7 1/2 divided by 2 1/2

Answer:

n($6.50) + m($5.50) + k($6.00) = p

Step-by-step explanation:

n = matinee ticket

m = drink

k = popcorn

p = total cost

Without knowing the exact amount that was bought we must put a variable to show an unknown number. All of this together makes an algebraic expression.

how many 2/3 are in 9/8?

or another way to put it will be

how many times does 2/3 go into 9/8?

well, is simply a division.

[tex]\cfrac{9}{8}\div \cfrac{2}{3}\implies \cfrac{9}{8}\cdot \cfrac{3}{2}[/tex]

Answer:

It's C. for plato. It's a rectangle

Step-by-step explanation:

Answer:

its a rectangle .

Step-by-step explanation:

Answer: z < -8

Step-by-step explanation:

Since z is negative, divide both sides by -1, which leaves you with z > -8.

Multiplying or dividing an inequality by a negative number flips the sign, thus the answer is z < 8.

Correct me if I am incorrect.

Answer:

1000

Step-by-step explanation:

Answer: 1000, hope it helps.

Step-by-step explanation:

Answers:

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

Explanation:

For equation A, I'll transform the right hand side into a similar form as the left side. Throughout the steps below, the left hand side stays the same.

[tex]\sqrt{448x^c} = 8x^3\sqrt{7x}\\\\\sqrt{448x^c} = \sqrt{(8x^3)^2}\sqrt{7x}\\\\\sqrt{448x^c} = \sqrt{64x^{3*2}}\sqrt{7x}\\\\\sqrt{448x^c} = \sqrt{64x^{6}}\sqrt{7x}\\\\\sqrt{448x^c} = \sqrt{64x^{6}*7x}\\\\\sqrt{448x^c} = \sqrt{64*7x^{6+1}}\\\\\sqrt{448x^c} = \sqrt{448x^{7}}\\\\[/tex]

Therefore, c = 7

Notice that 7/2 = 3 remainder 1. The quotient 3 is the exponent for the term outside the root for [tex]8x^3\sqrt{7x}[/tex] while the remainder 1 is the exponent for the x term inside the root.

---------------------------------------

We do the same idea for equation B.

[tex]\sqrt[3]{576x^{d}} = 4x\sqrt[3]{9x^{2}}\\\\\sqrt[3]{576x^{d}} = \sqrt[3]{(4x)^3}\sqrt[3]{9x^{2}}\\\\\sqrt[3]{576x^{d}} = \sqrt[3]{64x^3}\sqrt[3]{9x^{2}}\\\\\sqrt[3]{576x^{d}} = \sqrt[3]{64x^3*9x^{2}}\\\\\sqrt[3]{576x^{d}} = \sqrt[3]{64*9x^{3+2}}\\\\\sqrt[3]{576x^{d}} = \sqrt[3]{576x^{5}}\\\\[/tex]

This must mean d = 5

Note: 5/3 = 1 remainder 2, which means [tex]\sqrt[3]{x^5} = x^1\sqrt[3]{x^2} = x\sqrt[3]{x^2}[/tex]

Answer: betina or adelita

Step-by-step explanation: hope it helps

Answer:

x=109 degrees

Step-by-step explanation:

By alternate interior angles, the measure of angle ADE is the same as that of EAB, both of which are 38.

Because ADE is an isosceles triangle, the measure angle EAD is equal to that angle EDA; let that measure be x.

Because the angles of a triangle add up to 180, x+x+38=180 -> 2x+38=180 -> 2x=142 -> x=71

That means that angle EDA is 180 degrees

Because x is supplementary to angle EDA, the measure of angle x is 180-71=109 degrees

Answer:

64 ft

Step-by-step explanation:

A square has to be equal on all 4 sides for it to be considered a square. Considering that it is 8 feet on one side it would be 8 feet on all the others as well. From there you multiply 8*8 and receive the solution of 84 feet for the area of the garden.

Answer:

3/2

Step-by-step explanation:

Since the shape is an equilateral triangle, all the angles are equal measure, 60° and all the sides are also of equal measure that was given, root3. So half of the triangle has length (root3)/2. The perpendicular drawn in the interior is also an angle bisector. The triangles created are 30°-60°-90° triangles. The sides of this special right triangle are in the ratio

s : 2s : sroot3

The longest side of the 30-60-90 triangle is given. The shortest side is half the length of the longest side. The length of the long leg is the short leg × root3

In this diagram the short leg is (root3)/2 .

(root3)/2 × root3 = 3/2

See image.

[tex]\begin{cases} f(x) = x^4+14x\\ g(x) = 6-x \end{cases}\qquad \qquad h(x) = f(x) + g(x) \\\\\\ h(x) = (x^4+14x)+(6-x)\implies h(x) = x^4+14x-x+6 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = x^4+13x+6~\hfill[/tex]

now, if we graph h(x), Check the picture below, we can see that horizontally the line keeps on moving towards the left, going going and going towards -infinity, and it also keeps on moving towards the right, going going and going towards +infinity, and since the horizontal area used by the function is the domain of it, the domain for h(x) will be (-∞ , +∞).

