All categories

2 years ago

Answer the question now fast

Mathematics

1 answer:

beks73 [17]2 years ago

3 0

Answer:

second and third choice

Step-by-step explanation:

supplamentary = add up to be 180

You might be interested in

Explain how you can find the sign of the product of two or more rational numbers.

makkiz [27]

Rational and product

6 0

3 years ago

Math 8: Linear Functions, Part 2

Studentka2010 [4]

Answer:

$\boxed{\text{1. y + 5 = -4(x - 3); \qquad 2. y - 8 = x + 1}}$

Step-by-step explanation:

Question 1

The point-slope formula for a straight line is

y – y₁ = m(x – x₁)

x₁ = 3; y₁ = -5; m = -4

Substitute the values

$\boxed{\textbf{y + 5 = -4(x - 3)}}$

The diagram shows the graph of equation 1 (red) with slope -4 passing through (3,-5).

Question 2

x₁ = -1; y₁ = 8; m = 1

Substitute the values

$\boxed{\textbf{y - 8 = x + 1}}$

The diagram shows the graph of equation 2 (green) with slope 1 passing through (-1,8).

3 0

3 years ago

Juan and his father went on a driving trip.

Jet001 [13]

Answer:

162

Step-by-step explanation:

260/2=130

130+32=162

8 0

3 years ago

Read 2 more answers

Pls help !!! pleaseee.

ladessa [460]

I guess B. That statement is grammatically incorrect.

8 0

2 years ago

A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline.

Serjik [45]

Answer:

Step-by-step explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

$t=\sqrt{\frac{2h}{a}}$ ,where h=height of release

a=acceleration

$a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}$

Where I=moment of inertia

a for hoop

$a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}$

$a=\frac{g\sin \theta }{2}$

a for Uniform solid cylinder

$a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}$

$a=\frac{2g\sin \theta }{3}$

a for spherical shell

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}$

$a=\frac{3g\sin \theta }{5}$

a for Uniform Solid

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}$

$a=\frac{5g\sin \theta }{7}$

time taken will be inversely proportional to the square root of acceleration

$t_1=k\sqrt{2}=1.414k$

$t_2=k\sqrt{\frac{3}{2}}=1.224k$

$t_3=k\sqrt{\frac{5}{3}}=1.2909k$

$t_4=k\sqrt{\frac{7}{5}}=1.183k$

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

3 0

3 years ago