I need help asap. everythings in the image

If the height of ball is shown by h(t)=52t-16[tex]t^{2}[/tex] then the ball will reach after 1.625 seconds.

Given that height in t seconds is shown by h(t)=52t-16[tex]t^{2}[/tex].

We are required to find the time after which the ball will attain its maximum height.

The maximum height is the y coordinate of vertex of the parabola. Then we can use the following value of t.

h(t)=52t-16[tex]t^{2}[/tex]

Differentiate with respect to t.

dh/dt=52-32t

Again differentiate with respect to t.

[tex]d^{2}h/dt^{2}[/tex]=-32t

Because tim cannot be negative means the height is maximum.

Put dh/dt=0

52-32t=0

-32t=-52

t=52/32

t=1.625

Hence if the height of ball is shown by h(t)=52t-16[tex]t^{2}[/tex] then the ball will reach after 1.625 seconds.

Question is incomplete as the right and complete equation is

h(t)=52t-16[tex]t^{2}[/tex] .

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See attachment for the graph of the inequality expression x + 2y > 6

The condition is given as:

{(x, y): x + 2y > 6}

This means that

x + 2y > 6

The above is an inequality expression, where the variables are x and y

So, we simply plot the graph of the inequality expression x + 2y > 6

See attachment for the graph of the inequality expression x + 2y > 6

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Answer:

0.7

Step-by-step explanation:

Each increment is 0.1, so count the approximate number of increments needed to get from the orange circle on the left and the one on the right and multiply by 0.1 to get the difference.

by solving a system of equations, we conclude that there are 7 students with blue eyes in the class.

We know that there are 40 students in the first-grade class, and we also know that:

Let's define the variables.

We know that:

H = 3*B

We also know that there are 3 students that have both blue eyes and blond hair.

And there are 15 students with none of these traits, so:

40 - 15 = 25

There are 25 students that have blue eyes, blond hair, or both.

Because we know that 3 students have both traits, we can write:

B + H - 3  =25

(Where we subtract 3 because we don't want to add these students twice).

Then we created a system of two equations:

H = 3*B

B + H - 3  =25

Replacing the first equation into the second one, we get:

B + (3*B) - 3 = 25

4*B = 25 + 3 = 28

B = 28/4 = 7

Then, by solving that system of equations, we conclude that there are 7 students with blue eyes in the class.

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Answer:

40

Step-by-step explanation:

[tex]\frac{RW}{24}=\frac{30}{18} \\ \\ RW=40[/tex]

Usando la distribución hipergeométrica,  la probabilidad de que ambas estén numeradas con el valor 1, es: 0.0222 = 2.22%.

La fórmula es:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]

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

Los parámetros son:

Los valores de los parámetros son:

N = 10, k = 2, n = 2.

La probabilidad de que ambas estén numeradas con el valor 1, es P(X = 2), entonces:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 2) = h(2,10,2,2) = \frac{C_{2,2}C_{8,0}}{C_{10,2}} = 0.0222[/tex]

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Answer: The slope is -2 and the y-intercept is (0,1)

Step-by-step explanation:

Project's cost at time 0 is $46,272.84

What is profitability index?

Profitability index is the present value of future cash flows divided by the project cost at time 0.

The present value of each future cash flow can be determined using the present value formula of a single cash flow as  below;

PV=FV/(1+r)^N

FV=future cash inflows in years 1-4

r=discount rate=8.5%

N=the year of cash flows, 1 for year 1 , 2 for year 2, 3 for year 3 and 4 for year 4.

PV of future cash flows=$12,150/(1+8.5%)^1+$12,600/(1+8.5%)^2+$15,700/(1+8.5%)^3+$10,200/(1+8.5%)^4

PV of future cash flow=$41,553.01

Profitability index=PV of future cash flows/Project cost

Profitability index= .898

Project cost=unknown(assume it is X)

.898=$41,553.01 /X

X=$41,553.01/ .898

X=project's cost=$46,272.84

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Answer:15

Step-by-step explanation: 20 divided by 8=2.5 x 6=15

Answer:

The ratio is 3 : 2.