Answer: store b because you can get more ground beat.

Step-by-step explanation:

Answer:

32. A.

Step-by-step explanation: because i did 4/5 of 40 and got 32.

Answer:

y = -8

Step-by-step explanation:

-7y + 11 = 75 + y

Bring the "y" variable on one side, and rest on the other.

Rearranging, we get,

-7y - y = 75 - 11

-8y = 64

-y = 8

y = -8

Hope it helps :)

Add the sales tax to the bill:

Bill with tax = 14.60 x 1.09 = $15.95

Multiply the total bill by the tip percentage:

Tip = 15.95 x 0.19 = 3.02

Tip = $3.02

Answer:

elga is 32 and alvin is 49

it will be equal to the photo

median 5

Answer:

[tex]6 + 5 = 11 + 3 = 14 + 2 = 16[/tex]

[tex]6 \div 5 \div 3 \div 2 = 0.2[/tex]

[tex] 6 \times 5 \times 3 \times 2 = 180[/tex]

Step-by-step explanation:

If its addition add them, multiplication multiply them, division divided them.

Answer:

Step-by-step explanation:

i need help sorry no answer got u

Answer:

The exterior and interior angles must add up to 180 degrees. Thus, 180 divided by five gives the exterior angle as 36 degrees (and hence the interior angles as 144 degrees).

Answer:

10 Sides

Step-by-step explanation:

Int Angle + Ext Angle = 180

4Ext + Ext = 180

Ext = 36 degrees

-----

360/36 = 10 sides

Answer:

three hundred twenty one and fifty one hundredths

Answer:

A

Step-by-step explanation:

A. 2 toys/5 minutes

y/x

Step-by-step explanation:

Why use brainly for this

Answer:

$87.50

Step-by-step explanation:

Answer:

$50.00 is the discount price

Step-by-step explanation:

3/5 are done, that means 2/5 are still to do.

that means these 30 km are these 2/5.

2/5 x = 30

2x = 150

x = 75

the total distance is 75 km.

Answer:

75 km

Step-by-step explanation:

2/5 = 30 km

→ Find 1/5

1/5 = 15 km

→ Now find 5/5 or 1

1 = 75 km

Answer:

(a) The particle is moving to the right in the interval  [tex](0 \ , \ \displaystyle\frac{\pi}{2}) \ \cup \ (\displaystyle\frac{3\pi}{2} \ , \ 2\pi)[/tex] , to the left in the interval [tex](\displaystyle\frac{\pi}{2}\ , \ \displaystyle\frac{3\pi}{2})[/tex], and stops when t = 0, [tex]\displaystyle\frac{\pi}{2}[/tex], [tex]\displaystyle\frac{3\pi}{2}[/tex] and [tex]2\pi[/tex].

(b) The equation of the particle's displacement is [tex]\mathrm{s(t)} \ = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(t)} \ + \ 3[/tex]; Final position of the particle [tex]\mathrm{s(2\pi)} \ = \ 3[/tex].

(c)  The total distance traveled by the particle is 9.67 (2 d.p.)

Step-by-step explanation:

(a) The particle is moving towards the right direction when v(t) > 0 and to the left direction when v(t) < 0. It stops when v(t) = 0 (no velocity).

Situation 1: When the particle stops.

[tex]\-\hspace{1.7cm} v(t) \ = \ 0 \\ \\ 5 \ \mathrm{sin^{2}(t)} \ \mathrm{cos(t)} \ = \ 0 \\ \\ \-\hspace{0.3cm} \mathrm{sin^{2}(t) \ cos(t)} \ = \ 0 \\ \\ \mathrm{sin^{2}(t)} \ = \ 0 \ \ \ \mathrm{or} \ \ \ \mathrm{cos(t)} \ = \ 0 \\ \\ \-\hspace{0.85cm} t \ = \ 0, \ \displaystyle\frac{\pi}{2}, \ \displaystyle\frac{3\pi}{2} \ \ \mathrm{and} \ \ 2\pi[/tex].

Situation 2: When the particle moves to the right.

[tex]\-\hspace{1.67cm} v(t) \ > \ 0 \\ \\ 5 \ \mathrm{sin^2(t) \ cos(t)} \ > \ 0[/tex]

Since the term [tex]5 \ \mathrm{sin^{2}(t)}[/tex] is always positive for all value of t of the interval [tex]0 \ \leq \mathrm{t} \leq \ 2\pi[/tex], hence the determining factor is cos(t). Then, the question becomes of when is cos(t) positive? The term cos(t) is positive in the first and third quadrant or when [tex]\mathrm{t} \ \epsilon \ (0, \ \displaystyle\frac{\pi}{2}) \ \cup \ (\displaystyle\frac{3\pi}{2}, \ 2\pi)[/tex] .  

