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3 years ago

I’m sure this is easy but I hateeee geometry. help ASAP plsssss

Mathematics

1 answer:

hram777 [196]3 years ago

8 0

Given points

(-4,3)
(2,5)

Distance:-

$\\ \sf\longmapsto \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\\ \sf\longmapsto \sqrt{(2+4)^2+(5-3)^2}$

$\\ \sf\longmapsto \sqrt{(6)^2+(2)^2}$

$\\ \sf\longmapsto \sqrt{36+4}$

$\\ \sf\longmapsto \sqrt{40}$

$\\ \sf\longmapsto 6.2$

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If the center of the small gear is rigidly attached to the center of a bicycle wheel with a radius of 10 inches, how many inches

kakasveta [241]

The distance traveled is 62.8 inches.

<h3>How many inches will the bicycle travel with a single rotation of the large gear?</h3>

We know that the gear is rigidly attached to the center of the wheel. This means that in one complete rotation of the gear, we also have a complete rotation of the wheel.

Then the distance that the bicycle travel with a single rotation of the gear is equal to the circumference of the wheel, which is:

C = 2*pi*R

Where pi = 3.14 and R is the radius, which we know is 10inches.

Then:

C = 2*3.14*10in = 62.8in

The distance traveled is 62.8 inches.

If you want to learn more about circles:

brainly.com/question/14283575

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3 years ago

A line parallel to a triangle's side splits AB into lengths of x - 7 and x - 3. The other side, AC, is split into lengths of x a

Nesterboy [21]

Answer:

Step-by-step explanation:

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4 years ago

The slope of the line

Fantom [35]

Answer: -2

Explanation: On the image we can see it goes down 4 units and right 2. Giving us the slope $\frac{-4}{2}$

Simplified that is -2 which is the slope of the line

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3 years ago

find the equation of the pair of lines perpendicular to the lines pair represented by the equation ax^2-2hxy+by^2=0 and passing

Naddika [18.5K]

The equation of the pair of lines perpendicular to the lines given equation is $b x^{2}-2 h x y+a y^{2}=0$ .

Solution:

Given equation is $a x^{2}+2 h x y+b y^{2}=0$ .

Let $m_1$ and $m_2$ be the slopes of the given lines.

Sum of the roots = $-\frac{\text {coefficient of } x y}{\text {coefficient of } y^{2}}$

$$m_1+m_2=\frac{-2h}{b}$ – – – – – (1)

Product of the roots = $-\frac{\text {coefficient of } x^2}{\text {coefficient of } y^{2}}$

$$m_1 \cdot m_2=\frac{a}{b}$ – – – – – (2)

The required lines are perpendicular to these lines.

Slopes of the required lines are $$-\frac{1}{m_{1}} \text { and }-\frac{1}{m_{2}}$

Required lines also passes through the origin,

therefore their y-intercepts are 0.

Hence their equations are:

$$y=-\frac{1}{m_{1}}x \text { and }y=-\frac{1}{m_{2}}x$

Do cross multiplication, we get

$m_1y=-x \ \text{and} \ m_2y=-x$

Add x on both sides of the equation, we get

$x+m_1y=0 \ \text{and} \ x+m_2y=0$

Therefore, the joint equation of the line is

$\left(x+m_{1} y\right)\left(x+m_{2} y\right)=0$

$x^2+m_2xy+m_1xy+m_1m_2y^2=0$

$x^{2}+\left(m_{1}+m_{2}\right) x y+m_{1} m_{2} y^{2}=0$

Substitute (1) and (2), we get

$$x^{2}+\left(\frac{-2 h}{b}\right) x y+\left(\frac{a}{b}\right) y^{2}=0$

To make the denominator same, multiply and divide first term by b.

$$ \frac{b}{b} x^{2}+\left(\frac{-2 h}{b}\right) x y+\left(\frac{a}{b}\right) y^{2}=0$

$$\frac{bx^2-2hxy+ay^2}{b} = 0$

Do cross multiplication, we get

$b x^{2}-2 h x y+a y^{2}=0$

Hence equation of the pair of lines perpendicular to the lines given equation is $b x^{2}-2 h x y+a y^{2}=0$ .

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4 years ago

Find the area of a triangle with base 21 inches and the height 8 inches

ankoles [38]

Answer:

84in²

Step-by-step explanation:

this is very easy

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