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3 years ago

In ΔOPQ, q = 6.5 inches, o = 8.6 inches and ∠P=55°. Find ∠Q, to the nearest 10th of an degree

1 answer:

4 0

Hey, I am Triparna. And I wanted to say that the question is incomplete. It said q= 6.5 inches; what q? Is is the length or something else. I say that this question is incomplete.

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The photo shows a small coin. The scale from the actual coin to the photo is 8 mm to 2 cm In the photo, the distance across the

alekssr [168]

Answer:

1.3cm

Step-by-step explanation:

The scale from the actual coin to the photo is 8 mm to 2 cm In the photo,

converting 8mm to cm = 0.8 cm

using ratio and proportion,

if 2cm = 3.25 cm

then 0.8cm = ? less

if less then more devides.

thus ; (0.8cm ÷2cm ) × 3.25cm =

( 0.4) × 3.25cm = 1.3cm

therefore the distance across the actual coin is 1.3cm.

3 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:

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

The square root of t is greater than 2 and less than 3.5. How many integer values of t satisfy this condition?

Jlenok [28]

$2$

8 integers

4 0

3 years ago

Read 2 more answers

Determine the center and scale factor for the dilation shown below

iren2701 [21]

A center at q scale factor of 1/2

3 0

3 years ago

PLEASE HELP ASAP!!! I NEED CORRECT ANSWERS ONLY PLEASE!!!

Zolol [24]

Answer:

Step-by-step explanation:

Triangle DEF is a right angle triangle.

From the given right angle triangle,

DE represents the hypotenuse of the right angle triangle.

With ∠E as the reference angle,

EF represents the adjacent side of the right angle triangle.

DF represents the opposite side of the right angle triangle.

To determine EF, we would apply

trigonometric ratio

Cos θ = opposite side/hypotenuse. Therefore,

Cos 49 = EF/8

EF = 8Cos49 = 8 × 0.6561

EF = 5.2488

Rounding to the nearest tenth, it becomes 5.2

3 0

3 years ago