An airplane is flying at a speed of 300 mi/h at an altitude of one mile. The plane passes directly above a radar station at time
t = 0.

Answer:

19 tables

Step-by-step explanation:

Let X represent the number of tables that have 1 student and 3 seniors

Let Y represent the number of tables that have 2 student2 and 4 seniors

Therefore number students total

1X + 2Y =23   (1)

Number of seniors total

3X + 4Y = 61   (2)

Multiply equation (1) by 2 giving

2X + 4Y = 46  (3)

Subtract (3) from (2)

3X + 4Y - (2X+4y) = 61-46

3X + 4Y - 2X -4Y  = 15

3X-2X = 15

X = 15

Plugging this value into equation (1) gives

1(15) + 2Y = 23

15 + 2Y = 23

Subtract 15 from both sides

2Y = 23 - 15

2Y = 8

Divide by 2 both sides

Y = 4

Thus total number of tables used = X +Y = 15+4 = 19

Answer:

17.7 m

Step-by-step explanation:

57 m / 30 m/s = time in the air = 1.9 s

vertical drop for 1.9 s

 vertical d = 1/2 a t^2 = 17. 7 m

The logarithmic expression is equal to:

Log₂(11)*log₆(2) = 1.34

Here we have the logarithmic expression:

Log₂(11)*log₆(2)

Now, remember that we can rewrite:

Logₐ(b) = Ln(b)/Ln(a)

Using that we can rewrite the expression:

Log₂(11)*log₆(2) = (Ln(11)/Ln(2))*(Ln(2)/Ln(6)) = Ln(11)/Ln(6)

Now using a calculator you can find these two natural logarithms:

Ln(11)/Ln(6) = 1.34

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The line has a slope of 9 and y-intercept of -9

A linear equation is a equation that has a linear degree of 1.

Given the inequality expression 9x-y<9

Write in slope-intercept form

-y < -9x + 9

y > 9x - 9

From the equation, we can see that the line has a slope of 9 and y-intercept of -9

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The solution by substitution of values to 4 ) is ( - 8 ) / 14 and 6 ) is - 213 .

4 ) Using substituition

By replacing values of c and d in given equation

( 5 [tex]c^{2}[/tex] - [tex]d^{2}[/tex] + 3 ) / ( 2 c - 4 d )

c = - 1 and d = - 4

( 5 [tex]( - 1 )^{2}[/tex] - [tex]( - 4 )^{2}[/tex] + 3 ) / ( 2 ( - 1 ) - 4 ( - 4 ) )

= ( 5 - 16 + 3 )  / ( - 2 + 16 )

= ( - 8 ) / 14

6 ) Using substituition

By replacing values of x and y in given equation

3 [tex]| x + y |^{2}[/tex] - [tex]( x y )^{2}[/tex]

x = 3 and y = - 5

3 [tex]| 3 + ( - 5 ) |^{2}[/tex] - [tex]( 3 * ( - 5 ) )^{2}[/tex]

( 3 * 4 ) - 225

= - 213

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The quadrilateral A'B'C'D' is the result of reflecting and translating the quadrilateral ABCD.

In this problem we find a representation of a quadrilateral and its image on a Cartesian plane. That image is the result of the following rigid transformations, that is, transformations applied on geometric loci such that Euclidean distance is conserved at every point of the locus.  

After a careful inspection, we find that the following two rigid transformations were done:

Now let is prove this steps:

Given: A(x, y) = (- 5, - 7), B(x, y) = (- 5, - 2), C(x, y) = (- 3, - 4), D(x, y) = (- 3, - 7)

Reflection:

A''(x, y) = (- 5, - 7) + (0, 14)

A''(x, y) = (- 5, 7)

B''(x, y) = (- 5, - 2) + (0, 4)

B''(x, y) = (- 5, 2)

C''(x, y) = (- 3, - 4) + (0, 8)

C''(x, y) = (- 3, 4)

D''(x, y) = (- 3, - 7) + (0, 14)

D''(x, y) = (- 3, 7)

Translation

A'(x, y) = (- 5, 7) + (7, 0)

A'(x, y) = (2, 7)

B'(x, y) = (- 5, 2) + (7, 0)

B'(x, y) = (2, 2)

C'(x, y) = (- 3, 4) + (7, 0)

C'(x, y) = (4, 4)

D'(x, y) = (- 3, 7) + (7, 0)

D'(x, y) = (4, 7)

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Given that the total length of the piece of steel is 25in, the numerical lengths of the first, second and third piece of the steel are 3in, 6in and 16in respectively.

