 \\ \\ \bold \purple{hope \: it \: helps \: ♡}[/tex]

Answer:

Step-by-step explanation:

Linear equations can have the form y = mx + b

where m is the slope of the line and equals Δy/Δx

and b is the y intercept.

Just for fun, let's choose the first and last sets of data points.

Δy = (275 - 135) = 140

Δx = (5 - 1) = 4

so slope m = 140/4 = 35    and the units on the slope would be $/month

Answer:

The slope of the line is 35

Step-by-step explanation:

We'll select [tex](1,135)[/tex] and [tex](2,170)[/tex] for this problem.

The slope is defined as the rate of change. Its formula is [tex]Slope=m=\frac{Rise}{Run}=\frac{y_2-y_1}{x_2-x_1}[/tex] where our two points are [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex].

Therefore, we have [tex]m=\frac{y_2-y_1}{x_2-x_1}=\frac{170-135}{2-1}=\frac{35}{1}=35[/tex] as our slope.

Answer:

£55

Step-by-step explanation:

50% is just another way to say 1/2. This means that 1/2 of £110 is £55.

Answer:

the answer would be 12 I say this because 72 divied by 6 would equal 12

Answer:

12 dollars

Set up a proportion.

6/72=1/x

Solve by cross-multiplying (my preferred method) and then you will end up with the final answer of x=12.

12 is the answer :)

PLEASE MARK BRAINLIEST!

THANK YOU & HAVE A WONDERFUL DAY :))

Answer:

Segment ED has the same length as EA

The line segment which is equal to EA is ED. Therefore, option B is the correct answer.

In the given diagram, a line perpendicular to line AB passes through point C.

A tangent to a circle is a line which intersects the circle at only one point. The common point between the tangent and the circle is called the point of contact.

The length of two tangents drawn from an external point to a circle is equal.

From the given figure we can see there are three circles, two large circles and one small circle.

Line segment EA is tangent to a small circle.

The line segment which is equal to EA is ED because ED is another tangent to a small circle from the same external point.

The line segment which is equal to EA is ED. Therefore, option B is the correct answer.

To learn more about the tangent to a circle visit:

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Answer:

  x +2y +4z = 27

Step-by-step explanation:

The parallel plane will have the same coefficients of x, y, z as the given plane. We notice those have a common factor of 2, so the equation can be reduced to ...

  x +2y +4z = constant

This equation is satisfied for every point on the line, so we have ...

  (3 +2t) +2(t) +4(6 -t) = constant . . . . . substituting for x, y, z

  3 +2t +2t +24 -4t = constant

  27 = constant

The equation of the desired plane is ...

  x +2y +4z = 27

The normal to the given plane is (2, 4, 8), and the plane we want is parallel to this one so it has the same normal vector.

When t = 0, the given line, and thus the plane we want, passes through the point (3, 0, 6).

Then the equation of the plane is given by

(2, 4, 8) • (x - 3, y - 0, z - 6) = 0

2 (x - 3) + 4y + 8 (z - 6) = 0

2x + 4y + 8z = 54

or

x + 2y + 4z = 27

There's something very off about this question.

In spherical coordinates,

x² + y² + z² = ρ²

so that

f(x, y, z) = 6 - (x² + y² + z²)

transforms to

g(ρ, θ, φ) = 6 - ρ²

When transforming to spherical coordinates, we also introduce the Jacobian determinant, so that

dV = dx dy dz = ρ² sin(φ) dρ dθ dφ

Since we integrate over a sphere with radius 4, the domain of integration is the set

E = {(ρ, θ, φ) : 0 ≤ ρ ≤ 4 and 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π}

so that the integral is

[tex]\displaystyle \int_{\phi=0}^{\phi=\pi} \int_{\theta=0}^{\theta=2\pi} \int_{\rho=0}^{\rho=4} (6 - \rho^2) \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi[/tex]

Computing the integral is simple enough.

[tex]\displaystyle = \int_{\phi=0}^{\phi=\pi} \int_{\theta=0}^{\theta=2\pi} \int_{\rho=0}^{\rho=4} (6 \rho^2 - \rho^4) \sin(\phi) \, d\rho \, d\theta \, d\phi[/tex]

[tex]\displaystyle = 2\pi \int_{\phi=0}^{\phi=\pi} \int_{\rho=0}^{\rho=4} (6 \rho^2 - \rho^4) \sin(\phi) \, d\rho \, d\phi[/tex]

[tex]\displaystyle = 2\pi \left(\int_{\phi=0}^{\phi=\pi} \sin(\phi) \, d\phi\right) \left(\int_{\rho=0}^{\rho=4} (6 \rho^2 - \rho^4) \, d\rho\right)[/tex]

[tex]\displaystyle = 2\pi \cdot 2 \cdot \left(-\frac{384}5\right) = \boxed{-\frac{1536\pi}5}[/tex]

but the mass can't be negative...

