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Mrrafil [7]
3 years ago
14

What are the slope and the y-intercept of a linear function that is represented by the table

Mathematics
2 answers:
Katena32 [7]3 years ago
4 0

Answer:

i think its the first one but i'm not sure

Step-by-step explanation:

Natasha_Volkova [10]3 years ago
4 0

Answer: The slope is  \frac{-2}{5} and the y-intercept is \frac{-1}{3}

Step-by-step explanation:

We know that the intercept form of a linear function is given by :-

y=mx+c, where m is the slope of the line and c is the y intercept point.

The slope of a function is given by :-

m=\dfrac{\text{change in y}}{\text{change in x}}\\\\\\\Rightarrow\ m=\dfrac{\frac{-2}{15}-\frac{-1}{30}}{\frac{-1}{2}-\frac{-3}{4}}\\\\\\\Rightarrow\ m=\dfrac{\frac{-3}{30}}{\frac{1}{4}}=\dfrac{-2}{5}

Put the value of m in the general intercept equation , we get

y=\frac{-2}{5}x+c

Put x=\frac{-1}{2} and y=\frac{-2}{15} in it , we get

\frac{-2}{15}=\frac{-2}{5}\times \frac{-1}{2}+c\\\\\Rightarrow\ c=\frac{-2}{15}\times\frac{5}{1}=\frac{-1}{3}

Hence, The slope is  \frac{-2}{5} and the y-intercept is \frac{-1}{3}

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[4]
Tanya [424]

Answer:

Angle ACD = 38°

Step-by-step explanation:

The full, correct question is presented the attached image to this solution.

Given

Point O is the centre if the circle

Points A, B, C and D are points on the circle

Angle AOB = 140°

Angle OAC = 14°

Angle AOB = 2 × (Angle ACB) [angle subtended at the centre of the circle is twice the angle subtended at the circumference of the circle)

140° = 2 × (Angle ACB)

Angle ACB = (140°/2) = 70°

(Angle AOB) + (Angle OAB) + (Angle ABO) = 180° [sun of angles in a triangle is 180°]

But Angle OAB = Angle ABO = a [base angles of an iscosceles triangle are equal since OA and OB are both radii for the circle O)

(Angle AOB) + (Angle OAB) + (Angle ABO) = 180°

140° + a + a = 180°

2a = 40°

a = (40/2) = 20°

Angle OAB = Angle ABO = 20°

Angle CAB = (Angle OAC) + (Angle OAB) = 14° + 20° = 34°

Triangle ADC is an iscosceles triangle, hence,

Angle DAC = Angle ACD = x [base angles of an iscosceles triangle are equal]

But

(Angle DCB) + (Angle DAB) = 180° [Opposite angles of a cyclic quadilateral (a quadilateral inscribed in a circle) sum up to give 180°]

Angle DCB = (Angle DCA) + (Angle ACB) = (x + 70°)

Angle DAB = (Angle DAC) + (Angle CAB) = (x + 34°)

(Angle DCB) + (Angle DAB) = 180°

(x + 70°) + (x + 34°) = 180°

2x + 104° = 180°

2x = 180° - 104° = 76°

x = (76°/2) = 38°

Angle DAC = Angle ACD = 38°

Angle ACD = 38°

Hope this Helps!!!

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