Given the data in the question, let x represent the length of the first piece.

To determine the length of each piece, we equate the sum of the pieces to the total length of the steel ( 25in ).

x + 2x + ( 5x + 1 ) = 25

3x + 5x + 1 = 25

8x + 1 = 25

8x = 25 - 1

8x = 24

x = 24/8

x = 3

Hence,

Length of first piece = x = 3 inch

Length of second piece = 2x = 2(3) = 6 inch

Length of third piece = 5x + 1 = 5(3) + 1 = 15 + 1 = 16 inch

Given that the total length of the piece of steel is 25in, the numerical lengths of the first, second and third piece of the steel are 3in, 6in and 16in respectively.

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Answer: 59 degrees warmer and -2 is farther away form 0 than 61 degrees.

Step-by-step explanation: 59+2=61 so, that's how I got the answer. The rest is really easy.

The absolute value of 61°F is greater than the absolute value of 2°F. Therefore, the temperature of 61°F in April is the farthest away from 0°F.

To determine how much warmer the location is in April than in December, you can calculate the difference between the two temperatures:

Temperature difference = April temperature - December temperature

Temperature difference = 61°F - (-2°F)

Temperature difference = 61°F + 2°F

Temperature difference = 63°F

So, the location is 63 degrees Fahrenheit warmer in April than in December.

To determine which temperature is the farthest away from 0°F, you can calculate the absolute value of the difference between each temperature and 0°F and then compare them:

Absolute value of (61°F - 0°F) = 61°F

Absolute value of (-2°F - 0°F) = 2°F

The absolute value of 61°F is greater than the absolute value of 2°F. Therefore, the temperature of 61°F in April is the farthest away from 0°F.

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Answer:

43,200

Step-by-step explanation:

4.32 x 10∧4

10 x 10 x 10 x 10

4.32 x 10000

43,200

7y+2=3x-5 is the given equation. Finding the point that does not lie on the specified line, 7y+2=3x-5, is the next step.

To determine whether or not the point meets the provided equation, we will simply enter in the supplied point.

Check for (2,-1/7)

Connect it to 7y+2=3x–5.

7(-1/7)+2=3(2)-5

-1+2=6-5

1=1

Which is correct; as a result, this point is on the line 7y+2=3x-5.

Verify (3,-2/7)

Connect it to 7y+2=3x–5.

7(-2/7)+2=3(3)-5

-2+2=09-5

1=4

Which is False; as a result, the supplied line 7y+2=3x-5 does not include this point.

Connect it to 7y+2=3x–5.

7(-1)+2=3(0)-5

-7+2=0-5

-5=-5

Which is correct; as a result, this point is on the line 7y+2=3x-5.

Verify (1, -4, 7)

Connect it to 7y+2=3x–5.

7(-4/7)+2=3(1)-5

-4+2=3-5

-2=-2

This is why I'm making this point

lies on the line 7y+2=3x-5 that is presented.

Verify (-1, -10/7)

Connect it to 7y+2=3x–5.

7(-10/7)+2=3(-1)-5

-10+2=-3-5

-8=-8

Which is correct; as a result, this point is on the line 7y+2=3x-5.

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The total spent amount on water is $220

The total area is 37.5=75/2

The number of lots is 75/2/3/4=50 nos

Total rent is 50×50=2500

So 3/5 will go to david 2500×3/5=1500

2/5 will go to john 2500×2/5=1000

So 60% Renters choose 100 gallons

Total renters is 50

60% renters is 50×60%=30 nos

40% renters choose 70 gallons

40% of 50 is 20 nos

The tank fills twice

100×30+70×20=4400 to be filled

Per gallon cost is $0.05

So the total cost is 4400×0.5=$220

So they spent $220 on water

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Answer:

321,800 cm. (Rounded)

Step-by-step explanation:

2 × 160,900 = 321,800

There are 160,900 centimeters in one mile, so you must multiply that by 2 and round.

