Consider the generic chemical equation:

2 A + 3 B ------> 3 C

If 6 moles of A and 2 moles of B are reacted, what is the maximum number of moles of C that can be formed?

Answer:

(-4,-3)

Step-by-step explanation:

When a point (x,y) is reflected over the x-axis, you take the opposite so it becomes (x,-y).

(-4,3) --> (-4,-3)

It seems like you want to find the sum of 38 to 115:

[tex] \displaystyle \large{38 + 39 + 40 + 41 + ... + 114 + 115}[/tex]

If we notice, this is arithmetic series or the sum of arithmetic sequences.

To find the sum of the sequences, there are three types of formulas but I will demonstrate only one and the best for this problem.

[tex] \displaystyle \large{S_n = \frac{n(a_1+a_n) }{2} }[/tex]

This formula only applies to the sequences that have the common difference = 1.

Given that a1 = first term of sequence/series, n = number of terms and a_n = last term

We know the first term which is 38 and the last term is 115. The problem here is the number of sequences.

To find the n, you can use the following formula.

[tex] \displaystyle \large{n = (a_n - a_1) + 1}[/tex]

Substitute an = 115 and a1 = 38 in the formula of finding n.

[tex] \displaystyle \large{n = (115 - 38) + 1} \\ \displaystyle \large{n = (77) + 1} \\ \displaystyle \large{n = 78}[/tex]

Therefore the number of sequences is 78.

Then we substitute an = 115, a1 = 38 and n = 78 in the sum formula.

[tex] \displaystyle \large{S_{78} = \frac{78(38+115) }{2} } \\ \displaystyle \large{S_{78} = \frac{39(38+115) }{1} } \\ \displaystyle \large{S_{78} = 39(153) } \\ \displaystyle \large \boxed{S_{78} = 5967}[/tex]

Hence, the sum is 5967.

Answer: Screenshot attached

Step-by-step explanation:

Answer:

Function C is the only one that matches both definitions for the domain and range.

Step-by-step explanation:

First let's understand what the domain and range definitions mean in plain English.

The following statement means x is defined such that x can be any real number except for 1.

[tex]\{x | x \ne 1, x \epsilon R \}[/tex]

The following statement means y is defined such that y can be any real number except for 3.

[tex]\{y | y \ne 3, y \epsilon R \}[/tex]

So let's see what functions match those definitions. Take a look at the denominator of the functions in question 7, if we plug in x = 1 for any of them, we would be dividing by zero, which we know we can't do since the function would not be defined at that value. Therefore the domain definition is correct for all functions, we cannot use x = 1 for any of them.

Let's move on to the range. Our goal is to find a function where plugging in y = 3 would give us an undefined result. Simply plugin y = 3 into all of the given equations and simplify.

Therefore the function C is the only one that matches both definitions for the domain and range.

Answer:

SA = 153.9m^2

Step-by-step explanation:

SA = 4[tex]\pi[/tex][tex]r^{2}[/tex]

r = 3.5

SA = 4[tex]\pi[/tex][tex](3.5)^{2}[/tex]

SA = 4[tex]\pi[/tex](12.25)

SA = 49[tex]\pi[/tex]

SA = 153.9m^2

Answer:

B

Step-by-step explanation:

I took a test in school and this was the answer...at least for my class.

Answer:

30×5=150

so 150+10=160

thus his payment in the 30th installment is

rs.160

Answer:

D) 50% of the students surveyed spend at least two hours online each day.

50% of the students surveyed spend at least two hours online each day.

Hence option d is correct.

Given that,

Number of students who spend at least two hours online = 100

Total number of students surveyed = 200

To calculate the percentage of students who spend at least two hours online each day,

we can use the formula:

percentage = (number of students who spend at least two hours online / total number of students surveyed) x 100%

Plugging in the values from the problem, we get:

percentage = (100 / 200) x 100% = 50%

Therefore, we can conclude that 50% of the students surveyed spend at least two hours online each day.

Learn more about the percent visit:

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Answer:

B

Step-by-step explanation:

xy=6

y=6/x (1)

substitute (1) into 3y=x-3 and solve

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Answer:

Step-by-step explanation:

Answer:

-4×2-2xy+5y2

Step-by-step explanation:

(6-5xy+2y2)-(14-3xy-3y2)

=6-5xy+2y2-14+3xy+3y2

=(6-14)+(-5xy+3xy)+(2y2+3y2)

=-4×2-2xy+5y2

Answer:

−4x2 − 2xy + 5y2

Step-by-step explanation:

Answer:

297 square feet

Step-by-step explanation:

Let w represent the width of the garden.