Step-by-step explanation:

42 cm : 28 cm

= 21 : 14 (Divide both sides by 2)

= 3 : 2 (Divide both sides by 7)

The ratio between the length and width of the rectangle is 3:2.

We have,

The concept used here is the concept of ratio.

In mathematics, a ratio is a comparison of two quantities.

It expresses how many times one quantity contains another or how many times it is greater or smaller than the other quantity.

The ratio between the length and width of the rectangle can be found by dividing the length by the width:

Ratio = Length / Width

= 42 cm / 28 cm

= 7 x 6 / 7 x 4

Cancel out the common factor.

= 6/4

= 2 x 3 / 2 x 2

Cancel out the common factor.

= 3/2

Thus,

The ratio between the length and width of the rectangle is 3:2.

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Answer: 5 units

Step-by-step explanation:

x+2+2x-7=13

solve for x

3x=18

x=6

∴SR: 2x-7

=2(6)-7

=12-7

=5

Answer:

SR = 5

Step-by-step explanation:

[tex]TR=x+2+2x-7=3x-5[/tex]

[tex]3x-5=13[/tex]

[tex]3x=13+5[/tex]

[tex]x=18/3=6[/tex]

[tex]SR=2x-7=2(6)-7=12-7=5[/tex]

Hope this helps

Step-by-step explanation:

Let ABC be the equilateral triangle. Let AE, BD and CF be the medians. A meridian divides a side into two equal parts. Hence, proved that medians of an equilateral triangle are equal .

5x-30y=-35

Divide both the side by 5 and we get

x-6y = -7

x = 6y -7

Answer:

x=-7+6y

You simply need to add 30y and then divide by 5 to isolate the variable.

What is ordinary number?

1 : a number designating the place (such as first, second, or third) occupied by an item in an ordered sequence — see Table of Numbers. 2 : a number assigned to an ordered set that designates both the order of its elements and its cardinal number.

Answer: A

Step-by-step explanation:

The total value of the prizes is [tex]1000+500+2(50)=1600[/tex].

The total cost of the tickets is [tex]1000(4.00)=4000[/tex].

So, the total loss is $2400.

Dividing this by 1000 tickets gives $-2.40.

The correct answer for the solution of  inequality  is [tex]x\geq 8[/tex].

Let the unknown number be "[tex]x[/tex]."

The statement  "a number plus four" can be written  as "[tex]x+4[/tex]."

The phrase "is greater than or equal to twelve" can be expressed as "[tex]\geq 12.[/tex]

Put these together, the inequality becomes:

[tex]x + 4\geq 12[/tex]

solve for "[tex]x[/tex],"  isolate the variable on one side of the inequality:

Subtract -[tex]4[/tex] from both side of the expression:

[tex]x + 4 - 4 \geq 12 - 4[/tex]

[tex]x\geq 8[/tex]

The correct expression for the  inequality is  [tex]x\geq 8[/tex]

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Answer:

[tex]\lim_{x\rightarrow +\infty } x-1+\frac{1+ln^{2}x}{x} = + \infty[/tex]

[tex]\lim_{x\rightarrow 0 } x-1+\frac{1+ln^{2}x}{x} = + \infty[/tex]

Step-by-step explanation:

[tex]\lim_{x\rightarrow +\infty } x-1+\frac{1+ln^{2}x}{x}[/tex]

[tex]= [\lim_{x\rightarrow +\infty } (x-1)]+[ \lim_{x\rightarrow +\infty } (\frac{1+ln^{2}x}{x})][/tex]

=          +∞             +          0

= +∞

[tex]\lim_{x\rightarrow +\infty } x-1+\frac{1+ln^{2}x}{x}[/tex]

[tex]= [\lim_{x\rightarrow 0 } (x-1)]+[ \lim_{x\rightarrow 0 } (\frac{1+ln^{2}x}{x})][/tex]

=            -1             +      +∞  

= +∞  

Total 8 triangles are formed in figure and 4 edges used to make triangles and The area of all triangle is 1 square unit.