Chances are good that this question was recycled without carefully changing all the parameters. Going through the same steps as above, the mass of a spherical body with radius R and mass density given by

[tex]\delta(x, y, z) = k - (x^2 + y^2 + z^2)[/tex]

for some positive number k is

[tex]\dfrac{4\pi r^3}{15} \left(5k - 3r^2\right)[/tex]

so in order for the mass to be positive, we must have

5k - 3r² ≥ 0   ⇒   k ≥ 3r²/5

In this case, k = 6 and r = 4, but 3•4²/5 = 9.6.

[tex] \sqrt{128} \\ = \sqrt{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2} \\ = \sqrt{ {2}^{2} \times {2}^{2} \times {2}^{2} } \\ = 2 \times 2 \times 2 \\ = 8[/tex]

[tex] \sqrt{48} \\ = \sqrt{2 \times 2 \times 2 \times 2 \times 3} \\ = \sqrt{ {2}^{2} \times {2}^{2} \times 3 } \\ = 2 \times 2 \sqrt{3} \\ = 4 \sqrt{3} [/tex]

[tex] \sqrt{300} \\ = \sqrt{2 \times 2 \times3 \times 5 \times 5} \\ = \sqrt{ {2}^{2} \times 3 \times {5}^{2} } \\ = 2 \times 5 \sqrt{3} \\ = 10 \sqrt{3} [/tex]

Hope you could get an idea from here.

Doubt clarification - use comment section.

Answer:

See explanation

Step-by-step explanation:

Dilation of 1.5 about the origin. (Each new point is 1.5 times as far from the origin.)

G (-1, 3)  G' (-1.5, 4.5)

F (-2, -2)  F' (-3, -3)

G'x / Gx = -1.5/-1 = 1.5

G'y / Gy = 4.5 / 3 = 1.5

F'x / Fx = -3 / -2 = 1.5

F'y / Fy = -3 / -2 = 1.5

150−28w

i think i dont know

The expression which is equivalent to 1/4 (5x + 6) is; Choice A: {5(1/4)x} + {6(1/4)x}.

According to the question:

In a bid to expand the expression; we must multiply each term in the parenthesis by (1/4);

In essence; we have;

Read more on multiplication;

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Answer:

Cosine Formula

Thus, the cosine of angle α in a right triangle is equal to the adjacent side's length divided by the hypotenuse. To solve cos, simply enter the length of the adjacent and hypotenuse and solve.

[tex]slope = m = \cfrac{rise}{run} \implies \cfrac{ f(x_2) - f(x_1)}{ x_2 - x_1}\impliedby \begin{array}{llll} average~rate\\ of~change \end{array}\\\\[-0.35em] \rule{34em}{0.25pt}\\\\ f(x)= 2x^3+4\qquad \begin{cases} x_1=4\\ x_2=6 \end{cases}\implies \cfrac{f(6)-f(4)}{6-4} \\\\\\ \cfrac{[2(6)^3+4]~~ -~~[2(4)^3+4]}{2}\implies \cfrac{436~~ -~~132}{2}\implies \cfrac{304}{2}\implies 152[/tex]

Answer:

D: [tex]\frac{3}{(6-6)}[/tex]

Step-by-step explanation:

Option A equates to [tex]-\frac{0}{2}=0[/tex]

Option B equates to [tex]\frac{(-4+0)}{8}=\frac{-4}{8}=-\frac{1}{2}[/tex]

Option C equates to [tex]0\div11 =0[/tex]

Option D equates to [tex]\frac{3}{(6-6)}=\frac{3}{0}[/tex] which cannot be defined as division by 0 is impossible

Answer:

Yes

Step-by-step explanation:

1.To know this answer, we need to know some vocabulary that's used in t. Most people don't know how to solve questions if they have hard Vocabulary and what the question is Asking

Mean=The average of all his throws

It's asking us to calculate his total score  

2. Find Clues to help you solve word problems

Finding clues in Word Problems is a very important part a lot of people skip but this helps you solve the question easier

5 throws and Mean Score is 32 are the clues

3. Guess and check

The method to this thingy is called guess and check. There is another method which is slightly harder and i just don't want to do it.

