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3 years ago

Describe how to find all the points on a baseball field that are equidistant from second base and third base

Mathematics

1 answer:

BlackZzzverrR [31]3 years ago

8 0

This might help explain how to do a very similar problem.... http://www.mathopenref.com/constbisectline.html

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What's the Answer? Provide steps please

dalvyx [7]

Answer:

0.24930286...

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Graph a line with a slope of -3 that contains the point (4, -2)

Crank

Answer:

Step-by-step explanation:

Given that

Slope m=-3

A point (x, y)=(4,-2)

Equation of a line can generally be written as

y=mx+c

Where, m is the slope

And c is the intercept on y-axis

Since we are given m=-3

Then, y=mx+c becomes

y=-3x+c

To get c, let substitute the point given, i.e x=4 and y=-2

y=-3x+c

-2=-3(4)+c

-2=-12+c

-2+12=c

c=-2+12

c=10

y=mx+c, then m=-3 and c=10

Then, the equation the line becomes y = -3x + 10

This is the required line equation

5 0

3 years ago

Rationalise the denominator.

grin007 [14]

Let's see what to do buddy...

_________________________________

To do this, we have to multiply the face and denominator of the fraction by the denominator conjunction, which is :

$(5 -\sqrt{2})$

°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°

Reminder :

$(a + b)(a - b) = {a}^{2} - {b}^{2} \\$

°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°

So we have :

$\frac{1}{ 5-\sqrt{2} } \times \frac{ 5+\sqrt{2} }{ 5 +\sqrt{2} } = \frac{1 \times ( 5 +\sqrt{2}) }{(5- \sqrt{2} )( 5 +\sqrt{2}) } = \\$

The other is in the photo.

_________________________________

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

7 0

3 years ago

Special Right Trianglws

Masteriza [31]

Answer:

x = $2\sqrt7$

y = $2\sqrt7$

Step-by-step explanation:

Hello!

This is a right isosceles triangle (45°-45°-90°). You can solve for the missing angle by simply subtracting 45° and 90° from 180° (sum of angles in a triangle equals 180°)

Since the triangle is isosceles, x and y are congruent because they are the legs opposite to the congruent angles.

In a right isosceles triangle, the hypotenuse is always the measure of one leg multiplied by $\sqrt2$ .

So, let's use leg y, $y\sqrt2 = 2\sqrt{14}$

Solve for y:

$y\sqrt2 = 2\sqrt{14}\\\\y = $\frac{2\sqrt{14}}{\sqrt2}$\\\\y = 2\sqrt7$

y = $2\sqrt7$ , which means that x is also $2\sqrt7$

3 0

3 years ago

What is the area of this whole figure in square units?

IrinaK [193]

Well, it is a good and interesting question

Such areas could be calculated in a single shoot and also by dividing the whole shape into other shapes and the total area would be the sum of these areas

I prefer the way of a single shoot ...

The figure represents a trapezoid:
its lower base = 21 units
its upper base = 12 units
the normal height between them = 8 units

NOW

The area of a trapezoid = [(sum of bases' lengths) ÷ 2] * height
= [(12 + 21) ÷ 2] * 8 = 132 sq units

Hope that helps

7 0

3 years ago