According to the statement

we have given that a figure and we have to the number of triangle present in it and the area of each square and the number of edges used in all triangles.

So, For this purpose, we know that the

A triangle is a polygon with three edges and three vertices. It is one of the basic shapes in geometry.

So,

A. The number of triangles present in figure:

There is a rectangle provided with two diagonals and due to this structure

Firstly 4 triangles are formed with 2 diagonals (big size).

And 4 triangles are formed in diagonals because diagonals cut each other (small size).

So, Total 8 triangles are formed in figure.

B. Edges used to form triangle :

there are 4 edges used to form triangles because of presence of overall shape of rectangle because triangles are formed in a rectangle shape.

So, 4 edges used to make triangles.

C. Area of the triangle:

If the area is 1 square unit and area of all triangles are same

So, The area of all triangle is 1 square unit.

So, Total 8 triangles are formed in figure and 4 edges used to make triangles and The area of all triangle is 1 square unit.

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The total income and expenditure in September are "Income is R 106 360.00 and expenditure is R 4 850.00." Option C. This is further explained below.

Generally, the equation for  Insurance and investment  is  mathematically given as

II= 42% of the balance

II = 0.42*0.64A

II= 0.2688A

Generally, the equation for Total After All other exp is  mathematically given as

Texp = 0.64A – 0.2688A

Texp= 0.3712A

Total After All other exp = 111210

subbing ahead and equating we have

111210 = 0.3712 A

Therefore

A = 299596

The total income is 299596

Therefore, the calculation for the expense is given as

Expense = 0.25A + 0.11A + 0.2688A + 4850

Expense= 0.6288A + 4850 = (0.6288*299596) + 4850

Expense= 193236

The Expense is 299596

In conclusion,  At the end of September, Agri-Small Business's minimal spending on gasoline for their fleet amounted to R 4 850.00, and they still had a balance of R106 360.00 in their account. The majority of the remaining revenue was allocated to insurance and investments by the business, which accounted for 42% of the total. Wages and salaries accounted for 25% of the company's income in September. The overall revenue and expenses for the month of September were as follows: the income was R 106 360.00, and the expenses were R 4 850.00. Option C.

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Answer:

no habla espanol

Step-by-step explanation:

The probability of being dealt no pair is 0.5011 or 50.11% if the tandard deck of playing cards has 52 cards-4 suits.

It is defined as the ratio of the number of favourable outcomes to the total number of outcomes, in other words, the probability is the number that shows the happening of the event.

It is given that:

In 5-card poker, find the probability of being dealt the following hand. Refer to the table.

From the table:

Total number of outcomes = 2598960

Total number of favourable outcomes = 1302540

The probability of being dealt no pair:

P(no pair) = 1302540/2598960

P(no pair) = 0.5011

In percentage:

P(no pair) = 50.11%

Thus, the probability of being dealt no pair is 0.5011 or 50.11% if the tandard deck of playing cards has 52 cards-4 suits.

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Answer:

4.12 miles

Step-by-step explanation:

Use distance formula

d^2 = (-8- -4)^2 + ( -9 - -10)^2

d^2 =   16   +   1

d^2 = 17

d = sqrt 17 = 4.12 miles

The equation of the line that passes through the given points is

y = 1/3x + 5.

The formula for the equation of a line passing through two points (x1, y1) and (x2, y2) is

[tex](y-y1) = \frac{(y2-y1)}{(x2-x1)} (x-x1)[/tex]

Where the fraction (y2 - y1)/(x2 - x1) is the slope of the line denoted by 'm'.

It is given that, a line passes through the points (-3,4) and (6,7).

