Which of the following is used only in non-Euclidean geometry?
A.
triangle
B.
circle
C.
great circle
D.
line

Answer:

C. interquartile range

Step-by-step explanation:

using this to show im right

"The interquartile range (IQR) is the difference between the upper (Q3) and lower (Q1) quartiles, and describes the middle 50% of values when ordered from lowest to highest."

Hope This Helped

Answer:

Step-by-step explanation:

           [tex]\sf (\sqrt{x+1}) + (\sqrt{x-1})[/tex]

[tex]\sf\dfrac{\sqrt{x+1} +\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}} = \dfrac{(\sqrt{x+1} +\sqrt{x-1})(\sqrt{x+1} +\sqrt{x-1})}{(\sqrt{x+1} -\sqrt{x-1})(\sqrt{x+1} +\sqrt{x-1})}\\[/tex]

           When doing so, in the numerator, we get in the form of the identity (a +b)² and in the denominator a² - b²

                            [tex]\sf =\dfrac{(\sqrt{x+1} +\sqrt{x-1})^2}{(\sqrt{x+1})^2 - (\sqrt{x-1})^2}\\\\=\dfrac{(\sqrt{x+1})^2+(\sqrt{x-1})^2+2*\sqrt{x+1}*\sqrt{x-1}}{x + 1 - (x - 1)}\\\\=\dfrac{x+1 +x -1 + 2*\sqrt{(x+1)(x-1)}}{x + 1 - x + 1}\\\\=\dfrac{x+ x + 1 - 1 +2\sqrt{x^2-y^2}}{x-x+1+1}\\\\=\dfrac{2x+2\sqrt{x^2-y^2}}{2}\\\\= \dfrac{2x}{2}+\dfrac{2\sqrt{x^2-y^2}}{2}\\\\=x + \sqrt{x^2-y^2}[/tex]

The equation of the line parallel to y=-2x-5 and passing through the point (-1,3) is y = -2x + 1.

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

Given that:-

The equation of the new line will be:-

y - y₁ = m ( x - x₁ )

Since the line is parallel to the line y = -2x - 5 so slope will be same m = -2 now the coordinates are ( -1, 3).

y - ( 3) = -2 [( x - (-1) ]

y - 3 = -2x - 2

y = -2x + 3 -2

Y = -2x + 1

Therefore the equation of the line parallel to y=-2x-5 and passing through the point (-1,3) is y = -2x + 1.

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Answer:

Step-by-step explanation:

Formula:

Lateral surface area = perimeter of the base × height

………………………………………………………………

perimeter of the base = 2×(1.5+3) = 9

height = 4.

then

Lateral surface area = 9 × 4 = 36

Area of two bigger rectangles

Area of two smaller rectangles

LSA

To factorize something:

 ⇒ need to simplify the equation to its most basic form

Let's solve:

  [tex]2x^2+6x=2(x^2+3x)=2x(x+3)[/tex]

Thus the factorized form is ⇒ [tex]2x(x+3)[/tex]

Answer: 2x(x+3)

Hope that helps!

Answer:

Solution:

→ 2x(x + 3)

Step-by-step explanation:

Method:

→ Common factor

→ Rewrite in factored form

Answer:

By proportional method

[tex] \frac{4}{6} = \frac{5}{x} [/tex]

4x= 30

x= 7.5

#1

Volume

#2

Radius=10/2=5cm

Volume

#3

Volume

#4

#5

radius=8/2=4

Volume

Answer:

1)  314 cm³

2)  523.33 cm³

3)  87.92 in³

4)  11 cm

5)  502.4 cm³

Step-by-step explanation:

[tex]\textsf{Volume of a cylinder}=\sf \pi r^2 h \quad\textsf{(where r is the radius and h is the height)}[/tex]

Given:

Substitute the given values into the formula:

[tex]\begin{aligned}\implies \textsf{Volume} & =3.14 \cdot 5^2 \cdot 4\\& = 3.14 \cdot 25 \cdot 4\\& = 3.14 \cdot 100\\& = 314 \: \sf cm^3\end{aligned}[/tex]

[tex]\textsf{Volume of a sphere}=\sf \dfrac43 \pi r^3\quad\textsf{(where r is the radius)}[/tex]

Given:

Substitute the given values into the formula:

