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3 years ago

what is the shape of a cross section that is perpendicular to the base of a cube A. rectangle B. triangle C.square D. circle

Mathematics

1 answer:

777dan777 [17]3 years ago

8 0

Answer:

2qwertfghtrew3eret

Step-by-step explanation:

3ewrew

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Given the following exponential function, identify whether the change represents growth or decay, and determine the percentage r

IceJOKER [234]

Answer:

decay

2% decrease

Step-by-step explanation:

The growth factor is the base of the exponent: 0.98. Its relation to the rate of change is ...

growth factor = 1 + rate of change

0.98 = 1 + (-0.02)

So, the rate of change is -0.02 = -2%. Since the rate of change is a 2% decrease, it represents decay.

4 0

2 years ago

Cos^(2)x=(cscxcosx)/(tanx+cotx)

bixtya [17]

$\cos^2 x = \dfrac{\csc x \cos x}{\tan x + \cot x}$

$\cos^2 x = \dfrac{\csc x \cos x}{\dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x}}$

$\cos^2 x = \dfrac{\csc x \cos x}{\frac{\sin x}{\cos x} \times 
\frac{\sin x}{\sin x} + \frac{\cos x}{\sin x} \times \frac{\cos x}{\cos 
x}}$

$\cos^2 x = \dfrac{\csc x \cos x}{\frac{\sin^2 x}{\sin x \cos x} + \frac{\cos^2 x}{\sin x \cos x}}$

$\cos^2 x = \dfrac{\csc x \cos x}{\frac{1}{\sin x \cos x}}$

$\cos^2 x = \csc x \cos x \sin x \cos x$

$\cos^2 x = \dfrac{1}{\sin x} \cos x \sin x \cos x$

$\cos^2 x = \cos^2 x$

6 0

3 years ago

X+2y+1=0,2x-3y-12=0 solve by substitution method

Kitty [74]

Answer:

Step-by-step explanation:

x=-1-2y

-2-4y-3y-12=0

-7y=14

y=-2

x=3

3 0

2 years ago

Carlos is adding the mixed numbers 1 1/5 and 2 1/8 by changing each number into an improper fraction. Which pair of improper fra

irga5000 [103]

Answer

6/5 and 17/8

Step-by-step explanation:

1 1/5 = 6/5

1= 5/5 + 1/5 = 6/5

2 1/8 = 17/8

2 = 16/8 + 1/8 = 17/8

3 0

3 years ago

Hi . I'm kinda stuck on this question . <br>Help please. <br>Workings please

Gre4nikov [31]

Answer:

$\large \boxed{\text{Q31, D. 6.00 F; 32. C. N21.75; Q33. A. 168.00 F}}$

Step-by-step explanation:

Your graph is hard to read, so I re-drew it as best I could.

You must first develop an equation relating francs and naira.

The line goes through the origin, so the general equation is

y = kx

One of the points is (40,240).

$\begin{array}{rcl}F & = &k \times N\\240 & = &k \times 40\\k & = & \dfrac{240}{40}\\\\k&=&\textbf{6 F/N}\\\end{array}\\\text{The equation is $\large \boxed{\mathtbf{F = 6 N}}$}$

Q 31.

$\begin{array}{rcl}F & = &6N\\& = & 6 \times 1.00\\& = & \mathbf{6.00}\end{array}\\\text{The equivalent value of N1.00 is $\large \boxed{\textbf{6.00 F}}$}$

Q 32.

$\begin{array}{rcl}F & = &6N\\130.5& = & 6 N\\N&=& \dfrac{130.5}{6}\\\\& = & \mathbf{21.75}\\\end{array}\\\text{The equivalent value of 130.5 F is $\large \boxed{\textbf{N21.75}}$}$

Q 33.

$\begin{array}{rcl}F & = &6N\\& = & 6 \times 28.00\\& = & \mathbf{168.00}\end{array}\\\text{An amount of N28.00 is $\large \boxed{\textbf{168.00 F}}$}$

6 0

2 years ago