So, the equation of the line is

[tex](y-y1) = \frac{(y2-y1)}{(x2-x1)} (x-x1)[/tex]

On substituting x1 = -3, y1 = 4, x2 = 6, and y2 = 7

(y - 4) = [(7 - 4)/(6 + 3)](x + 3)

⇒ (y - 4) = (3/9)(x + 3)

⇒ (y - 4) = 1/3(x + 3)

⇒ y- 4 = 1/3x + 1

⇒ y = 1/3x + 1 + 4

∴ y = 1/3x + 5

Thus, the equation of the line passing through the points (-3,4) and (6,7) is y = 1/3x + 5. Where the slope m = 1/3.

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Answer:

A convenient formula to use is

S = ((1 + i)^n - 1) / i where S is the value of 1$ deposited for n periods at an interest rate of i

in this case n = 12 * 28 = 336 periods of deposit at an interest rate of

.0021 / 12 = .00175 = i

S = (1.00175^336 - 1) / .00175 = 456.8338  the value of 1$ after 336 periods

350 * 456.8338 = 159891.81   the value of 350 deposited monthly

Note that 350 * 336 would be 117,600

One must be careful to distinguish the above formula from

(1 - (1 + i)^-n) / i which gives the value of 1$ when the borrower is "paying" an interest rate of i - this would be the case for a mortgage - or what is  the value of 1$ paid for n periods when paying an interest rate of i

Answer:

The Co-ordinate of C is (4/3, -1/2)

Step-by-step explanation:

We have two equation one is of straight line equation which is:

                                              y=2x-3           (i)

Other equation is of quadratic function which is:

                                         y=-3x^2+5         (ii)

Put the value of y from equation (i) in equation (ii)

So, we have:

                                     2x-3=-3x^2+5

                                      3x^2+2x-8=0

By factorization:

                                    3x^2+6x-4x-8=0

                                   3x(x+2)-4x(x+2)=0

                                        (x+2)(3x-4)=0

              x+2=0                             ;                      3x-4=0

              x=-2                                ;                       x=4/3

Put first x=-2 in equation (i)

                                                y=2(-2)-3

                                                  y=-4-3

                                                    y=-7

Now Put x=4/3 in equation (i)

                                             y=2(4/3)-3

                                                 y=8/3-3

                                                   y=-1/2

So, we have two Order pair One is (-2 , -7) and Second one is (4/3 , -1/2)

Hence the Co-ordinate of C is:

                                               C=(4/3 , -1/2)

Answer:

Point C:  (3, 3)

Point D:  (3, -22)

Step-by-step explanation:

If the distance between points C and D is 25 units, the y-value of point D will be 25 less than the y-value of point C.  The x-values of the two points are the same.

Therefore:

[tex]\textsf{Equation 1}: \quad y=2x-3[/tex]

[tex]\textsf{Equation 2}: \quad y-25=-3x^2+5[/tex]

As the x-values are the same, substitute the first equation into the second equation and solve for x to find the x-value of points C and D:

[tex]\implies 2x-3-25=-3x^2+5[/tex]

[tex]\implies 3x^2+2x-33=0[/tex]

[tex]\implies 3x^2-9x+11x-33=0[/tex]

[tex]\implies 3x(x-3)+11(x-3)=0[/tex]

[tex]\implies (x-3)(3x+11)=0[/tex]

[tex]\implies x=3, -\dfrac{11}{3}[/tex]

From inspection of the given graph, the x-value of points C and D is positive, therefore x = 3.

To find the y-value of points C and D, substitute the found value of x into the two original equations of the lines:

[tex]\begin{aligned} \textsf{Point C}: \quad 2x-3 & =y\\2(3)-3 & =3\\ \implies & (3, 3)\end{aligned}[/tex]

[tex]\begin{aligned} \textsf{Point D}: \quad -3x^2+5 & = y \\ -3(3)^2+5 & =-22\\ \implies & (3, -22)\end{aligned}[/tex]

Therefore, point C is (3, 3) and point D is (3, -22).

Using the quadratic formula, we solve for [tex]z^2[/tex].

[tex]z^4 - 5(1+2i) z^2 + 24 - 10i = 0 \implies z^2 = \dfrac{5+10i \pm \sqrt{-171+140i}}2[/tex]

Taking square roots on both sides, we end up with

[tex]z = \pm \sqrt{\dfrac{5+10i \pm \sqrt{-171+140i}}2}[/tex]

Compute the square roots of -171 + 140i.

[tex]|-171+140i| = \sqrt{(-171)^2 + 140^2} = 221[/tex]

[tex]\arg(-171+140i) = \pi - \tan^{-1}\left(\dfrac{140}{171}\right)[/tex]

By de Moivre's theorem,

[tex]\sqrt{-171 + 140i} = \sqrt{221} \exp\left(i \left(\dfrac\pi2 - \dfrac12 \tan^{-1}\left(\dfrac{140}{171}\right)\right)\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= \sqrt{221} i \left(\dfrac{14}{\sqrt{221}} + \dfrac5{\sqrt{221}}i\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= 5+14i[/tex]

and the other root is its negative, -5 - 14i. We use the fact that (140, 171, 221) is a Pythagorean triple to quickly find

[tex]t = \tan^{-1}\left(\dfrac{140}{171}\right) \implies \cos(t) = \dfrac{171}{221}[/tex]

as well as the fact that

[tex]0<\tan(t)<1 \implies 0along with the half-angle identities to find

[tex]\cos\left(\dfrac t2\right) = \sqrt{\dfrac{1+\cos(t)}2} = \dfrac{14}{\sqrt{221}}[/tex]

[tex]\sin\left(\dfrac t2\right) = \sqrt{\dfrac{1-\cos(t)}2} = \dfrac5{\sqrt{221}}[/tex]

(whose signs are positive because of the domain of [tex]\frac t2[/tex]).

This leaves us with

[tex]z = \pm \sqrt{\dfrac{5+10i \pm (5 + 14i)}2} \implies z = \pm \sqrt{5 + 12i} \text{ or } z = \pm \sqrt{-2i}[/tex]

Compute the square roots of 5 + 12i.

[tex]|5 + 12i| = \sqrt{5^2 + 12^2} = 13[/tex]

[tex]\arg(5+12i) = \tan^{-1}\left(\dfrac{12}5\right)[/tex]

By de Moivre,

[tex]\sqrt{5 + 12i} = \sqrt{13} \exp\left(i \dfrac12 \tan^{-1}\left(\dfrac{12}5\right)\right) \\\\ ~~~~~~~~~~~~~= \sqrt{13} \left(\dfrac3{\sqrt{13}} + \dfrac2{\sqrt{13}}i\right) \\\\ ~~~~~~~~~~~~~= 3+2i[/tex]

and its negative, -3 - 2i. We use similar reasoning as before:

[tex]t = \tan^{-1}\left(\dfrac{12}5\right) \implies \cos(t) = \dfrac5{13}[/tex]

[tex]1 < \tan(t) < \infty \implies \dfrac\pi4 < t < \dfrac\pi2 \implies \dfrac\pi8 < \dfrac t2 < \dfrac\pi4[/tex]

[tex]\cos\left(\dfrac t2\right) = \dfrac3{\sqrt{13}}[/tex]

[tex]\sin\left(\dfrac t2\right) = \dfrac2{\sqrt{13}}[/tex]

Lastly, compute the roots of -2i.

[tex]|-2i| = 2[/tex]

[tex]\arg(-2i) = -\dfrac\pi2[/tex]

[tex]\implies \sqrt{-2i} = \sqrt2 \, \exp\left(-i\dfrac\pi4\right) = \sqrt2 \left(\dfrac1{\sqrt2} - \dfrac1{\sqrt2}i\right) = 1 - i[/tex]

as well as -1 + i.