Since the length is 24 ft longer than the width, it can be represented by w + 24.

Use the perimeter formula, p = 2l + 2w. Plug in the perimeter and w + 24 as l, then solve for w:

p = 2l + 2w

84 = 2(w + 24) + 2w

84 = 2w + 48 + 2w

84 = 4w + 48

36 = 4w

9 = w

So, the width is 9 ft. Add 24 to this to find the length:

24 + 9 = 33

So, the length is 33 ft.

Find the area of the garden with the formula A = lw:

A = lw

A = (33)(9)

A = 297

The area of the garden is 297 square feet.

Answer: Length 132cm, Breadth 48cm

Explanation:

Perimeter = 360cm

Let the length and breadth be x

ATQ

2(11x + 4x) = 360

22x + 8x = 360

30x = 360

x = 360/30

x = 12

Length = 11×12

= 132 cm

Breadth = 4×12

= 48cm

Must click thanks and mark brainliest

Refer to the attaachment

Answer:

percentage of paper used= 18%

Number of papers printed= 90

Total number of papers in the paper package= 'y'

total number of papers

= [tex]\frac{18}{100}[/tex] × y =90

transpose to the other side,

y= 90 ×[tex]\frac{100}{18}[/tex]

y=10 x100

y=1000

hence total number of papers in the package is 1000

Hope this helps

Please mark me as brainliest

Answer:

[tex]27,\!405[/tex] if:

Step-by-step explanation:

[tex]\displaystyle \genfrac{(}{)}{0}{}{30}{4} = \frac{30 \times 29 \times 28 \times 27}{4 \times 3 \times 2 \times 1} = 27,\!405[/tex].

Assume for now that the ordering of the four cards does matter. Hands like [tex]\verb!A!\, \verb!B!\, \verb!C!\, \verb!D![/tex] and [tex]\verb!A!\, \verb!B!\, \verb!D!\, \verb!C![/tex] would then be considered different from one another.

There would [tex]30[/tex] choices for the first card. Since the first card was not returned to the pile, there would be only [tex]29[/tex] choices for the second card. Likewise, there would be [tex]28[/tex] choices for the third card and [tex]27[/tex] for the fourth.

By this reasoning, there would be [tex]30 \times 29 \times 28 \times 27 = 657,\!720[/tex] different ways to draw a hand of four cards from this set when the ordering of these four cards do matter.

However, in many card games, once a hand of cards is drawn, the ordering of cards within that hand does not matter. In other words, hands like [tex]\verb!A!\, \verb!B!\, \verb!C!\, \verb!D![/tex] and [tex]\verb!A!\, \verb!B!\, \verb!D!\, \verb!C![/tex] would not be considered as distinct from one another.

In that case, the [tex]30 \times 29 \times 28 \times 27 = 657,\!720[/tex] ways of drawing cards would include a large number of duplicates.

There are be [tex]4 \times 3 \times 2 \times 1 = 24[/tex] ways to arrange a hand of four cards when the order matter. Hence, when the ordering within a hand no longer matters, each hand of four cards would have been counted [tex]24[/tex] times among those [tex]30 \times 29 \times 28 \times 27 = 657,\!720[/tex] ways.

Therefore, when the ordering of cards within a set does not matter, [tex]\displaystyle \frac{30 \times 29 \times 28 \times 27}{4 \times 3 \times 2 \times 1} = 27,\!405[/tex] would give the number of distinct ways to draw a hand of four out of this set of thirty distinct cards.

the models aren't given..

Answer: no models given

Step-by-step explanation:

9514 1404 393

Answer:

  125.6 cm²

Step-by-step explanation:

The relevant area formula is ...

  Area = (1/2)ac·sin(B)

  Area = (1/2)(13 cm)(21 cm)·sin(67°) ≈ 125.6 cm²

Step-by-step explanation:

5.25....................

Answer:

I think

The choose B. center.

I hope I helped you^_^

Answer:

A=13×13

=169sqr.units

The equation of the line through (0, 1) and (c, 0) is

y - 0 = (0 - 1)/(c - 0) (x - c)   ==>   y = 1 - x/c

Let L denote the given lamina,

L = {(x, y) : 0 ≤ x ≤ c and 0 ≤ y ≤ 1 - x/c}

Then the center of mass of L is the point [tex](\bar x,\bar y)[/tex] with coordinates given by

[tex]\bar x = \dfrac{M_x}m \text{ and } \bar y = \dfrac{M_y}m[/tex]

where [tex]M_x[/tex] is the first moment of L about the x-axis, [tex]M_y[/tex] is the first moment about the y-axis, and m is the mass of L. We only care about the y-coordinate, of course.