The class width of the histogram is 5000.

It should be noted that the class width is the distance that can be found between the lower limits of consecutive classes.

The range is the difference between the maximum and minimum data entries.

class width of a histogram is given by the equation

Therefore, class width:

= maximum of a class - minimum of a class

In the given histogram, first-class start at 2500 and ends at 7500..Therefore considering first class,.minimum =2500 and maximum=7500.

Therefore, class width will be:

= 7500-2500

= 5000

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The perimeter of the triangle will be 26 + 2c + 20b + 11d.

It should be noted that the perimeter of the triangle will be the addition of all the sides.

This will be:

26 + 5c + 10b + 7d + 4d - 3c

Collect like terms

26 + 5c - 3c + 10b + 11d

= 26 + 2c + 20b + 11d

Therefore, the perimeter of the triangle will be 26 + 2c + 20b + 11d.

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A triangle has side lengths of (26 +5c) centimeters, (10b + 7d) centimeters, and

(4d-3c) centimeters. Which expression represents the perimeter, in centimeters,

of the triangle?

4c + 26d

b + d + 26

26 + 2c + 20b + 11d.

Answer: -13 °C.

Step-by-step explanation:

the domain of the following graph= [-5,10)

Domain :

The domain of a function is the set of its possible inputs, i.e., the set of input values where for which the function is defined. In the function machine metaphor, the domain is the set of objects that the machine will accept as inputs.

For example, when we use the function notation f: R→R, we mean that f is a function from the real numbers to the real numbers. In other words, the domain of f is the set of real numbers R (and its set of possible outputs or codomain is also the set of real numbers R).

To,  identify the domain and range of functions by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the

x-axis. The range is the set of possible output values, which are shown on the

y-axis. Keep in mind that if the graph continues beyond the portion of the graph we can see, the domain and range may be greater than the visible values.

We can observe that the horizontal extent of the graph is -5 to 10, so the domain of f is [-5,10)

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Answer:

numeric

Step-by-step explanation:

Answer:

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

Parallel lines have equal slopes.

Find the slope of the first line, mn:

Find the slope of the second line, ab:

Compare the slopes and find the value of k:

Answer:

[tex]k=\dfrac{3}{2}[/tex]

Step-by-step explanation:

[tex]\boxed{\begin{minipage}{4.3 cm}\underline{Slope formula}\\\\$m=\dfrac{y_2-y_1}{x_2-x_1}$\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ \\ are points on the line.\end{minipage}}[/tex]

Segment MN

[tex]\implies \textsf{slope}=\dfrac{y_N-y_M}{x_N-x_M}=\dfrac{1-0}{5-0}=\dfrac{1}{5}[/tex]

Segment AB

[tex]\implies \textsf{slope}=\dfrac{y_B-y_A}{x_B-x_A}=\dfrac{k-2 \frac{1}{4}}{-2\frac{1}{4}-1\frac{1}{2}}=\dfrac{k-2 \frac{1}{4}}{-2\frac{1}{4}-1\frac{2}{4}}=\dfrac{k-2 \frac{1}{4}}{-3\frac{3}{4}}[/tex]

Parallel lines have the same slope.  

Therefore, equate the found slopes for the two segments and solve for k:

[tex]\implies \dfrac{k-2 \frac{1}{4}}{-3\frac{3}{4}}=\dfrac{1}{5}[/tex]

Cross multiply:

[tex]\implies 5 \left(k-2 \frac{1}{4}\right)=-3\frac{3}{4}[/tex]

[tex]\implies 5k-10 \frac{5}{4}=-3\frac{3}{4}[/tex]

[tex]\implies 5k-11 \frac{1}{4}=-3\frac{3}{4}[/tex]

Add 11 ¹/₄ to both sides:

[tex]\implies 5k-11 \frac{1}{4}+11 \frac{1}{4}=-3\frac{3}{4}+11 \frac{1}{4}[/tex]

[tex]\implies 5k=(-3+11)+\left(-\frac{3}{4}+\frac{1}{4}\right)[/tex]

[tex]\implies 5k=8-\frac{1}{2}[/tex]

[tex]\implies 5k=7\frac{1}{2}[/tex]

Divide both sides by 5:

[tex]\implies 5k \div 5=7\frac{1}{2} \div 5[/tex]

[tex]\implies 5k \div 5=\dfrac{15}{2} \div 5[/tex]

[tex]\implies k=\dfrac{3}{2}[/tex]

Therefore, the value of k for which the two segments are parallel is ³/₂.

