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2 years ago

HELPPPPPPPPP i have the most trouble with math what’s the answer??

Mathematics

1 answer:

kiruha [24]2 years ago

4 0

Answer:

option B

Step-by-step explanation:

Given in the question an expression

$\sqrt\frac{48}{x}$

Whole numbers are positive numbers, including zero, without any decimal or fractional parts.

Possible range of domain will be 1 ≤ x ≤ 48

We know that perfect square between 1 and 48 are

1 , 4 , 16 , 25 , 36

x = 3

√48/3 = √16 = 4

x = 12

√48/12 = √4 = 2

x = 48

√48/48 = √1 = 1

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Find the area of the shaded shape pls quickly

professor190 [17]

${\qquad\qquad\huge\underline{{\sf Answer}}}$

Let's solve ~

Area of shaded region is equal to :

Area of rectangle - Area of two triangles ~

1. Area of rectangle :

$\qquad \sf \dashrightarrow \: length \times width$

$\qquad \sf \dashrightarrow \: 20 \times 15$

$\qquad \sf \dashrightarrow \: 300 \: \: cm {}^{2}$

2. Area of first triangle :

$\qquad \sf \dashrightarrow \: \cfrac{1}{2} \times (base) \times (height)$

$\qquad \sf \dashrightarrow \: \cfrac{1}{2} \times (20 - 18) \times (15)$

$\qquad \sf \dashrightarrow \: \cfrac{1}{2} \times (2) \times (15)$

$\qquad \sf \dashrightarrow \: 15 \: \: cm {}^{2}$

3. Area of second triangle :

$\qquad \sf \dashrightarrow \: \cfrac{1}{2} \times (base) \times (height)$

$\qquad \sf \dashrightarrow \: \cfrac{1}{2} \times (6) \times(20)$

$\qquad \sf \dashrightarrow \: 3 \times 20$

$\qquad \sf \dashrightarrow \: 60 \: \: cm {}^{2}$

Area of shaded region is :

$\qquad \sf \dashrightarrow \: 300 - (15 + 60)$

$\qquad \sf \dashrightarrow \: 300 - 75$

$\qquad \sf \dashrightarrow \: 225 \: \: cm {}^{2}$

3 0

9 months ago

Read 2 more answers

? What is the asthenosphere? Select all that apply.

Basile [38]

Answer:part of the earths crust and part of the earths core

Step-by-step explanation:

6 0

1 year ago

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A cylinder and a cone have the same volume. The cylinder has radius x and height y. The cone has radius 2x. Find the height of t

n200080 [17]

$\bf \textit{volume of a cylinder}\\\\
V_c=\pi r^2h\qquad 
\begin{cases}
r=radius\\
h=height\\
------\\
r=x\\
h=y
\end{cases}\implies V_c=x^2y\pi 
\\\\\\
\textit{volume of a cone}\\\\
V_n=\cfrac{\pi r^2h}{3}\qquad r=2x\implies V_n=\cfrac{\pi \cdot (2x)^2h}{3}
\\\\\\
V_n=\cfrac{\pi \cdot 2^2x^2h}{3}\implies V_n=\cfrac{4x^2h\pi }{3}\\\\
-------------------------------\\\\$

$\bf V_c=V_n\implies x^2y\pi =\cfrac{4x^2h\pi }{3}\implies 3x^2y\pi =4x^2h\pi 
\\\\\\
\cfrac{3x^2y\pi }{4x^2\pi }=h\implies \boxed{\cfrac{3y}{4}=h}$

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2 years ago

What numbers would be in the in the 4 sections of a 2 x 2 array that shows the problem 18 x 17?

kenny6666 [7]

In the 2 x 2 array, the sum of all squares is 18 x 17 = 306.

The number in the 4 sections are 100, 80, 70, and 56.

A 2 x 2 array that shows the problem 18 x 17.

2 x 2 array has four blockers or magic quadrants, the matrix diagram is a simple square divided into four equal quadrants.

18 = 10 +8

17 = 10 +7

18 x 17 = 306

In the 2 x 2 array : 100+80+70+56 = 18 x 17 = 306

Therefore, In the 2 x 2 array, the sum of all squares is 18 x 17 = 306.

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KonstantinChe [14]

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Step-by-step explanation:

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