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3 years ago

PLEASE HELP!!!!! HURRY

Mathematics

1 answer:

svp [43]3 years ago

7 0

Its c first find area of the whole thing with piece cut out.(240) then of the cut out piece (12) 240-12=228 C

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If the common difference in an arithmetic sequence is 4, and the 20th term is 36, what is the first term

zhuklara [117]

D=2

a+29d=36

then a=-22

3 0

3 years ago

David and Mark are marking exam papers. Each set takes David 24 minutes and Mark 1 hour. Express the times David and Mark take a

attashe74 [19]

Answer:

2:5

Step-by-step explanation:

david=24mins

mark = 1hour

change 1 hour to mins

DAVID=24

MARK=60

24:60

THEN SIMPLIFY

24/60

=2/5

2:5

8 0

3 years ago

hi i need help once again lol A company asked its employees how much time they spend reading and responding to email. The line p

PSYCHO15rus [73]

Answer:

Step-by-step explanation:

you multiply each dot by the fraction of hours it is under. you add everything up and you should get 66/8 which is equivalent to 8 and 1/4

8 0

3 years ago

A car can travel 120km ik 2 hours How long will it take the car to travel 300km

ehidna [41]

It would take the car 5 hours to travel 300km.

Equation used to solve: (120÷2)×6

3 0

3 years ago

Read 2 more answers

Cosθ=−2√3 , where π≤θ≤3π2 .

Alex787 [66]

Answer:

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we know that

Applying the trigonometric identity

$sin^2(\theta)+ cos^2(\theta)=1$

we have

$cos(\theta)=-\frac{\sqrt{2}}{3}$

substitute

$sin^2(\theta)+ (-\frac{\sqrt{2}}{3})^2=1$

$sin^2(\theta)+ \frac{2}{9}=1$

$sin^2(\theta)=1- \frac{2}{9}$

$sin^2(\theta)= \frac{7}{9}$

$sin(\theta)=\pm\frac{\sqrt{7}}{3}$

Remember that

π≤θ≤3π/2

Angle θ belong to the III Quadrant

That means ----> The sin(θ) is negative

$sin(\theta)=-\frac{\sqrt{7}}{3}$

step 2

Find the sec(β)

Applying the trigonometric identity

$tan^2(\beta)+1= sec^2(\beta)$

we have

$tan(\beta)=\frac{4}{3}$

substitute

$(\frac{4}{3})^2+1= sec^2(\beta)$

$\frac{16}{9}+1= sec^2(\beta)$

$sec^2(\beta)=\frac{25}{9}$

$sec(\beta)=\pm\frac{5}{3}$

we know

0≤β≤π/2 ----> II Quadrant

sec(β), sin(β) and cos(β) are positive

$sec(\beta)=\frac{5}{3}$

Remember that

$sec(\beta)=\frac{1}{cos(\beta)}$

therefore

$cos(\beta)=\frac{3}{5}$

step 3

Find the sin(β)

we know that

$tan(\beta)=\frac{sin(\beta)}{cos(\beta)}$

we have

$tan(\beta)=\frac{4}{3}$

$cos(\beta)=\frac{3}{5}$

substitute

$(4/3)=\frac{sin(\beta)}{(3/5)}$

therefore

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

step 4

Find sin(θ+β)

we know that

$sin(A + B) = sin A cos B + cos A sin B$

In this problem

$sin(\theta + \beta) = sin(\theta)cos(\beta)+ cos(\theta)sin (\beta)$

we have

$sin(\theta)=-\frac{\sqrt{7}}{3}$

$cos(\theta)=-\frac{\sqrt{2}}{3}$

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

$cos(\beta)=\frac{3}{5}$

substitute the given values in the formula

$sin(\theta + \beta) = (-\frac{\sqrt{7}}{3})(\frac{3}{5})+ (-\frac{\sqrt{2}}{3})(\frac{4}{5})$

$sin(\theta + \beta) = (-3\frac{\sqrt{7}}{15})+ (-4\frac{\sqrt{2}}{15})$

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

8 0

3 years ago