How many square feet of tile is needed for a room of 24 foot x 24 foot

Answer: 2.7545

The question was a little off, but I can understand enough

Step-by-step explanation:

Grade       Grade Point(x)       Course Credit(y)        Product(xy)

A                     4                            3                                12

B                      3                           4                                12

B                     3                            4                                12

B                      3                          4                                 12

C                     2                           2                                 4

D                      1                           3                                 3

Total                                              20                             55

This means, the GPA of the student is :

[tex]GPA=\frac{55}{20}[/tex]

[tex]2.75[/tex]

he value of x and y from the given figure are 49/5 and 49/15

From the given similar triangles, the following expression is true

21/30 = 7/15 = k

Also, x/21 = y/7 = 7/15

Equate

x/21 = 7/15

15x = 7 * 21

5x = 7 * 7

x = 49/5

Similarly

y/7 = 7/15

15y = 49

y = 49/15

Hence the value of x and y from the given figure are 49/5 and 49/15

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The integral of [tex]f(x)[/tex] over the interval [tex][-a,a][/tex] corresponds to the area of a semicircle of radius [tex]a[/tex], so

[tex]\displaystyle \int_{-a}^a f(x) \, dx = \frac{\pi a^2}2[/tex]

The integral over [tex][a,3a][/tex] corresponds to the negative area of a trapezoid with height [tex]a[/tex] and base lengths [tex]2a[/tex] and [tex]a[/tex], so

[tex]\displaystyle \int_a^{3a} f(x) \, dx = -\frac{a(2a+a)}2 = -\frac{3a^2}2[/tex]

Then combining the integrals, we have

[tex]\displaystyle \int_{-a}^{3a} f(x) \, dx = \frac{\pi a^2}2 + \left(-\frac{3a^2}2\right) = \frac{(\pi-3)a^2}2[/tex]

When [tex]a=9[/tex], the integral evaluates to

[tex]\displaystyle \int_{-9}^{27} f(x) \, dx = \frac{(\pi-3)\times81}2 \approx \boxed{5.7345}[/tex][/tex]

The zeroes of the quadratic function f(x) = 6x² + 12x – 7 will be given below. Then the correct option is A.

The quadratic function is given as ax² + bx + c = f(x).

For finding the zeroes of the polynomial we have to find the roots of the polynomial , The value of f(x) = 0

The given function is f(x) = 6x² + 12x – 7

Then the zeroes of the quadratic function will be

[tex]roots = \dfrac {-b \pm \sqrt{b^2 -4ac}}{2a}\\\\[/tex]

Here a =6 , b = 12 ,c = -7

Therefore the zeroes are

[tex]\rm roots = \dfrac {-12 \pm \sqrt{12^2 -4 * 6 * (-7)}}{2 *6}\\\\[/tex]

[tex]\rm root_1 = -1 + \sqrt{\dfrac {13}{6}}\\\\\rm root_2 = -1 - \sqrt{\dfrac {13}{6}}[/tex]

Therefore the correct option is A.

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The graphed trigonometric function is the one in option a.

The key parts that we can see on the graph are:

From this we know that the correct option is either a or b.

Now we can evaluate options a and b in x = 0 and see which one is equal to zero.

for a:

y = sin(0 - pi/2) + 1 = sin(-pi/2) + 1 = -1 + 1 = 0

for b:

y = cos(0 + pi/2) + 1 = 0 + 1 =  1

Then we can see that option a is the correct option.

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Answer:

19 small frames and 3 large frames

Step-by-step explanation:

follows the steps in the attached

Answer:

J Trump has to be a great day of school and I have a great

Step-by-step explanation:

you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you for your loss of school and I have a great day of school ii

Answer:

y≤x²+4x-1.

Step-by-step explanation:

1) to define the equation of the given graph; it is y=x²+4x-1 (its vertex is (-2;-5));

2) to define inequation accorging to the given graph (the area outside of the parabola means ≤).

Answer:

vector VW= w - v

vector WV= v - w

Step-by-step explanation:

there are no figures to solve the question so i'll just write the formula..

vector VW= w - v

vector WV= v - w

The amount of the  interest revenue that  Princess will record in 2022 on this lease is:$413,553.82.

First step

Lease value payments after first payment:

Lease value payments=  $5,588,516- $716,000

Lease value payments= $4,872,516

Second step

Lease value payments after second payment:

Lease value payments=$4,872,516+ ($4,872,516×9%) - $716,000

Lease value payments=$4,872,516+$438,526.44-$716,000

Lease value payments=$4,595,042.44

Third step

2022 Interest revenue=$4,643,787.6×9%

2022 Interest revenue=$413,553.8196

2022 Interest revenue=$413,553.82 (Approximately)

Therefore the amount of the  interest revenue that  Princess will record in 2022 on this lease is:$413,553.82.

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Answer:

Step-by-step explanation:

Jakub wants to build a new shed.

The area of the shed is 37 m².

The width of the shed is 5 m.

What is the length of the shed?

to find the side, having the area, you need to use the inverse formula Area = W * L

so

L = A: W

L = 37 : 5 = 7.4 m

If the area of the shed is 37 m²,width of the shed is 5 m as a result the length of the shed will be 7.4 m.

It is defined as two-dimensional geometry in which the angle between the adjacent sides is 90 degrees. It is a type of quadrilateral. The area occupied by the rectangle in two-dimensional planner geometry.

The area of a rectangle can be calculated using the following formula:

Rectangle area = length x width

It is given that, the area of the shed is 37 m². The width of the shed is 5 m.

Since the shed is rectangular. The length of width is found as,

Area = length × width

A = l × w

l=A/w

l=37/5

l=7.4 m

Thus, if the area of the shed is 37 m², the width of the shed is 5 m as a result the length of the shed will be 7.4 m.

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The sector area and the arc length are 34.92 square inches and 13.97 inches, respectively

For a sector that has a central angle of θ, and a radius of r;

The sector area, and the arc length are:

[tex]A = \frac{\theta}{360} * \pi r^2[/tex] --- sector area

[tex]L = \frac{\theta}{360} * 2\pi r[/tex] ---- arc length

Here, the given parameters are:

Central angle, θ = 160

Radius, r = 5 inches

The sector area is

[tex]A = \frac{\theta}{360} * \pi r^2[/tex]

So, we have:

[tex]A = \frac{160}{360} * \frac{22}{7} * 5^2[/tex]

Evaluate

A = 34.92

The arc length is:

[tex]L = \frac{\theta}{360} * 2\pi r[/tex]

So, we have:

[tex]L = \frac{160}{360} * 2 * \frac{22}{7} * 5[/tex]

L = 13.97

Hence, the sector area and the arc length are 34.92 square inches and 13.97 inches, respectively

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The school bus will take 5.34 seconds to travel 95.1 m if the bus is moving 26.8 m/s on flat ground when it begins to accelerate at 4.73 m/s².

Any equation of the form [tex]\rm ax^2+bx+c=0[/tex] where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.

As we know, the formula for the roots of the quadratic equation is given by:

[tex]\rm x = \dfrac{-b \pm\sqrt{b^2-4ac}}{2a}[/tex]

We have:

A school bus is moving 26.8 m/s on flat ground when it begins to accelerate at 4.73 m/s².

From the second equation of motion:

[tex]\rm s = u + \dfrac{1}{2}at^2[/tex]

We have given:

s = 95.1 m

u = 26.8 m/s

a = 4.73 m/s²

[tex]\rm 95.1 = 26.8 + \dfrac{1}{2}(4.73)t^2[/tex]

The above equation is a quadratic equation:

After simplification:

[tex]\rm \dfrac{4.73}{2}t^2=68.3[/tex]

[tex]\rm t^2=28.879[/tex]

t = 5.373 seconds ≈ 5.34 seconds

Thus, the school bus will take 5.34 seconds to travel 95.1 m if the bus is moving 26.8 m/s on flat ground when it begins to accelerate at 4.73 m/s².

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[tex]z_{2}-3i=(-4+i)-3i=-4-2i\\\\z_{1}+2=(3-4i)+2=5-4i\\\\(z_{1}+2)-(z_{2}-3i)=(5-4i)-(-4-2i)=9-4i+4+2i=9-2i[/tex]

which corresponds with point C

The expression which is the result of subtracting (z2-3i) from (z1+2) is 9-2i.

Given On a coordinate plane, the y-axis is labeled imaginary and the x-axis is labeled real. Point A is (1, -2), B is (1, -6), C is (9,- 2), D is (9, - 6), z 1 is (3, - 4), and z 2 is (- 4, 1).

Coordinate plane is a two dimensional plane formed by the intersection of two number lines. One is horizontal line and one is vertical. On horizontal line x values are denoted and on vertical line y values are written. There can be three dimensional plane also.

z2-3i=(-4+i)-3i=-4-2i

z1+2=(3-4i)+2=5-4i

(z1+2)-(z2-3i)=(5-4i)-(-4-2i)

=9-4i+4+2i

=9-2i

Hence for the expression (z1+2)-(z2-3i)=9-2i

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Answer:

See proof below

Step-by-step explanation:

Important points

Onto

For a function with a given co-domain to be "onto," every element of the co-domain must be an element of the range.

However, the co-domain here is suggested to be [tex]\mathbb R[/tex], whereas the range of f is not [tex]\mathbb R[/tex] (proof below).

Proof (contradiction)

Suppose that f is onto [tex]\mathbb R[/tex].

Consider the output 7 (a specific element of [tex]\mathbb R[/tex]).

Since f is onto [tex]\mathbb R[/tex], there must exist some input from the domain [tex]\mathbb R[/tex], "p", such that f(p) = 7.