[tex]\begin{aligned}\implies \textsf{Volume} & =\dfrac{4}{3} \cdot 3.14 \cdot 5^3 \\& =\dfrac{4}{3} \cdot 3.14 \cdot 125 \\& =\dfrac{500}{3} \cdot 3.14 \\& = 523.33\: \sf cm^3\:(2\:dp)\end{aligned}[/tex]

[tex]\textsf{Volume of a cylinder}=\sf \pi r^2 h \quad\textsf{(where r is the radius and h is the height)}[/tex]

Given:

Substitute the given values into the formula:

[tex]\begin{aligned}\implies \textsf{Volume} & =3.14 \cdot 2^2 \cdot 7\\& = 3.14 \cdot 4 \cdot 7\\& = 3.14 \cdot 28\\& = 87.92\: \sf in^3\end{aligned}[/tex]

[tex]\textsf{Volume of a square pyramid}=\sf \dfrac{1}{3} a^2h \quad\textsf{(where a is the base edge and h is the height)}[/tex][tex]\textsf{Area of base of square pyramid}=\sf a^2 \quad\textsf{(where a is the base edge)}[/tex]

Given:

[tex]\implies 81=a^2[/tex]

[tex]\implies a=\sqrt{81}[/tex]

[tex]\implies a=9\: \sf cm[/tex]

Substitute the given values into the formula and solve for h:

[tex]\begin{aligned}\implies \textsf{297} & =\dfrac{1}{3} \cdot 9^2 \cdot h\\\\297 & =\dfrac{81}{3} h\\\\891 & =81 h\\\\h & = 11 \: \sf cm\end{aligned}[/tex]

[tex]\textsf{Volume of a cylinder}=\sf \pi r^2 h \quad\textsf{(where r is the radius and h is the height)}[/tex]

Given:

Substitute the given values into the formula:

[tex]\begin{aligned}\implies \textsf{Volume} & =3.14 \cdot 4^2 \cdot 10\\& = 3.14 \cdot 16 \cdot 10\\& = 3.14 \cdot 160\\& = 502.4\: \sf cm^3\end{aligned}[/tex]

Answer:

She spent 17/30 of her money and still has 13/30 of her money

Step-by-step explanation:

The first step to solve is to find a common denominator. The common denominator of this problem is 30. So 1/6 turns into 5/30, and 2/5 becomes 12/30. Then you add them together to get 17/30. If you subtract 17 from 30 you will get your answers.

Answer:

5.6

Step-by-step explanation:

If you add all the numbers together then divide it by 15 you would get 5.6

Answer:

280°

Step-by-step explanation:

using the relationship

π radians = 180° , then substituting gives

[tex]\frac{14\pi }{9}[/tex] = [tex]\frac{14(180)}{9}[/tex] = 14 × 20 = 280°

Answer:

What is 2 raised to the power of 3?

2 x 2 x 2

= 8

Step-by-step explanation:

Hope this helps.

Answer:

(4, 3)

Step-by-step explanation:

Answer:

2

Step-by-step explanation:

Step-by-step explanation:

Movement from one place to another

Answer:

His bank account now has 352 dollars in it.

Step-by-step explanation:

55+364-67=352

Answer:

$352

Step-by-step explanation:

1st balance: $55

amount deposited: $364

total: $55+$364=$419

amount withdrawed: $67

amount left in account: $419-$67=$352

ans=$352

The volume of the given cone.

[tex]\large\boxed{Formula:V= \frac{1}{3}\pi{r}^{2}h}[/tex]

[tex]\large\boxed{\red \pi \red = \red 3 \red . \red 1 \red 4}[/tex]

Let's solve!

Substitute the values according to the formula.

[tex]V= \frac{1}{3}×3.14×{3}^{2}×8[/tex]

[tex]\large\boxed{V= 75.36 \: {yd}^{3}}[/tex]

Therefore, the volume of the given cone is 75.36 cubic yards.

Answer: aₙ = -11 + 7(n - 1)

Step-by-step explanation: The equation of an arithmetic sequence is

aₙ = a₁ + d(n - 1).

a₁ is the first term.

d is the common difference.

Answer:

m=200

Step-by-step explanation:

percent means 100,so we duvude 300:by 100 then m=200

Answer:

6 + 7/10x + 2/y

Step-by-step explanation:

Hope this helps :)

Answer: some perfect cubes are

8 (2^3)

27 (3^3)

64 (4^3)

there are many perfect cubes

maybe there is more to this question?

