3 x 5 + 9(2+ 13)

Arrange these lunch buyers from greatest to least assuming they buy lunch 5 days per week: Alice spends $3 per day, Betty spends

ValentinkaMS [17]

Alice spends $3 per day, so for in a week and upto 5 weeks, Alice spend =$3*5*5 = $75

Betty spends $ 25 every 2 weeks, so for 5 weeks , she spends= 25+25+12.5

= $62.5

Cindy spends $ 75 per month, and on an average there are less then 5 weeks in a month . SO for 5 week, it must be greater than $75 .

So the order is

Cindy>Alice>Betty

3 0

2 years ago

HhhhhhhhhhhhhhhhhhhhhhheeeeeeeeeeeeeeeelpppPP!!!!!!!!<br><br> pls.

soldi70 [24.7K]

N = 2
M + 5

Hope that helped :)

3 0

2 years ago

If<br> R = (5+T)P/3<br> then what is p?

Illusion [34]

Answer:

Step-by-step explanation:

R = (5+T)P/3

3R = (5+T)P

3R/ (5+T) = P

$P = \frac{3R}{(5+T) }$

4 0

3 years ago

Solve for the brainlist

Sergio039 [100]

Answer:

x = 10.0

Step-by-step explanation:

Using the geometric mean formula:

x projection

----------------- = ---------------

entire base x

Here, the projection is 5 and the entire base is (20 + 5). So:

x 5

------------ = ------------

25 x

Cross multiply:

x² = 25 (5)

x² = 100

√x² = √100

x = 10

3 0

2 years ago

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side

lorasvet [3.4K]

Answer:

Step-by-step explanation:

1.

cot x sec⁴ x = cot x+2 tan x +tan³x

L.H.S = cot x sec⁴x

=cot x (sec²x)²

=cot x (1+tan²x)² [ ∵ sec²x=1+tan²x]

= cot x(1+ 2 tan²x +tan⁴x)

=cot x+ 2 cot x tan²x+cot x tan⁴x

=cot x +2 tan x + tan³x [ ∵cot x tan x $=\frac{ \textrm{tan x }}{\textrm{tan x}}$ =1]

=R.H.S

2.

(sin x)(tan x cos x - cot x cos x)=1-2 cos²x

L.H.S =(sin x)(tan x cos x - cot x cos x)

= sin x tan x cos x - sin x cot x cos x

$=\textrm{sin x cos x }\times\frac{\textrm{sin x}}{\textrm{cos x} } - \textrm{sinx}\times\frac{\textrm{cos x}}{\textrm{sin x}}\times \textrm{cos x}$

= sin²x -cos²x

=1-cos²x-cos²x

=1-2 cos²x

=R.H.S

3.

1+ sec²x sin²x =sec²x

L.H.S =1+ sec²x sin²x

= $1+\frac{{sin^2x}}{cos^2x}$ [ $\textrm{sec x}=\frac{1}{\textrm{cos x}}$ ]

=1+tan²x $[\frac{\textrm{sin x}}{\textrm{cos x}} = \textrm{tan x}]$

=sec²x

=R.H.S

4.

$\frac{\textrm{sinx}}{\textrm{1-cos x}} +\frac{\textrm{sinx}}{\textrm{1+cos x}} = \textrm{2 csc x}$

L.H.S= $\frac{\textrm{sinx}}{\textrm{1-cos x}} +\frac{\textrm{sinx}}{\textrm{1+cos x}}$

$=\frac{\textrm{sinx(1+cos x)+{\textrm{sinx(1-cos x)}}}}{\textrm{(1-cos x)\textrm{(1+cos x})}}$

$=\frac{\textrm{sinx+sin xcos x+{\textrm{sinx-sin xcos x}}}}{{(1-cos ^2x)}}$

$=\frac{\textrm{2sin x}}{sin^2 x}$

= 2 csc x

= R.H.S

5.

-tan²x + sec²x=1

L.H.S=-tan²x + sec²x

= sec²x-tan²x

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

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

$=\frac{cos^2x}{cos^2x}$

=1

8 0

3 years ago