The sun is about 93 x 10 6 miles from earth. What’s is this distance written as a whole number

The property that they would never be affected by is the most frequent monthly rent. Option C is correct.

Given that,
Kate collected a sample of monthly rents and recorded the following numbers: $500, $525, $550, $600, $600, $750, $775, $800, $1350. Kate drops the unusually high rent of $1350 from the data set, which of the following would not be affected is to be determined.

The statistic is the study of mathematics that deals with relations between comprehensive data.

Here,
If the rent of $1350 is dropped by Kate, definitely affect the average, highest monthly rent, and the spread of the monthly sheet, but it won't affect the most frequent monthly of rent because the most frequent rent is $600.

Thus,  the property that they would never be affected by is the most frequent monthly rent. Option C is correct.

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Answer:

2 sorry i don't know that's what answer i got

[tex]\boxed{\sf \displaystyle{\lim_{x\to a}}(x^n)=a^n}[/tex]

[tex]\\ \sf\longmapsto \displaystyle{\lim_{x\to \infty}}(\sqrt{3x}-\sqrt{x-5})[/tex]

[tex]\\ \sf\longmapsto \displaystyle{\lim_{x\to \infty}}\sqrt{3x}-\displaystyle{\lim_{x\to \infty}}\sqrt{x-5}[/tex]

[tex]\\ \sf\longmapsto \infty^{\dfrac{1}{2}}-\sqrt{\infty-5}[/tex]

[tex]\\ \sf\longmapsto \infty-\sqrt{\infty}[/tex]

[tex]\\ \sf\longmapsto\infty-\infty[/tex]

[tex]\\ \sf\longmapsto\infty[/tex]

George needs to fix the expression because if you did the equation as is, it wouldn't subtract the added value from 15.

George always requires $1.48.

The order of operations exists as a law that tells the correct sequence of measures for considering a math expression. We can determine the mandate utilizing PEMDAS: Parentheses, Exponents, Multiplication, and Division (from left to right), Addition, and Subtraction (from left to right).

Given: George tries to purchase a movie ticket ($9.00), a pretzel ($2.65), a drink ($1.35), and two veggie cups ($1.74 each)

9 + 2.65 + 1.35 + 1.74 + 1.74 = 16.48

Subtract 15 to 16.48

15 - 16.48 = 1.48

George always needs $1.48.

Therefore, the proper answer is $1.48.

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The range of equation is

A interval is a set of real numbers that contains all real numbers lying between any two numbers of the set.

Given function :  y ≤ –5

if, y > -5 then,

interval :(-5, ∞)

if y ≥ -5  then

interval :[-5, ∞)

if y < -5   then

interval: (-∞, -5)

if y ≤ -5   then

interval: (-∞, -5]

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Answer:

4d - 6

Step-by-step explanation:

-2-(-15d)+(-11d)+2-6

-2 + 15d - 11d + 2 - 6

15d - 11d - 2 + 2 - 6

4d - 6

Answer:

D. rectangles

Step-by-step explanation:

Perpendicular is when they cross but the base isn't perpendicular

Parallel is when two lines never touch, but in the base of a prism there are lines that touch

Intersecting is when two or more lines meet and cross over each other

2 3/4m for textbooks + 1 1/2m for Notebooks

2 3/4 + 11/2

11/4 + 3/2

11/4 + 6/4

17/4

4 1/4

5 1/2 - 4 1/4

11/2 - 17/4

22/4 - 17/4

5/4

1 1/4

1 1/4 meters of plastic cover will be left

HOPE THIS HELPS. GOODLUCK:)

Answer:

10,340 ft

Step-by-step explanation:

the peak is 10,740 ft and since the climber is at the peak, that's their elevation.

he goes down 400 ft so its 10,470 - 400 which is 10,340

Answer:

B. 10,340

Step-by-step explanation:

The peak is 10,740, since they descend 400, it is subtraction. 10,740 - 400 is 10,340.