So our simplified solutions to the quartic are

[tex]\boxed{z = 3+2i} \text{ or } \boxed{z = -3-2i} \text{ or } \boxed{z = 1-i} \text{ or } \boxed{z = -1+i}[/tex]

[tex]\boldsymbol{\sf{6-\dfrac{3}{4}x+\dfrac{1}{3}=\dfrac{1}{y}x+5 }}[/tex]

Convert 6 to the fraction 18/3.

                       [tex]\boldsymbol{\sf{\dfrac{18}{3} -\dfrac{3}{4}x+\dfrac{1}{3}=\dfrac{1}{y}x+5 }}[/tex]

Since the fractions 18/3 and 1/3 have the same denominator, we add their numerators to calculate them.

                       [tex]\boldsymbol{\sf{\dfrac{18+1}{3}-\dfrac{3}{4}x=\dfrac{1}{2}x+5 \ \longmapsto \ \ [Add \ 18+1] }}[/tex]

                              [tex]\boldsymbol{\sf{\dfrac{19}{3}-\dfrac{3}{4}x=\dfrac{1}{2}x+5 }}[/tex]

Subtract [tex]\bf{\frac{1}{2}x }[/tex] on both sides.

                               [tex]\boldsymbol{\sf{\dfrac{19}{3}-\dfrac{3}{4}x-\dfrac{1}{2}x=5 }}[/tex]

Combine [tex]\bf{-\frac{3}{4}x}[/tex] and [tex]\bf{-\frac{1}{2}x}[/tex] to get [tex]\bf{-\frac{5}{4}x}[/tex].

                                [tex]\boldsymbol{\sf{\dfrac{19}{3}-\dfrac{5}{4}x=5 }}[/tex]

Subtract 19x from both sides.

                                       [tex]\boldsymbol{\sf{-\dfrac{5}{4}x=5-\dfrac{19}{3} }}[/tex]

Convert 5 to the fraction 15/3.

                                      [tex]\boldsymbol{\sf{-\dfrac{4}{5}x=\dfrac{15}{3}-\dfrac{19}{3} }}[/tex]

Since the fractions 15/3 and 19/3 have the same denominator, we add their numerators to calculate them.

                             [tex]\boldsymbol{\sf{-\dfrac{5}{4}x=\dfrac{15-19}{3} \ \longmapsto \ \ [Subtract \ 15-19] }}[/tex]

                                 [tex]\boldsymbol{\sf{-\dfrac{5}{4}x=-\dfrac{4}{3} }}[/tex]

Multiply both sides by -4/3, the reciprocal of -4/3.

                                    [tex]\boldsymbol{\sf{x=-\dfrac{4}{5}\left(-\dfrac{4}{5}\right) }}[/tex]

Multiply -4/3 by -4/5 (to do this, multiply the numerator by the numerator and the denominator by the denominator).

                   [tex]\boldsymbol{\sf{x=\dfrac{-4(-4)}{3\times5} \ \ \longmapsto \ \ Multiply, \ numerator \ and \ denominator. }}[/tex]

                                 [tex]\red{\boxed{\boldsymbol{\sf{\blue{Answer \ \ \longmapsto \ \ \ \ x=\frac{16}{15} }}}}}[/tex]

The number -3 written as a logarithm with a base of 2 is log₂(0.125) or log₂(1/8)

As a general rule, logarithms are mathematical expressions that are written in the form log(x) or ln(x), for natural logarithms

The number is given as:

x = -3

The base of the logarithm is given as:

Base = 2

To rewrite the given number as a base of 2, we take the exponent of the number where the base is 2

This is represented as:

Number =2^-3

Apply the power rule of indices

Number =1/2^3

Evaluate the exponent

Number = 1/8

Evaluate the quotient

Number = 0.125

Hence, when the number -3 is rewritten as a logarithm with base 2, the equivalent logarithm expression is log₂(0.125) or log₂(1/8)

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