Let's just say his scores were... 5, 6, 4, 5,10Add them up and you get 30Then 30/5

Let's try bigger numbers

20+25+20+20

If you like this method, you may try it another method is called work Backwards

5x32=160

Then divide 160 by 5 to double check and you get 32!!

========================================

Therefore, the answer s 160 as his total score

hope this gives you a clue to what we are doing and when you have a question like this, I really hope you understand the concept

Hope this helped (:

Answer:

[tex]y = \frac{2}{3} x + 2[/tex]

Step-by-step explanation:

Start by finding the slope of the given line.

[tex]\boxed{ slope = \frac{y _{1} - y_2 }{x_1 - x_2} }[/tex]

Slope of given line

[tex] = \frac{5 - ( - 1)}{ - 2 - 2} [/tex]

[tex] = \frac{5 + 1}{ - 4} [/tex]

[tex] = \frac{6}{ - 4} [/tex]

[tex] = - \frac{3}{2} [/tex]

A perpendicular bisector cuts through the line at its midpoint perpendicularly.

The product of the slopes of two perpendicular lines is -1.

Let the slope of the perpendicular bisector be m.

[tex] - \frac{3}{2} m = - 1[/tex]

[tex]m = - 1 \div ( - \frac{3}{2} )[/tex]

[tex]m = - 1 \times ( - \frac{2}{3} )[/tex]

[tex]m = \frac{2}{3} [/tex]

[tex]y = \frac{2}{3}x + c[/tex], where c is the y-intercept.

To find the value of c, we need to substitute a pair of coordinates that lies on the perpendicular bisector into the equation. Since the perpendicular bisector passes through the midpoint of the given line, we can use the midpoint formula to find the coordinates.

[tex]\boxed{midpoint = ( \frac{x _{1} + x _2}{2} , \frac{y_1 + y_2}{2} )}[/tex]

Midpoint of given line

[tex] = ( \frac{ - 2+ 2}{2} , \frac{5 - 1}{2} )[/tex]

[tex] =( \frac{0}{2} , \frac{4}{2} )[/tex]

= (0, 2)

[tex]y = \frac{2}{3} x + c[/tex]

When x= 0, y= 2,

2= ⅔(0) +c

2= 0 +c

c= 2

Thus, the equation of the perpendicular bisector is [tex]y = \frac{2}{3} x + 2[/tex].

56 + 8x

Area of Rectangle = Length x Width

Area of Rectangle = 8(7 + x)

i think that the answer is 0 because f(2) means that you substitute 2 for x

2(2)=4

4-4=0

Answer:

Simple Interest: A=P(1+rt)

A=15000(1+(0.1*4.2))

A=$21,300

Compound Interest:A=P(1+r/n)^nt

A=15000(1+0.1/4.2)^1*4.2

A=$64,500

Step-by-step explanation:

Answer:

[tex]\huge\boxed{-y^2+2y}[/tex]

Step-by-step explanation:

[tex]\sqrt[3]y\cdot\left(7\sqrt[3]{8y^2}-\sqrt[3]{y^5}-4y\sqrt[3]{27y^2}\right)\\\\=(\sqrt[3]y)(7\sqrt[3]{8y^2})-(\sqrt[3]y)(\sqrt[3]{y^5})-(\sqrt[3]y)(4y\sqrt[3]{27y^2})\\\\=7\sqrt[3]{(y)(8y^2)}}-\sqrt[3]{(y)(y^5)}-4y\sqrt[3]{(y)(27y^2)}\\\\=7\sqrt{8y^3}-\sqrt{y^6}-4\sqrt{27y^3}\\\\=7\sqrt[3]{2^3y^3}-\sqrt{y^{2\cdot3}}-4\sqrt{3^3y^3}\\\\=7\sqrt[3]{(2y)^3}-\sqrt{(y^2)^3}-4\sqrt{(3y)^3}\\\\=7\cdot2y-y^2-4\cdot3y\\\\=14y-y^2-12y\\\\=-y^2+2y[/tex]

Used:

[tex]a(a+b)=ab+ac\\\\\sqrt[3]{a\cdot b}=\sqrt[3]a\cdot\sqrt[3]b\\\\\sqrt[3]{a^3}=a\\\\(a^n)^m=a^{n\cdot m}[/tex]