Substitute and solve to find values for "p".

[tex]f(x)=-3x^2+4\\f(p)=-3(p)^2+4\\7=-3p^2+4\\3=-3p^2\\-1=p^2[/tex]

Next, apply the square root property:

[tex]\pm \sqrt{-1} =\sqrt{p^2}[/tex]

By definition, [tex]\sqrt{-1} =i[/tex], so

[tex]i=p \text{ or } -i =p[/tex]

By the Fundamental Theorem of Algebra, any polynomial of degree n with complex coefficients, has exactly n complex roots.  Since the degree of f is 2, there are exactly 2 roots, and we've found them both, so we've found all of them.

However, neither [tex]i[/tex]  nor [tex]-i[/tex] are in [tex]\mathbb R[/tex], so there are zero values of p in [tex]\mathbb R[/tex] for which f(p)= 7, which is a contradiction.

Therefore, the contradiction supposition must be false, proving that f is not onto [tex]\mathbb R[/tex]

Answer:

1/7

Step-by-step explanation:

Alternative Form: 0.142857, 7^-1

Answer:

7

4

Step-by-step explanation:

The actual values are shown on the given graph as blue points.

The line of regression is shown on the given graph as the red line.

From inspection of the graph, in the year 2000 the actual rainfall was 43 cm, shown by point (2000, 43).  It appears that the regression line is at y = 50 when x is the year 2000.

⇒ Difference = 50 - 43 = 7 cm

In 2000, the actual rainfall was 7 centimeters below what the model predicts.

From inspection of the graph, in the year 2003 the actual rainfall was 44 cm,  shown by point (2003, 40).  It appears that the regression line is at y = 40 when x is the year 2003.

⇒ Difference = 44 - 40 = 4 cm

In 2003, the actual rainfall was 4 centimeters above what the model predicts.

Answer:

D.  [tex](4x^2 + 3)(x + 2)[/tex]

Step-by-step explanation:

Hello!

We can group the first two terms and the last two terms and factor each group.

Now we can combine the like factors:

The answer is Option D.

Answer:

Step-by-step explanation:

The domain is the set of x-values and the range is the set of y-values.

Answer:

c6

Step-by-step explanation:

justjustcancacanjustadda0totheproduct of5cxwewtogetthesameresultasx1totoandtimes

3) Number: 640,700

   the value of 4 in this number is 40,000

Number: 64,070

the value of 4 in this number is 4,000

So, times greater: 40,000 ÷ 4,000 = 10 times greater.

4)

The multiplication, 5 × 12 = 60. So, 5 × 120 = 600.

There is an extra zero at the very last of the result as '120' was multiplied with 5 instead of '12' which does not have zero at last which changes the result.

Similar cases with: 5 × 10 = 50. So, 5 × 100 = 500.

Answer:

the answer is c. the row is chili, the column is no cheese

If f(x)=x²+2 and g(x)=x²+5 then the domain of  the (fog) (x) is (-∞,∞).

Given the f(x)=x²+2 and g(x)=x²+5

We have to find the domain of  the (fog) (x)

The domain of a function is the set of all possible inputs for the function. For example, the domain of f(x)=x² is all real numbers, and the domain of g(x)=1/x is all real numbers except for x=0.

then,(fog) (x)= (x²+5)²+ 2

                    =x⁴+25+10x²+2

                    = x⁴+27+10x²

Now on any real value of x the value of (fog) (x) exists therefore (fog) (x) can take all the values of the real number.

So domain (fog) (x)= (-∞,∞).

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thomas age:44

huilan age: 29

Step-by-step explanation:

h = t + 15

t + t + 15 = 73

2t = 73 - 15

2t = 58

t = 29

The graph of the function is shown in the attached image.

Answer:

The balance after 37 years is $3869.99

Step-by-step explanation:

Given:

The Amount (A),

A = P (1 + [tex]\frac{R}{100n}[/tex][tex])^{nt}[/tex]        if compound n times per year

Amount (A) at 37 years,

A = ($2,520) (1 + [tex]\frac{1.16}{100(12)}[/tex][tex])^{12 x 37}[/tex]

A = ($2,520) (1 + [tex]\frac{1.16}{1,200}[/tex][tex])^{444}[/tex]

A = ($2,520) (1.00096667 [tex])^{444}[/tex]

A = ($2,520) (1.5357)

A = $3869.9944

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The stock worth now, to the nearest cent, is $58.14.

The current worth of stock = initial price of stock+increase in the price of stocks

If you bought stock last year for a price of $66, and it has gone down 13.9% since then,

Let X be the worth of stock;

66 ×  100 = (13.5% + 100%) × X

6600 = 113.5X

X = 6600/113.5

X = 58.14458

X = $58.14

Hence, The stock worth now, to the nearest cent is $58.14.

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