All categories

3 years ago

Please help I am confused

Mathematics

1 answer:

anzhelika [568]3 years ago

7 0

X=11
So do I think hope this helps!!

You might be interested in

Find the measure of each angle using the triangle sum theorem and exterior angle theorem

Paladinen [302]

(please open my photo for reference as you read, I am a visual learner/explainer so it will make the most sense that way)

So the first thing you want to do is look at the exterior angle 130°. A straight line is 180°, and every Triangle's angular sum is 180°. How I think of it is that every straight line has a mini protractor on either side. It makes it a bit easier to understand.

180 - 130 = 50

You now know that two of the angles are 50°.

You now have two of the measurements for the triangle farthest to the left.

75° and 50°

75 + 50 = 125

180 - 125 = 55

a = 55°

Now that you have all the measurements for the first triangle, let's move onto the next one.

With two measurements for the second triangle, all you need to do is find their sum and subtract that from 180 and you will have the third measurement!

50 + 60 = 110

180 - 110 = 70

b = 70°

Finally, for the last triangle, you already have two of the measurements 60° and 85°.

85 + 60 = 145

180 - 145 = 35

c = 35°

Sorry if this explanation is a bit messy, it's hard to describe certain things without a letter or some kind of name to differentiate between them verbally.

I hope this helps! <3

3 0

3 years ago

The description below represents Function A and the table represents Function B: Function A The function is 2 more than 5 times

Vlad [161]

Answer:

The slope of Function A is greater than the slope of function B while the y-intercept of Function B is greater than the y-intercept of function A.

Step-by-step explanation:

Function A is f(x) = 5x + 2

The slope is 5 and the y-intercept is (0, 2).

Function B. The slope is (5-2) / (0- -1) = 3.

also (8-5) / (1 - 0) = 3

The slope is 3.

The y-intercept is at (0, 5) because one point on the graph is (0, 5).

7 0

3 years ago

What is the standard form equation of the line shown below?

ch4aika [34]

(-1,1)(1,4)
slope = (4 - 1) / (1 - (-1) = 3/(1 + 1) = 3/2

y = mx + b
slope(m) = 3/2
use either of ur points....(1,4)...x = 1 and y = 4
now sub and find b, the y int
4 = 3/2(1) + b
4 = 3/2 + b
4 - 3/2 = b
8/2 - 3/2 = b
5/2 = b

equation is y = 3/2x + 5/2...but we need it in standard form

y = 3/2x + 5/2
-3/2x + y = 5/2.....multiply both sides by -2
3x - 2y = -5 <==== ur standard form equation

3 0

3 years ago

A department store's survey suggests that 76% of shoppers buy food, 49% buy clothes, and 28% buy both food and

Artyom0805 [142]

Answer:

0.57

Step-by-step explanation:

Given that:

food = f ; c = clothes

P(f) = 0.76

P(C) = 0.49

P(fnC) = 0.28

Suppose a shopper is selected from the store at random and learn that they buy clothes. What is the probability that the shopper also buys food?

P(f Given C) = P(f | C)

P(f | c) = p(fnC) / p(C)

P(f | c) = 0.28 / 0.49

P(f | c) = 0.5714

P(f | c) = 0.57

7 0

3 years ago

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

2 years ago