Answer:

51 cm²

Step-by-step explanation:

Let's ASSUME that the vertical measurements are taken perpendicular to the roughly horizontal baselines.

The three yellow triangles are all similar

the upper small triangle has height 3 and base 6 cm. The ratio of the lengths is 6/3 = 2

This means the medium (overlapping) triangle has lengths 10 - 3 = 7 and 7(2) = 14 cm

Area of the large triangle is ½(14 + 6)(10) = 100 cm²

Area of the overlapping yellow triangle is ½(7)(14) = 49 cm²

Exposed area of the blue triangle is 100 - 49 = 51 cm²

Both triangles have an area of 100 square centimeters and the area of the blue triangle showing is 51 square centimeters.

Based on the information given from statement, we can determine the missing length of the yellow triangle by means of simmilarity ratios, that is:

[tex]\frac{3\,cm}{6\,cm} = \frac{10\,cm}{x}[/tex] (1)

Where [tex]x[/tex] is the base of the yellow triangle, in centimeters.

Now we proceed to find the length base:

[tex]x = \frac{6\,cm}{3\,cm}\times 10\,cm[/tex]

[tex]x = 20\,cm[/tex]

The area of the triangle can be found by means of the following expression:

[tex]A = \frac{1}{2}\cdot b\cdot h[/tex] (2)

Where:

If we know that [tex]b = 20\,cm[/tex] and [tex]h = 10\,cm[/tex], then the area of the triangle is:

[tex]A = \frac{1}{2}\cdot (20\,cm)\cdot (10\,cm)[/tex]

[tex]A = 100\,cm^{2}[/tex]

The area of the yellow triangle is 100 square centimeters and the area of the blue triangle is also 100 square centimeters due to congruency. The dimensions of the section of the yellow triangle overlapping the blue triangle are [tex]b = 14\,cm[/tex] and [tex]h = 7\,cm[/tex]. The area of this section is:

[tex]A' = \frac{1}{2}\cdot (14\,cm)\cdot (7\,cm)[/tex]

[tex]A' = 49\,cm^{2}[/tex]

And the area of the rest of the blue triangle is:

[tex]A'' = A - A'[/tex] (2)

[tex]A'' = 100\,cm^{2}-49\,cm^{2}[/tex]

[tex]A'' = 51\,cm^{2}[/tex]

In a nutshell, both triangles have an area of 100 square centimeters and the area of the blue triangle showing is 51 square centimeters.

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Answer:

The answer is (5, 2)

Step-by-step explanation:

I. If move left = -x

    move right = +x

  So -2 - 3 =  -5 => move left

        -5 + 10 = 5 => move right

I suggest that move left and right is x cordinate

The answer is (5,2)

Is that correct?

Prime number or u also can write palindromic number

Answer:

15647569

Step-by-step explanation:

1+1

Answer:

1) = -10

3) = 4

Step-by-step explanation:

Chow,...!

[tex]x^2\gt2x-2[/tex]

[tex]x^2-2x+2\gt0[/tex]

You are essentially asking at this point where is the quadratic function greater than 0. To get the interval find the roots. Between the roots the function is guaranteed to be LESS THAN zero. Everywhere else the function is greater than 0.

[tex]x^2-2x+2=0[/tex]

Use Vieta's rule to get the roots,

[tex](x-1)(x-1)=0[/tex]

Both roots are the same so no value is less than 0 which means all values except at the point of root are strictly greater, the point of root is at [tex]x_{1,2}=1[/tex] which means the set for which the quadratic function is greater than 0 is [tex]\mathbb{R}-\{1\}[/tex] "all real numbers except 1".