Step-by-step explanation:

A perfect cube of a number is a number that is equal to the number, multiplied by itself, three times. If x is a perfect cube of y, then x = y3. Therefore, if we take the cube root of a perfect cube, we get a natural number and not a fraction. Hence, 3√x = y. For example, 8 is a perfect cube because 3√8 = 2.

Answer:

a. D. f⁻¹(x) = x - 21, for all x

b. f(f ⁻¹(x)) = f(x - 21) = x and f ⁻¹(f(x)) = f ⁻¹(x + 21) = x

Step-by-step explanation:

See attached picture.

The required answers are:

a. The inverse of the function is f1(x) = x - 21

b. Verified by showing that f(f(x)) ≠ x and f¯¹(f(x)) = x.

a. To find the inverse function of f(x) = x + 21, we can interchange x and y and solve for y:

x = y + 21

Now, let's solve for y:

y = x - 21

Therefore, the equation for f1(x), the inverse function, is f1(x) = x - 21.

b. To verify that the equation is correct, we need to show that f(f(x)) = x and f¯¹ (f(x)) = x.

First, let's calculate f(f(x)):

f(f(x)) = f(x + 21) = (x + 21) + 21 = x + 42

Since f(f(x)) = x + 42, we can see that f(f(x)) is not equal to x. Therefore, f(f(x)) ≠ x.

Now, let's calculate f¯¹(f(x)):

f¯¹(f(x)) = f¯¹(x + 21) = (x + 21) - 21 = x

Since f¯¹(f(x)) = x, we can see that f¯¹(f(x)) is equal to x.

Thus, we have verified that the equation for the inverse function, f1(x) = x - 21, is correct by showing that f(f(x)) ≠ x and f¯¹(f(x)) = x.

Therefore, the required answers are:

a. The inverse of the function is f1(x) = x - 21

b. Verified by showing that f(f(x)) ≠ x and f¯¹(f(x)) = x.

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Answer:

B

Step-by-step explanation:

look at the graph. y is negative from x>2, so interval is 2<x<infinity.

in other case y=0 for x=2, y<0 for infinity<x<2

Answer:

The intervel that describes where the graph is negative is C :)

I hope this helps you!

Step-by-step explanation:

Answer:

12.6

Step-by-step explanation:

4.2×3=12.6

I hope this answer can help you

Answer:

1.4 kg

Step-by-step explanation:

Given :

⇒ Watermelon (Peter) = 3 × Papaya (Paul)

⇒ Watermelon (Peter) = 4.2 kg

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

Solving :

⇒ 4.2 = 3 × Papaya (Paul)

⇒ Papaya (Paul) = 1.4 kg

Therefore the mass of the papaya is 1.4 kg.

Answer:

A. [tex]y\leq \frac{1}{2}x+2[/tex]

Step-by-step explanation:

First, let's find the slope of the line, so we can eliminate some answers.

From point (0,2) we can go up 1 and over to the right 2 to the point of (2,3). Slope is rise over run so 1 over 2.

This eliminates C and D.

The last part is to recognize that y being less than the equation will result in the shaded region being below the line. If y was greater than the equation then the shaded region would be above the line.

Therefore, the correct answer would be A. [tex]y\leq \frac{1}{2}x+2[/tex].

Hope this helps! If you have questions about my work please let me know down in the comments!

Answer:

[tex]y\leq \frac{1}{2}x+2[/tex]

Step-by-step explanation:

The first part of finding our answer is finding the slope of the line. The image attached shows this.

[tex]\frac{rise}{run} = \boxed{\frac{1}{2}}[/tex]

This means the slope is [tex]\frac{1}{2}[/tex], removing the bottom 2 answer choices.

Now we have to choose between the [tex]\leq[/tex] and [tex]\geq[/tex] sign. Since the shaded area is below the line, it will be the [tex]\leq[/tex]. (If it was the [tex]\geq[/tex] sign, the shaded area would be above the line.) This marks the first answer as the correct answer.

Answer:

The formation of the equation is odd, but I assume it's supposed to be:
4k - 10k = -8k

-6k = -8k
-6k + 8k = -8k + 8k
2k = 0
[tex]\frac{2k}{2}[/tex] = [tex]\frac{0}{2}[/tex]
k = 0