IT'S NOT 3???????????????

Answer:no it 11

Step-by-step explanation:

1+1 = 11

I’m kidding btw

The acceleration of the object is found as 14 m/s².

Newton's second law states that acceleration of an object is determined by two variables: the net force acting on it and its mass.

The formula for the Newton's second law is;

F = ma

F = acting force in Newton

m = mass of the body in Kg.

a is the acceleration in m/s².

The mass of 7 kg accelerates with 10m/s².

Force acting on it is;

F = ma

F = 7×10

F = 10 N.

The same force is acting on the body of mass 5 kg.

F = ma

a = F/m

a = 70/5

a = 14 m/s².

Therefore, the acceleration of the body is found as 14 m/s².

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Answer: Sue fills 32/3 (Decimal form = 10.6667 containers)

Step-by-step explanation: 8 divided by 3/4 equals 32/3. If you need it in decimal form it's 10.6667

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Answer:

10 2/3 containers BUT can only fully fill 10 containers

Step-by-step explanation:

to solve this equation you need to divide 8 by 3/4

8 / 3/4

that is

8/1 / 3/4

when dividing fractions you must flip the second number and multiply both numbers together

8/1 * 4/3

32/3 is your answer

or

10 2/3

she can fully fill 10 containers

Answer:

1,4

Step-by-step explanation:

7/5

Esto es una división, solo divide y ya

The left over sand is a 3/10 bag of the colored sand.

According to the statement

We have to find that the how much sand does she use for the project.

So, For this purpose, we know that the

Left over is a something that remains unused or unconsumed

From the given information:

amber has a 4/5 pound bag of colored sand she use 1/2 of the bag for an art project

then it means

Total sand is = 4/5 pound bag of colored sand

Used sand is = 1/2 of the bag of colored sand

Then

Left over sand is = Total sand - Used sand

Left over sand is = 4/5 - 1/2

Left over sand is = 8 - 5/10

Left over sand is = 3/10.

So, The left over sand is a 3/10 bag of the colored sand.

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Answer:

Step-by-step explanation:

Solve for x

7/8 of x is 14

7/8x = 14

x = 14 : 7/8

x = 14 * 8/7

x =16

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

check

7/8 * 16 = 14

14 = 14

the answer is good

Answer:

400 students, 1/8 are left handed

Step-by-step explanation:

Answer:

The two solutions are

5 and 9

-5 and -9

Step-by-step explanation:

Let x and y be the numbers

x-y = 4

x*y = 45

Using substitution

x = 4+y

(4+y)y =45

4y + y^2 = 45

Subtract 45 from each side

y^2 +4y -45 =0

Factor

(y - 5) ( y+9) = 0

Using the zero product property

y -5 =0    y+9 =0

y=5    y = -9

Now we can find x

y=5

x = 4+y

x =4+5

x=9

y = -9

x = 4+y

x = 4+-9

x = -5

The two solutions are

5 and 9

-5 and -9

Answer:

The first set is a set of linear equations.

The way to figure this out is pretty easy. If you want to see it visually, go search up desmos graphing calculator and put in these equations.

A linear equation is a function that has a constant slope, meaning that the rate it increases or decreases will never change. The first one is a set of linear equations because it is 2 equations with constant slopes, meaning that the slopes will never change no matter what y and x are.

The second set is not, because while the first equation is linear, the second is an inequality. While it is a straight line, it doesn't count as a linear equation.

The third set, both equations have exponents on the x, which means that the slope will change depending on x. This means that both of these are not linear equations.

The only set that is a linear set is the one that has only linear equations.