To convert this into interval notation use all numbers up to 1 but not including and all numbers after 1 but not including,

[tex](-\infty,1)\cup(1,\infty)[/tex]

Hope this helps :)

Answer:

[tex]{ \underline{ \sf{x \: is \: \: \{ \frac{ - 5 + i \sqrt{35} }{10} \} } \: \: and \: \: \{ \frac{ - 5 - i \sqrt{35} }{10} \}} }[/tex]

Step-by-step explanation:

[tex]5 {x}^{2} + 4x + 3 = 0 [/tex]

from quadratic formular:

[tex]{ \sf{x = \frac{ - b± \sqrt{ {b}^{2} - 4ac } }{2a} }} [/tex]

general equation:

[tex] {ax}^{2} + bx + c = 0[/tex]

a is 5, b is 4 and c is 3:

substitute in formular:

[tex]{ \sf{x = \frac{ - 5± \sqrt{ {5}^{2} - (4 \times 5 \times 3) } }{(2 \times 5)} }} \\ \\ { \sf{x = \frac{ - 5± \sqrt{ - 35} }{10} }}[/tex]

but from complexes, i² = -1

[tex]{ \sf{x = \frac{ - 5± \sqrt{35 {i}^{2} } }{10} }} \\ \\ = { \sf{x = \frac{ - 5±i \sqrt{35} }{10} }}[/tex]

Answer:

a. solution

Area = 50 m^2

Area of square = l^2

or, 50 m^2 = l^2

or, √50m^2 = l

or, 5√2 m= l

Hence, The exact length of the playground is 5√2 m.

⇛ -12 + 8

⇛ -4

‌

‌

‌

[tex] \: [/tex]

Hey there!

The word “sum” simply means “to add” or “addition” in mathematical terms

So, we have to ADD -12 from 8 from each other.

Positives are on the RIGHT side of the number line whereas negatives are on the LEFT side of the number line

Start at -12 and go UP 8 spaces to the right.

-12 + 8

= -4

Answer: -4

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

Answer:

can you answer this first

The discussion is shown below:

In computing, data is information that has been translated into a form that is efficient for movement or processing.

A hall way with laser imagine that you must cross the hall way while staying at equal distance from each laser.

So, from the question we can consider the a point o which is in between laser 1 and laser 2.

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Answer:

1/24 chance

Step-by-step explanation:

To begin you would need to find the probability of picking a green sock which would be 3/9; and you could simplify this to 1/3.

After that you need to find the probability of picking a purple sock out of the 8 left, which would provide you with 1/8. To find the probability of both you would multiply 1/8 by 1/3 to find the combined probability which would leave you with 1/24.

Answer:

1/4

Step-by-step explanation:

its 0.25 which is rational. and its between the two.

Answer:

its c

Step-by-step explanation:

9514 1404 393

Answer:

  x = 4

Step-by-step explanation:

The parallel lines cut the transversals proportionally, so we have ...

  (10 -x)/x = 3/2

  2(10 -x) = 3x . . . . . . . multiply by 2x

  20 -2x = 3x . . . . . . . eliminate parentheses

  20 = 5x . . . . . . . . . . add 2x

  4 = x . . . . . . . . . . . . divide by 5

9514 1404 393

Answer:

  81

Step-by-step explanation:

Rewriting the expression as powers of 3, we get ...

  [tex]\dfrac{27^x}{3^y}=\dfrac{(3^3)^x}{3^y}=3^{3x-y}=3^4=\boxed{81}[/tex]

The argument; if he did not get a good wage, then he must not have studied medicine is valid.

Let P represent "He studies medicine"

Let Q represent "He gets a good job"

Let R represent "He gets a good wage"

The argument can be written as:

P → Q

Q → R

¬R

∴ ¬P (conclusion: he did not study medicine)

Assume for contradiction that the argument is invalid:

P → Q

Q → R

¬R

P (assumption for contradiction: he studied medicine)

Use modus ponens to derive Q from (5) and (1):

Q (from 1 and 5 using MP)

Use modus ponens again to derive R from (6) and (2):

R (from 2 and 6 using MP)

This contradicts ¬R in (3), so our assumption in (5) must be false. Therefore, we can conclude that the argument is valid:

¬P (from contradiction of assumption in (5))

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