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2 years ago

Y=3x-8 using function notation

Mathematics

1 answer:

maria [59]2 years ago

4 0

To write it in function notation you'll convert it into a function, substituting y by f(x):

y = 3x - 8 = f(x) = 3x - 8

Hope it helped,

BioTeacher101

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1/2 + 1/6 what is it pls help!!!!

ale4655 [162]

Answer:

4/6

Step-by-step explanation:

1/2 = 3/6. 3/6 + 1/6 = 4/6

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2 years ago

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F(x)=x^2(x+3)^2 <br><br> How do I find the roots of this equation?

Genrish500 [490]

= x^2 ( x+3 ) (x+3)
= set equal to 0

X^2= 0
X=0

X+3=0
X=-3

Answer is -3 and 0

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2 years ago

PLSSS HELP

iragen [17]

Answer:

Hey!

Step-by-step explanation:

Legit just like make a graph like this:

10 | 1

20 | 2

30 | 3

And so on and so forth

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2 years ago

jennifer bikes 7 miles south and then turns to bike 13 miles east how far away is she from where she started?

Marina CMI [18]

Answer:

She is about 14.765 miles ( $\sqrt{218}$ miles) from where she started

Step-by-step explanation:

There is a relation between the three sides of the right triangle

The side opposite to the right angle is called hypotenuse and it is the longest side

The other two sides called legs of the right angle

The relation between them is: (hypotenuse)² = (leg1)² + (leg2)²

∵ Jennifer bikes 7 miles south

∵ She turns to bike 13 miles east

∵ South and East are perpendicular

→ That means the distance from her start point to end point represents

a hypotenuse of a right triangle, whose legs are 7 and 13

∴ (hypotenuse)² = (leg1)² + (leg2)², where

hypotenuse is the distance between her start and end points

leg1 is her distance in south direction

leg2 is her distance in east direction

∵ Leg1 = 7 miles

∵ leg 2 = 13 miles

∴ (hypotenuse)² = (7)² + (13)²

∴ (hypotenuse)² = 49 + 169

∴ (hypotenuse)² = 218

→ Take √ for both sides

∴ hypotenuse = $\sqrt{218}$

∴ hypotenuse ≅ 14.76482306

∴ She is about 14.765 miles ( $\sqrt{218}$ miles) from where she started.

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2 years ago

The median AM of triangle ∆ABC is half the length of the side towards which it is drawn, BC . Prove that triangle ∆ABC is a righ

MaRussiya [10]

The problem statement tells us that

AM ≅ MB, ∴ ΔAMB is isosceles
AM ≅ MC, ∴ ΔAMC is isosceles

Base angles of an isosceles triangle are equal, so ∠MAB ≅ ∠MBA. ∠AMC is the exterior angle opposite those two, so it is equal to their sum, 2∠MAB.

The base angles in isosceles ΔAMC are equal to half the difference between the apex angle, ∠AMC, and 180°. That is,

... ∠MAC = (1/2)(180° -∠AMC) = (1/2)(180° -2∠MAB) = 90° -∠MAB

The angle at A of ΔABC is the sum of the two angles created by the median AM. That is ...

∠A = ∠MAC + ∠MAB

∠A = (90° -∠MAB) +∠MAB

∠A = 90°

_____

Maybe a shorter way to get there is to realize that ...

... MA ≅ MB ≅ MC

so M is the center of a circle with BC as a diameter and A a point on the circle. Angle BAC is inscribed in the semicircle and subtends an arc of 180°, so angle A is 90°.

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2 years ago