All categories

2 years ago

Please Help I Don't Understand!

Mathematics

1 answer:

asambeis [7]2 years ago

8 0

Answer:

5.3

Step-by-step explanation:

Since they are similar triangles, we assume the ratio between AM/AN is equal to the ratio between MB/NC.
AM/AN = 6/8 = 3/4
If MB is 4
3/4 = MB/NC = 4/NC
3/4 = 4/NC

3NC = 16
NC = 16/3
16/3 = 5.333333

You might be interested in

Define dot product, inner product, cross product.

sattari [20]

Answer:

Dot product. ... In Euclidean geometry, the dot product of the Cartesian coordinates of two vectors is widely used and often called "the" inner product (or rarely projection product) of Euclidean space even though it is not the only inner product that can be defined on Euclidean space; see also inner product space.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

PLEASE HELP WILL MARK BRAINLIEST THANK YOU

Alina [70]

Answer:

$\huge\boxed{\sf 46}$

Step-by-step explanation:

Formula for area of rectangle:

= length × width

Area of the smaller rectangle:

length = 7

width = 6

= 7 × 6

= 42

Area of the larger rectangle:

length = 7 + 4 = 11

width = 6 + 2 = 8

= 11 × 8

= 88

Area of the shaded region:

= Area of the larger rectangle - Area of the smaller rectangle

= 88 - 42

= 46

$\rule[225]{225}{2}$

Hope this helped!

5 0

3 years ago

Classify the following triangle. Check all that apply.

Pachacha [2.7K]

That is a Right Triangle

7 0

3 years ago

Y + 12 = -26 thank you if you answer me

tia_tia [17]

Answer:

-38

Step-by-step explanation:

1. Isolate y to one side.

y+12=-26

subtract 12 to both sides

y=-38

3 0

3 years ago

Read 2 more answers

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2

Alika [10]

As the ladder is pulled away from the wall, the area and the height with the

wall are decreasing while the angle formed with the wall increases.

The correct response are;

(a) The velocity of the top of the ladder = 1.5 m/s downwards

(b) The rate the area formed by the ladder is changing is approximately -75.29 ft.²/sec

Reasons:

The given parameter are;

Length of the ladder, l = 25 feet

Rate at which the base of the ladder is pulled, $\displaystyle \frac{dx}{dt}$ = 2 feet per second

(a) Let y represent the height of the ladder on the wall, by chain rule of differentiation, we have;

$\displaystyle \frac{dy}{dt} = \mathbf{\frac{dy}{dx} \times \frac{dx}{dt}}$

25² = x² + y²

y = √(25² - x²)

$\displaystyle \frac{dy}{dx} = \frac{d}{dx} \sqrt{25^2 - x^2} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}$

Which gives;

$\displaystyle \frac{dy}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times \frac{dx}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2$

$\displaystyle \frac{dy}{dt} = \mathbf{ \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2}$

When x = 15, we get;

$\displaystyle \frac{dy}{dt} = \frac{15 \times \sqrt{625-15^2} }{15^2- 625}\times2 = \mathbf{-1.5}$

The velocity of the top of the ladder = 1.5 m/s downwards

When x = 20, we get;

$\displaystyle \frac{dy}{dt} = \frac{20 \times \sqrt{625-20^2} }{20^2- 625}\times2 = -\frac{8}{3} = -2.\overline 6$

The velocity of the top of the ladder = $\underline{-2.\overline{6} \ m/s \ downwards}$

When x = 24, we get;

$\displaystyle \frac{dy}{dt} = \frac{24 \times \sqrt{625-24^2} }{24^2- 625}\times2 = \mathbf{-\frac{48}{7}} \approx -6.86$

The velocity of the top of the ladder ≈ -6.86 m/s downwards

(b) $\displaystyle The \ area\ of \ the \ triangle, \ A =\mathbf{\frac{1}{2} \cdot x \cdot y}$

Therefore;

$\displaystyle The \ area\ A =\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}$

$\displaystyle \frac{dA}{dx} = \frac{d}{dx} \left (\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}\right) = \mathbf{\frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250}}$

$\displaystyle \frac{dA}{dt} = \mathbf{ \frac{dA}{dx} \times \frac{dx}{dt}}$

Therefore;

$\displaystyle \frac{dA}{dt} = \frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250} \times 2$

When the ladder is 24 feet from the wall, we have;

x = 24

$\displaystyle \frac{dA}{dt} = \frac{(2 \times 24^2- 625)\cdot \sqrt{625-24^2} }{2\times 24^2 - 1250} \times 2 \approx \mathbf{ -75.29}$

The rate the area formed by the ladder is changing, $\displaystyle \frac{dA}{dt}$ ≈ -75.29 ft.²/sec

$\displaystyle sin(\theta) = \frac{x}{25}$

$\displaystyle \theta = \mathbf{arcsin \left(\frac{x}{25} \right)}$

$\displaystyle \frac{d \theta}{dt} = \frac{d \theta}{dx} \times \frac{dx}{dt}$

$\displaystyle\frac{d \theta}{dx} = \frac{d}{dx} \left(arcsin \left(\frac{x}{25} \right) \right) = \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625}}$

Which gives;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-x^2} }{x^2 - 625}\times \frac{dx}{dt}= \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625} \times 2}$

When x = 24 feet, we have;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-24^2} }{24^2 - 625} \times 2 \approx \mathbf{ 0.286}$

Rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 24 feet from the wall is $\displaystyle \frac{d \theta}{dt}$ ≈ 0.286 rad/sec

Learn more about the chain rule of differentiation here:

brainly.com/question/20433457

3 0

3 years ago