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Ganezh [65]
2 years ago
8

Determine if 0.69789978997899789 is rational or irrational

Mathematics
1 answer:
mylen [45]2 years ago
7 0
Don’t click the link it’s a scam
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June bought ​ 3 1 2 312 ​ gallons of white paint at $17.50 per gallon and ​ 1 3 4 134 ​gallons of blue paint at $19.00 per gallo
algol [13]
312* 17.5= 5460.00

134*19=2546

2546+5460= $8,006.00
5 0
3 years ago
Translate into an alegbraic expression and simplify if possible.
serg [7]

Answer:

(10-x)GB

Step-by-step explanation:

Music=10-x, simple arithmetic

6 0
3 years ago
Select the correct product.<br><br> (x 2 + 4x + 8)(2x - 1)
marusya05 [52]

Answer:

\large\boxed{(x^2+4x+8)(2x-1)=2x^3+7x^2+12x-8}

Step-by-step explanation:

Use FOIL: <em>(a + b)(c + d) = ac + ad + bc + bd</em>

(x^2+4x+8)(2x-1)\\\\=(x^2)(2x)+(x^2)(-1)+(4x)(2x)+(4x)(-1)+(8)(2x)+(8)(-1)\\\\=2x^3-x^2+8x^2-4x+16x-8\qquad\text{combine like terms}\\\\=2x^3+(-x^2+8x^2)+(-4x+16x)-8\\\\=2x^3+7x^2+12x-8

7 0
3 years ago
Read 2 more answers
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
3 years ago
What is the slope of the line that passes through the points (5,11) and (6,2)?
Leya [2.2K]

Answer:

-9/1

y2-y1/x2-x1

(2-11)/(6-5)

=-9

3 0
3 years ago
Read 2 more answers
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