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2 years ago

Which of the following items has the lowest unit price? 4 for $5.00 $1.22 each 6 for $7.44 3 for $3.60

Mathematics

1 answer:

aev [14]2 years ago

3 0

3 for 3.60 the unit price come out to 1.20

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2 years ago

Can anybody do all of number 1??

MAXImum [283]

Answer:

You use the formulas for the shapes

Cylinder= 2×π×r×h+2×π×r² π×r²×h

Cone= $\pi*r(r+\sqrt{ h^{2} +r^{2}} )$ π×r²× $\frac{h}{3}$

Square Prism= 2×a²+4×a×h a²×h

↑a=base edge

Step-by-step explanation:

Surface Area Volume

Cylinderical Head ≈282.74 ≈314.16

Conical Hat ≈170.13 ≈78.54

Right Cylinder Torso ≈471.24 ≈785.4

Right Cylinder Arm ≈56.55 ≈25.13

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3 years ago

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3 years ago

Read 2 more answers

) f) 1 + cot²a = cosec²a

notsponge [240]

Answer:

It is an identity, proved below.

Step-by-step explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).

$\displaystyle \large{\cot x=\frac{1}{\tan x}}\\\displaystyle \large{\csc x=\frac{1}{\sin x}}$

csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.

$\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}$

Another identity is:

$\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}$

Therefore:

$\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}$

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

$\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}$

Another identity:

$\displaystyle \large{\sin^2x+\cos^2x=1}$

Therefore:

$\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}\longrightarrow \boxed{ \frac{1}{\sin^2x}={\frac{1}{\sin^2x}}}$

Hence proved, this is proof by using identity helping to find the specific identity.

6 0

2 years ago