All categories

3 years ago

What is the difference of 9 5/9 - 6 5/6

1 answer:

6 0

Hi there, first we change the mixed numbers into improper fractions, 9 5/9=86/9 and 6 5/6=41/6. Second, we solve the problem now. 86*6-41*9÷9*6, 516-369/54=147/54. Third, we simplify 147/54 into 49/18. Fourth, we make 147/54 into a mixed number. 147/54 as a mixed number is 2 13/18

You might be interested in

if $6700 is invested at 4.6% interest compounded semiannually, how much will the investment be worth in 15 years???

ella [17]

$6700(1+ \frac{4.6}{100} )^{15}$

6 0

3 years ago

Read 2 more answers

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:

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

(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

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

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

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

-tan²x + sec²x=1

L.H.S=-tan²x + sec²x

= sec²x-tan²x

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

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

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

8 0

3 years ago

2. The width of a rectangle is 5 units less than the length. If the area is 150 square units, then find the dimensions of the re

xz_007 [3.2K]

Answer:

The length = x = 15 units

The width = x-5 = 10 units

Step-by-step explanation:

Let 'x' be the length of a rectangle

As the width of a rectangle is 5 units less than the length.

width of rectangle will be = w = x - 5 units

Area = 150 square units

Using the formula

Area = length × width

$150=\left(x\right)\left(x-5\right)$

$150=x^2-5x$

$x^2-5x=150$

Subtract 50 from both sides

$x^2-5x-150=150-150$

$x^2-5x-150=0$

$\left(x+10\right)\left(x-15\right)=0$

$ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

$x+10=0\quad \mathrm{or}\quad \:x-15=0$

$x=-10,\:x=15$

As length can not be negative. so

x = 15 units

Hence,

The length = x = 15 units

The width = x-5 = 15-5 = 10 units

5 0

2 years ago

SOMEONE HELP ASAPPPP PLEASEEE,PLEASE EXPLAIN HOW U GOT YOUR ANSWER, I NEED AN EXPLANATION IN ORDER TO COMPLETE THIS! NO LINKS OR

Sedbober [7]

1. 7,3
2.4,6
3. 6,8
4. 2,3

8 0

3 years ago

Read 2 more answers

Is it a function or not a function

BigorU [14]

Answer:

This is NOT a function. Because if you used the vertical line test it would hit that line 2 times.

Step-by-step explanation:

6 0

3 years ago