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2 years ago

The total cost of an order of shirts from a company consists of

Mathematics

1 answer:

Luba_88 [7]2 years ago

4 0

Answer:

The cost of each shirt is the same regardless of how many shirts are ordered. The company provides the following examples to customers to help them ... Using only this information, determine the cost of ordering 200 shirts. Also ... a shirt company charges a one-time setup fee plus a certain price per shirt.

Step-by-step explanation:

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Hello whats is the size of the angle

Gennadij [26K]

Answer:

x = 65°

Step-by-step explanation:

The interior angle of the triangle and 140° are adjacent and supplementary, thus

interior angle = 180° - 140° = 40°

The exterior angle of a triangle is equal to the sum of the 2 opposite interior angles.

105° is an exterior angle of the triangle, thus

x + 40° = 105° ( subtract 40° from both sides )

x = 65°

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2 years ago

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Analyze the diagram below and complete the instructions that follow.

Semmy [17]

B. All sides are equal and each side is at a 60 degree angle.

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2 years ago

Match each function with the correct translation of the parent function f(x) = |x|

GarryVolchara [31]

Answer:they are all straight across so drag each line to the one that is across from the equation

Step-by-step explanation:

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3 years ago

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Use the given information to find (a) sin(s+t), (b) tan(s+t), and (c) the quadrant of s+t. cos s = - 12/13 and sin t = 4/5, s an

Anton [14]

Answer:

Part a) $sin(s + t) =-\frac{63}{65}$

Part b) $tan(s + t) = -\frac{63}{16}$

Part c) (s+t) lie on Quadrant IV

Step-by-step explanation:

[Part a) Find sin(s+t)

we know that

$sin(s + t) = sin(s) cos(t) + sin(t)cos(s)$

step 1

Find sin(s)

$sin^{2}(s)+cos^{2}(s)=1$

we have

$cos(s)=-\frac{12}{13}$

substitute

$sin^{2}(s)+(-\frac{12}{13})^{2}=1$

$sin^{2}(s)+(\frac{144}{169})=1$

$sin^{2}(s)=1-(\frac{144}{169})$

$sin^{2}(s)=(\frac{25}{169})$

$sin(s)=\frac{5}{13}$ ---> is positive because s lie on II Quadrant

step 2

Find cos(t)

$sin^{2}(t)+cos^{2}(t)=1$

we have

$sin(t)=\frac{4}{5}$

substitute

$(\frac{4}{5})^{2}+cos^{2}(t)=1$

$(\frac{16}{25})+cos^{2}(t)=1$

$cos^{2}(t)=1-(\frac{16}{25})$

$cos^{2}(t)=\frac{9}{25}$

$cos(t)=-\frac{3}{5}$ is negative because t lie on II Quadrant

step 3

Find sin(s+t)

$sin(s + t) = sin(s) cos(t) + sin(t)cos(s)$

we have

$sin(s)=\frac{5}{13}$

$cos(t)=-\frac{3}{5}$

$sin(t)=\frac{4}{5}$

$cos(s)=-\frac{12}{13}$

substitute the values

$sin(s + t) = (\frac{5}{13})(-\frac{3}{5}) + (\frac{4}{5})(-\frac{12}{13})$

$sin(s + t) = -(\frac{15}{65}) -(\frac{48}{65})$

$sin(s + t) =-\frac{63}{65}$

Part b) Find tan(s+t)

we know that

tex]tan(s + t) = (tan(s) + tan(t))/(1 - tan(s)tan(t))[/tex]

we have

$sin(s)=\frac{5}{13}$

$cos(t)=-\frac{3}{5}$

$sin(t)=\frac{4}{5}$

$cos(s)=-\frac{12}{13}$

step 1

Find tan(s)

$tan(s)=sin(s)/cos(s)$

substitute

$tan(s)=(\frac{5}{13})/(-\frac{12}{13})=-\frac{5}{12}$

step 2

Find tan(t)

$tan(t)=sin(t)/cos(t)$

substitute

$tan(t)=(\frac{4}{5})/(-\frac{3}{5})=-\frac{4}{3}$

step 3

Find tan(s+t)

$tan(s + t) = (tan(s) + tan(t))/(1 - tan(s)tan(t))$

substitute the values

$tan(s + t) = (-\frac{5}{12} -\frac{4}{3})/(1 - (-\frac{5}{12})(-\frac{4}{3}))$

$tan(s + t) = (-\frac{21}{12})/(1 - \frac{20}{36})$

$tan(s + t) = (-\frac{21}{12})/(\frac{16}{36})$

$tan(s + t) = -\frac{63}{16}$

Part c) Quadrant of s+t

we know that

$sin(s + t) =negative$ ----> (s+t) could be in III or IV quadrant

$tan(s + t) =negative$ ----> (s+t) could be in III or IV quadrant

Find the value of cos(s+t)

$cos(s+t) = cos(s) cos(t) -sin (s) sin(t)$

we have

$sin(s)=\frac{5}{13}$

$cos(t)=-\frac{3}{5}$

$sin(t)=\frac{4}{5}$

$cos(s)=-\frac{12}{13}$

substitute

$cos(s+t) = (-\frac{12}{13})(-\frac{3}{5})-(\frac{5}{13})(\frac{4}{5})$

$cos(s+t) = (\frac{36}{65})-(\frac{20}{65})$

$cos(s+t) =\frac{16}{65}$

we have that

$cos(s+t)=positive$ -----> (s+t) could be in I or IV quadrant

$sin(s + t) =negative$ ----> (s+t) could be in III or IV quadrant

$tan(s + t) =negative$ ----> (s+t) could be in III or IV quadrant

therefore

(s+t) lie on Quadrant IV

4 0

2 years ago

PLEASE HELP!!! a 300 cm b 60 cm c 20 cm d 45 cm

brilliants [131]

Answer:

45 cm

Step-by-step explanation:

Let length of the field in the drawing = x

80 : 120 :: 30 : x

Product of extreme = Product of means

80 * x = 120 * 30

$x = \frac{120*30}{80}\\\\x = 15 * 3\\x = 45$

3 0

2 years ago