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3 years ago

Angles 4 and 5 form what type of angles?

Mathematics

1 answer:

VashaNatasha [74]3 years ago

3 0

Answer:

Vertical angles

Step-by-step explanation:

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6 × 3 − (4 × 3) + 16 ÷ 4 = 11

rjkz [21]

Answer:

Step-by-step explanation:

6x3 = 18

4x3 = 12

18 - 12 = 6

16 / 4 = 4

6 - 4 = 2

4 0

3 years ago

A garden has a width of 24 feet and a diagonal length of 51 feet. What is the perimeter of the garden ( in feet )?

bekas [8.4K]

Answer:

1,224 feet

Step-by-step explanation:

Multiply 51 feet times 24 feet and then you get 1,224 ft.

3 0

3 years ago

Can someone please help me with this math problem

IrinaK [193]

Answer:

y = 10x + 7

Step-by-step explanation:

5 0

2 years ago

Identify the equation of the circle that has its center at (16, 30) and passes through the origin

Veronika [31]

To solve this question, we have to find the equation of the circle with given center and where it passes. Doing this, we get that the equation of the circle is:

$(x - 16)^2 + (y - 30)^2 = 1156$

Equation of a circle:

The equation of a circle with center $(x_0, y_0)$ and radius r is given by:

$(x - x_0)^2 + (y - y_0)^2 = r^2$

Center at (16, 30)

This means that $x_0 = 16, y_0 = 30$

Thus

$(x - 16)^2 + (y - 30)^2 = r^2$

Passes through the origin:

We use this to find the radius squared, as this means that $x = 0, y = 0$ is part of the circle. Thus

$(x - 16)^2 + (y - 30)^2 = r^2$

$(0 - 16)^2 + (0 - 30)^2 = r^2$

$r^2 = 16^2 + 30^2 = 1156$

Thus, the equation of the circle is:

$(x - 16)^2 + (y - 30)^2 = 1156$

For another example to find the equation of a circle, you can look at brainly.com/question/23719612

4 0

3 years ago

A particle moves so that r(t) = ati + b sin atj. Show that the magnitude of the acceleration of the particle is proportional to

ValentinkaMS [17]

Answer:

Step-by-step explanation:

Given

Position of particle is $r(t)=at\hat{i}+b\sin (at)\hat{j}$

i.e. distance from x axis is $b\sin (at)---1$

Distance from y axis at

velocity is given by $v=\frac{\mathrm{d} r}{\mathrm{d} t}$

$v=a\hat{i}+ba\cos (at)\hat{j}$

Similarly acceleration is given by

$a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$a=0\hat{i}-a^2b\sin (at)\hat{j}$

Magnitude of acceleration is $=\sqrt{(-a^2b\sin (at))^2}$

$=a^2b\sin (at)----2$

From 1 and 2 we can see that

Magnitude of acceleration is proportional to distance from x axis

$a\propto distance\ from\ x-axis$

5 0

3 years ago

Read 2 more answers