Please helpppppppppp

Jimmy spent 45 minutes to finish his homework. There are 20 problems. What is the average amount of time he spends to finish eac

nexus9112 [7]

The average is (time)/(number of problems)=(45/60)/20=(3/4)/20=3/80 hours per problem. In minutes, the answer would be 3/80*60=9/4 minutes per problem.

6 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

What is the true solution to the equation below?

Naddik [55]

Yeeee

assuming your equaiton is
$2ln(e^{ln(2x)})-ln(e^{ln(10x)})=ln(30)$

remember some nice log rules
$log_a(b)=c$ translates to $a^c=b$
and
$a^{log_a(b)}=b$
and
$xlog_c(b)=log_c(b^x)$
and
$ln(x)=log_e(x)$
and
$log(a)-log(b)=log(\frac{a}{b})$
and
if $log(a)=log(b)$ then a=b

so

we can simplify a bit of stuff here

the $e^{ln(2x)} \space\ and \space\ the \space\ e^{ln(10x)}$ can be simplified to $2x \space\ and \space\ 10x$

so we gots now

$2ln(2x)-ln(10x)=ln(30)$
$ln((2x)^2)-ln(10x)=ln(30)$
$ln(4x^2)-ln(10x)=ln(30)$
$ln(\frac{4x^2}{10x})=ln(30)$
same base so
$\frac{4x^2}{10x}=30$
$\frac{2x}{5}=30$
times both sides by 5
$2x=150$
divide both sides by 2
$x=75$
answer is x=75

5 0

2 years ago

There are 112 seats in the school auditorium there are 7 seats in each row there are 70 people seated filling up full rows of se

irina [24]

6 rows
112/7=16
70/7=10
16-10=6

3 0

3 years ago

What is the perimeter of a rectangle with a length of

solong [7]

Answer:

6

Step-by-step explanation: