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2 years ago

What fraction of the game did Danrnels team score 2 or more runs

Mathematics

1 answer:

Paraphin [41]2 years ago

3 0

Answer:the anwser is 2/4ths

Step-by-step explanation:

your welcome

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Wich is greater, 0.04 or 0.4

neonofarm [45]

0.4 is greater because 0.04 is has a zero in front of it

7 0

3 years ago

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Find the slope of the line that passes through (-33, -27) and (-32, -25).

Snezhnost [94]

Answer: 2.

Step-by-step explanation:

You can find the slope between two points by using the "rise-over-run" method.

This method essentially calculates the slope as the quotient of the vertical span and horizontal span of the two points.

You can find it through this formula:

$s = \frac{y2 - y1}{x2 - x1}$

To find the slope, use two points, (x1, y1) and (x2 and y2).

For the points given, I will replace those values in the formula with (-33, -27) and (-32, -25).

$s = \frac{-27 - (-25)}{-33 - (-32)}\\\\s = \frac{-2}{-1}\\\\s = 2$

The slope of the line is 2.

6 0

2 years ago

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Planet X is 1/2 of a light-year away from Earth. Planet Y is 3/10 of a light-year away from Earth. How much farther away is Plan

kakasveta [241]

Answer:

1/2= 5/10

5/10 - 3/10 = 2/10 or 1/5

so planet X is 1/5 of a light-year closer

Step-by-step explanation:

8 0

2 years ago

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What is the definition of standard form of an equation?

Talja [164]

Standard form of equation is to be defined as: to write the equation of a line in standard form. Definitions: Standard Form: the standard form of a line is in the form Ax + By = C where A is a positive integer, and B, and C are integers

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3 years ago

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

3 years ago