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3 years ago

7

The table shows a linear function. Which equation represents the function.

1 answer:

mote1985 [20]3 years ago

7 0

Answer:

$\large\boxed{f(x)=\dfrac{3}{2}x-1}$

Step-by-step explanation:

$\text{The slope-intercept formula:}\\\\y=mx+b\\\\m\ -\ slope\\b\ -\ y-intercept\\\\\text{The formula of a slope:}\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\\\\\text{From the table we hve two points (0, -1) and (2, 2). Substitute:}\\\\m=\dfrac{2-(-1)}{2-0}=\dfrac{3}{2}$

$\text{Therefore the equation is}\ f(x)=\dfrac{3}{2}x-1\\\\\text{Check for the point (4, 5):}\\\\f(4)=\dfrac{3}{2}(4)-1=(3)(2)-1=6-1=5\qquad\ CORRECT$

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bixtya [17]

Answer:

The monthly payment is $35.10.

Step-by-step explanation:

p = 510

r = $13.9/12/100=0.011583$

n = $12+4=16$

The EMI formula is :

$\frac{p\times r\times(1+r)^{n} }{(1+r)^{n}-1 }$

Now putting the values in formula we get;

$\frac{510\times0.011583\times(1+0.011583)^{16} }{(1+0.011583)^{16}-1 }$

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4 years ago

If tan A = 4/3 and sin B = 45/53 and angles A and B are in Quadrant I, find the value of tan(A+B)

lbvjy [14]

Answer:

First, find tan A and tan B.

cosA=35 --> sin2A=1−925=1625 --> cosA=±45

cosA=45 because A is in Quadrant I

tanA=sinAcosA=(45)(53)=43.

sinB=513 --> cos2B=1−25169=144169 --> sinB=±1213.

sinB=1213 because B is in Quadrant I

tanB=sinBcosB=(513)(1312)=512

Apply the trig identity:

tan(A−B)=tanA−tanB1−tanA.tanB

tanA−tanB=43−512=1112

(1−tanA.tanB)=1−2036=1636=49

tan(A−B)=(1112)(94)=3316

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2 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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MrRissso [65]

Answer:

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Step-by-step explanation:

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