*Note that parentheses are used to demonstrate the interval of t in which cos(t) is strictly positive, implying that the endpoints of the interval are non-inclusive for the set of values for t.

Situation 3: When the particle moves to the left.

[tex]\-\hspace{1.67cm} v(t) \ < \ 0 \\ \\ 5 \ \mathrm{sin^2(t) \ cos(t)} \ < \ 0[/tex]

Similarly, the term [tex]5 \ \mathrm{sin^{2}(t)}[/tex] is always positive for all value of t of the interval [tex]0 \ \leq \mathrm{t} \leq \ 2\pi[/tex], hence the determining factor is cos(t). Then, the question becomes of when is cos(t) positive? The term cos(t) is negative in the second and third quadrant or  [tex]\mathrm{t} \ \epsilon \ (\displaystyle\frac{\pi}{2}, \ \displaystyle\frac{3\pi}{2})[/tex].

(b) The equation of the particle's displacement can be evaluated by integrating the equation of the particle's velocity.

[tex]s(t) \ = \ \displaystyle\int\ {5 \ \mathrm{sin^{2}(t) \ cos(t)}} \, dx \ \\ \\ \-\hspace{0.69cm} = \ 5 \ \displaystyle\int\ \mathrm{sin^{2}(t) \ cos(t)} \, dx[/tex]

To integrate the expression [tex]\mathrm{sin^{2}(t) \ cos(t)}[/tex], u-substitution is performed where

[tex]u \ = \ \mathrm{sin(t)} \ , \ \ du \ = \ \mathrm{cos(t)} \, dx[/tex].

[tex]s(t) \ = \ 5 \ \displaystyle\int\ \mathrm{sin^{2}(t) \ cos(t)} \, dx \\ \\ \-\hspace{0.7cm} = \ 5 \ \displaystyle\int\ \ \mathrm{sin^{2}(t)} \, du \\ \\ \-\hspace{0.7cm} = \ 5 \ \displaystyle\int\ \ u^{2} \, du \\ \\ \-\hspace{0.7cm} = \ \displaystyle\frac{5u^{3}}{3} \ + \ C \\ \\ \-\hspace{0.7cm} = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(t)} \ + \ C \\ \\ s(0) \ = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(0)} \ + \ C \\ \\ \-\hspace{0.48cm} 3 \ = \ 0 \ + \ C \\ \\ \-\hspace{0.4cm} C \ = \ 3.[/tex]

Therefore, [tex]s(t) \ = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(t)} \ + \ 3[/tex].

The final position of the particle is [tex]s(2\pi) \ = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(2\pi)} \ + \ 3 \ = \ 3[/tex].

(c)

[tex]s(\displaystyle\frac{\pi}{2}) \ = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(\frac{\pi}{2})} \ + \ 3 \\ \\ \-\hspace{0.85cm} \ = \ \displaystyle\frac{14}{3} \qquad (\mathrm{The \ distance \ traveled \ initially \ when \ moving \ to \ the \ right})[/tex]

[tex]|s(\displaystyle\frac{3\pi}{2}) - s(\displatstyle\frac{\pi}{2})| \ = \ |\displaystyle\frac{5}{3} \ (\mathrm{sin^{3}(\frac{3\pi}{2})} \ - \ \mathrm{sin^{3}(\displaystyle\frac{\pi}{2})})| \ \\ \\ \-\hspace{2.28cm} \ = \ \displaystyle\frac{5}{3} | (-1) \ - \ 1| \\ \\ \-\hspace{2.42cm} = \displaystyle\frac{10}{3} \\ \\ (\mathrm{The \ distance \ traveled \ when \ moving \ to \ the \ left})[/tex]

[tex]|s(2\pi) - s(\displaystyle\frac{3\pi}{2})| \ = \ |\displaystyle\frac{5}{3} \ (\mathrm{sin^{3}(2\pi})} \ - \ \mathrm{sin^{3}(\displaystyle\frac{3\pi}{2})})| \ \\ \\ \-\hspace{2.28cm} \ = \ \displaystyle\frac{5}{3} | 0 \ - \ 1| \\ \\ \-\hspace{2.42cm} = \displaystyle\frac{5}{3} \\ \\ (\mathrm{The \ distance \ traveled \ finally \ when \ moving \ to \ the \ right})[/tex].

The total distance traveled by the particle in the given time interval is[tex]\displaystyle\frac{14}{3} \ + \ \displaystyle\frac{5}{3} \ + \ \displaystyle\frac{10}{3} \ = \ \displaystyle\frac{29}{3}[/tex].

Answer:

a. 6

b. 21

c. 78

d. 13.5

e. 14

f. 24

Hope that helps!