Let ρ be the mass density of L. Then L has a mass of

[tex]\displaystyle m = \iint_L \rho \,\mathrm dA = \rho\int_0^c\int_0^{1-\frac xc}\mathrm dy\,\mathrm dx = \frac{\rho c}2[/tex]

Now we compute the first moment about the y-axis:

[tex]\displaystyle M_y = \iint_L x\rho\,\mathrm dA = \rho \int_0^c\int_0^{1-\frac xc}x\,\mathrm dy\,\mathrm dx = \frac{\rho c^2}6[/tex]

Then

[tex]\bar y = \dfrac{M_y}m = \dfrac{\dfrac{\rho c^2}6}{\dfrac{\rho c}2} = \dfrac c3[/tex]

but this clearly isn't independent of c ...

Maybe the x-coordinate was intended? Because we would have had

[tex]\displaystyle M_x = \iint_L y\rho\,\mathrm dA = \rho \int_0^c\int_0^{1-\frac xc}y\,\mathrm dy\,\mathrm dx = \frac{\rho c}6[/tex]

and we get

[tex]\bar x = \dfrac{M_x}m = \dfrac{\dfrac{\rho c}6}{\dfrac{\rho c}2} = \dfrac13[/tex]

The center of mass for a uniform triangular shape is on its centroid. The y-coordinate of the center of mass of the lamina is 1/3 (independent of c).

If the surface is plane triangle approximately and mass is uniformally distributed, then its center of mass will lie on the centroid of that triangle.

The point of intersection of a triangle's medians is its centroid (the lines joining each vertex with the midpoint of the opposite side).

If the triangle has its vertices as  [tex](x_1, y_1), (x_2, y_2) , \: (x_3, y_3)[/tex], then the coordinates of the centroid of that triangle is given by:

[tex](x,y) = \left( \dfrac{x_1 + x_2 + x_3}{3} + \dfrac{y_1 + y_2 + y_3}{3} \right)[/tex]

For this case, the  triangular lamina has vertices (0, 0), (0, 1) and (c, 0)

Assuming its mass is spread regularly, the coordinates of its center of mass would be:

[tex](x,y) = \left( \dfrac{x_1 + x_2 + x_3}{3} + \dfrac{y_1 + y_2 + y_3}{3} \right)\\\\(x,y) = \left( \dfrac{0+0+c}{3} + \dfrac{0+1+0}{3} \right) = (c/3, 1/3)[/tex]

Thus, the y-coordinate of the center of mass of the lamina is 1/3 (independent of c).

Learn more about centroid here:

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Answer:

wdym R6

Step-by-step explanation:

Answer:

12

Step-by-step explanation:

6*2

Answer:

0.1088 or 10.88%

Step-by-step explanation:

q = 1 - 0.35 = 0.65

P(X=2) = 12C2 × (0.35)² × (0.65)¹⁰

= 0.1088

For drawing the graph for y = 3/4x + 7, a person can draw a point at x of 0 and y of __7_, a second point by going over 3 and up __8.25__, and then draw a line through the points.

All the points (and only those points) which lie on the graph of the function satisfy its equation.

Thus, if a point lies on the graph of a function, then it must also satisfy the function.

For this case, the equation given to us is:

[tex]y = \dfrac{3}{4}x + 7[/tex]

Any equation of the form [tex]y = mx + c[/tex] where m and c are constants and x and y are variables is the equation of a straight line.

For a straight line to be characterized, only two points are sufficient.

For x = 0, the y-coordinate would be such that it would satisfy the equation [tex]y = \dfrac{3}{4}x + 7[/tex]

Putting x = 0, we get:

[tex]y = \dfrac{3}{4} \times 0 + 7 = 7[/tex]

Thus, y-coordinate of the point on this line whose x-coordinate is 0 is 7. Thus, (0,7) is one of the point's coordinate on the considered line.

Putting x = 3, we get:

[tex]y = \dfrac{3}{4} \times 3 + 7 = 9.25[/tex]

Thus, y-coordinate of the point on this line whose x-coordinate is 0 is 9.25 . Thus, (0,9.25) is another of the point's coordinate on the considered line.

Thus, for drawing the graph for y = 3/4x + 7, a person can draw a point at x of 0 and y of __7_, a second point by going over 3 and up __8.25__, and then draw a line through the points.

Learn more about points lying on graph of a function here:

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Answer:

8:15

Step-by-step explanation: