Identify the exact stamp using a catalog. This could involve learning about perforations, watermarks and colors as well as condition. The catalog will provide a value. The value is what a collector could expect to pay for a stamp in fine/very fine condition. If selling, most cases you would be lucky to get 75% of the catalog, unless it is very valuable, then an auction would be worth looking into. Dealers would be able to take a look at the overall collection and make an offer. Find a local dealer throught the American Stamp Dealers' Association. The most common American catalog for identification is Scott's. Others are Stanley Gibbons, and Minkus. ____ The USSR released 179 stamps in 1964.
To find the value of (\tan(15^\circ) \tan(195^\circ)), we can use the identity (\tan(195^\circ) = \tan(15^\circ + 180^\circ) = \tan(15^\circ)). Thus, (\tan(195^\circ) = \tan(15^\circ)). Consequently, (\tan(15^\circ) \tan(195^\circ) = \tan(15^\circ) \tan(15^\circ) = \tan^2(15^\circ)). The exact value of (\tan^2(15^\circ)) can be computed, but it is important to note that it will yield a positive value.
To find the value of (\tan(1190^\circ)), first reduce the angle by subtracting multiples of (360^\circ) until it falls within the range of (0^\circ) to (360^\circ). (1190^\circ - 3 \times 360^\circ = 110^\circ). Thus, (\tan(1190^\circ) = \tan(110^\circ)). The tangent of (110^\circ) is negative, specifically (-\tan(70^\circ)), which is approximately (-2.747).
To find the value of ( \cos^2 67^\circ - \sin^2 23^\circ ), we can use the identity ( \cos^2 \theta = 1 - \sin^2 \theta ). Since ( \sin 23^\circ = \cos 67^\circ ) (because ( 23^\circ + 67^\circ = 90^\circ )), we have ( \sin^2 23^\circ = \cos^2 67^\circ ). Thus, ( \cos^2 67^\circ - \sin^2 23^\circ = \cos^2 67^\circ - \cos^2 67^\circ = 0 ). Therefore, the value is ( 0 ).
The value of ( \sin 40^\circ \sin 50^\circ ) can be simplified using the product-to-sum identities. Specifically, it can be expressed as: [ \sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)] ] Substituting ( A = 40^\circ ) and ( B = 50^\circ ): [ \sin 40^\circ \sin 50^\circ = \frac{1}{2} [\cos(40^\circ - 50^\circ) - \cos(40^\circ + 50^\circ)] = \frac{1}{2} [\cos(-10^\circ) - \cos(90^\circ)] = \frac{1}{2} [\cos(10^\circ) - 0] = \frac{\cos(10^\circ)}{2} ] The approximate numerical value is about ( 0.4848 ).
The exact value of (\sin 165^\circ) can be calculated using the sine subtraction formula. Since (165^\circ = 180^\circ - 15^\circ), we have: [ \sin 165^\circ = \sin(180^\circ - 15^\circ) = \sin 15^\circ ] The value of (\sin 15^\circ) can be derived from the formula (\sin(45^\circ - 30^\circ)), which gives: [ \sin 15^\circ = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4} ] Thus, (\sin 165^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}).
The value of (\sin 135^\circ) can be determined using the unit circle or trigonometric identities. Since (135^\circ) is in the second quadrant, where sine is positive, we can express it as (\sin(180^\circ - 45^\circ)). Thus, (\sin 135^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}). Therefore, the value of (\sin 135^\circ) in radical form is (\frac{\sqrt{2}}{2}).
The exact value of (\tan 195^\circ) can be found using the tangent addition formula. Since (195^\circ) is in the third quadrant, where tangent is positive, we can express it as (\tan(180^\circ + 15^\circ)). This gives us (\tan 195^\circ = \tan 15^\circ), which is (\frac{\sin 15^\circ}{\cos 15^\circ}). Using the known values, (\tan 15^\circ = 2 - \sqrt{3}). Therefore, (\tan 195^\circ = 2 - \sqrt{3}).
To find the exact value of (\tan 150^\circ), you can use the fact that (150^\circ) is in the second quadrant, where the tangent function is negative. The reference angle for (150^\circ) is (180^\circ - 150^\circ = 30^\circ). Therefore, (\tan 150^\circ = -\tan 30^\circ). Since (\tan 30^\circ = \frac{1}{\sqrt{3}}), it follows that (\tan 150^\circ = -\frac{1}{\sqrt{3}}), or (-\frac{\sqrt{3}}{3}) when rationalized.
To find the value of ( \frac{19 \sin(50^\circ)}{\sin(40^\circ)} ), we can use the sine function values. Using the sine of complementary angles, ( \sin(50^\circ) = \cos(40^\circ) ). Therefore, ( \frac{19 \sin(50^\circ)}{\sin(40^\circ)} = \frac{19 \cos(40^\circ)}{\sin(40^\circ)} = 19 \cot(40^\circ) ). For an exact numerical value, you can compute ( 19 \cot(40^\circ) ) using a calculator.
The expression ( y \cos(\theta) ) will have its maximum value when ( \cos(\theta) ) reaches its maximum, which is 1. Since ( \theta ) is constant at 720 degrees, we can calculate ( \cos(720^\circ) ). The angle 720 degrees is equivalent to 0 degrees (since ( 720^\circ - 360^\circ = 360^\circ ), and ( 360^\circ - 360^\circ = 0^\circ )), thus ( \cos(720^\circ) = 1 ). Therefore, the maximum value of ( y \cos(θ) ) is simply ( y ) when ( \theta = 720 ) degrees.
The value of ( \tan 15^\circ ) can be calculated using the tangent subtraction formula: [ \tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} ] Substituting the known values ( \tan 45^\circ = 1 ) and ( \tan 30^\circ = \frac{1}{\sqrt{3}} ), we find: [ \tan(15^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} ] Thus, ( \tan 15^\circ = 2 - \sqrt{3} ) in its simplest fractional form.
A regular decagon can rotate onto itself at angles that are multiples of ( \frac{360^\circ}{10} ), which is ( 36^\circ ). This means it can rotate by ( 0^\circ ), ( 36^\circ ), ( 72^\circ ), ( 108^\circ ), ( 144^\circ ), ( 180^\circ ), ( 216^\circ ), ( 252^\circ ), ( 288^\circ ), and ( 324^\circ ). In total, there are 10 distinct angles (including ( 0^\circ )) at which the decagon can map onto itself.