How do you make a mixed fraction and write it out as a improper fraction?

Answer:

1/2

Step-by-step explanation:

We can use the slope formula to find the slope

m = ( y2-y1)/(x2-x1)

We have two points on the line

(-1,3) and ( 1,4)

m = ( 4-3)/(1 - -1)

   = (4-3)/(1+1)

    = 1/2

Answer:

Start where the line meets a point. Then go up and over until it meets another.

The answer is 1/2

So go up one and over two. Then other problems such as this one should be pretty straight forward

it will be equal to the photo

median 5

Answer:

D: {-2, 0, 2}

R: {-1, 1, 3

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.

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.

if we can assume the ball is being thrown from the ground upwards, then we can say that the inital height of it is 0, whilst its initial velocity is 80 ft/s, thus

[tex]~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&80\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&0\\ \qquad \textit{of the object}\\ h=\textit{object's height}\\ \qquad \textit{at "t" seconds} \end{cases} \\\\\\ h(t)=-16t^2+80t+0\implies h(t)=-16t^2+80t[/tex]

Answer:

what is the question?

Step-by-step explanation:

Answer:

elga is 32 and alvin is 49

Answer:

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

Step-by-step explanation:

Answer:

its a rectangle .

Step-by-step explanation:

Answer:

three hundred twenty one and fifty one hundredths

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

[tex]t = \sqrt \frac{ {2(x - ut)} }{a} \\ = > t = \sqrt{ \frac{2x - 2ut}{a} } \\ = > {t}^{2} = \frac{2x - 2ut}{a} \\ = > a {t}^{2} = 2x - 2ut \\ = > \frac{ { - at}^{2} }{2ut} = 2x \\ = > \frac{ - at}{2u} = 2x \\ = > \frac{ - at}{2u \times 2} = x \\ = > \frac{ - at}{4u} = x[/tex]

Hope you could get an idea from here.

Doubt clarification - use comment section.

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 :)

Answer:

6.15

Step-by-step explanation:

10 gallons is 80/8

1 5/8= 13/8

80 divided by 13 is 6.15384615385 but rounded 6.15

6.15 hours

if its not that then keep rounding to 6.2 or 6hrs

Answer:

1000

Step-by-step explanation:

Answer: 1000, hope it helps.

Step-by-step explanation:

Answer:

B. [tex]\frac{1}{4}[/tex]

Step-by-step explanation:

The whole beaker is 1. If you measure the beaker, you will notice you can fill up the beaker to the brim (whole, which is 1) if you use the current amount 4 times, and four times is [tex]\frac{1}{4}[/tex].

[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:

The bottom right is a linear equation.

Step-by-step explanation:

Answer:

right side down one

Step-by-step explanation:

as you know linear means supplementary having 180 °

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:

[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

Using the binomial distribution, it is found that there is a 0.8202 = 82.02% probability that the shipment will make it through the control check.

For each article, there are only two possible outcomes, either it meets the specifications, or it does not. The probability of an article meeting the specifications is independent of any other article, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

In this problem:

The probability is:

[tex]P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)[/tex]

Then:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 8) = C_{10,8}.(0.85)^{8}.(0.15)^{2} = 0.2759[/tex]

[tex]P(X = 9) = C_{10,9}.(0.85)^{9}.(0.15)^{1} = 0.3474[/tex]

[tex]P(X = 10) = C_{10,10}.(0.85)^{10}.(0.15)^{0} = 0.1969[/tex]

[tex]P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.2759 + 0.3474 + 0.1969 =  0.8202[/tex]

0.8202 = 82.02% probability that the shipment will make it through the control check.

For more on the binomial distribution, you can check https://brainly.com/question/24863377

Answer:

Please send a picture or explain properly

Answer:

3/8 bottle

Step-by-step explanation:

Fractions are just division.

Martin + 2sisters + 5bros

= 8 people

3bottles ÷ 8people

= 3/8 bottles per person

Martin is a person, so he gets 3/8 of a bottle, if they all share equally.

Using proportions, it is found that Martin will get 0.375 of a bottle.

The rule of three is:

1 person - x bottles

8 people - 3 bottles

Applying cross multiplication:

[tex]8x = 3[/tex]

[tex]x = \frac{3}{8}[/tex]

[tex]x = 0.375[/tex]

Martin will get 0.375 of a bottle.

To learn more about proportions, you can take a look at https://brainly.com/question/24372153

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:

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:

(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].