Can you help me with my math

A drawer is filled with 7 black shirts, 5 white shirts, and 10 gray shirts.

igomit [66]

Answer:

5/11

Step-by-step explanation:

7+5+10=22

fraction of grey t-shirts would be 10/22

simplify it and u get 5/11

6 0

4 years ago

75 1/3 percent as a fraction

katovenus [111]

Answer:

1.0/5 is the answer i got

6 0

3 years ago

Andrew uses the equation 25(x-4)=30 x, where x represents the number of sales, to determine the number of items he needs to sell

pshichka [43]

Hello,

Start off by distributing 25 to both x and -4

25 * X = 25x & 25 * -4 = -100

25x - 100 = 30

At this point, we add 100 to both sides so we can get 25 to be seperate from 100.

25x = 130

To get rid of the 25, divide it in both sides:

130/25 = 5.2

The answer would be 5.2

I hope this helps you with your assignment!

6 0

3 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

4 years ago

Find the x-intercepts if the parabola with vertex (-3,-18) and y-intercept (0,0). Write your answer

anzhelika [568]

Vertex form is
y = a(x - b)^2 + c

here a = -3 and b = -18 so we have

y = a(x + 3)^2 - 18

when x = 0 , y= 0 ( the y-intercept) so:-

0 = a(3^)2 - 18
9a = 18
a = 2

so the parabola is y = 2(x + 3)^2 - 18

x intercepts found as follows:-

2(x + 3)^2 - 18 = 0
(x + 3)^2 = 9
x + 3 = +/- sqrt9 = +/- 3

so x intercepts are 0 and -6 Answer

4 0

3 years ago