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3 years ago

If a ramp has a slope less than 1 but greater than 0, is it steeper than a ramp with a slope of 1

Mathematics

1 answer:

allochka39001 [22]3 years ago

7 0

No the larger the number the steeper the slope. 1 would be a steeper slope than a number less than 1.

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If there are 4 people, and they are sharing $3000.how much will each person get?

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2 years ago

What is the answer to this in cm

jasenka [17]

The hardest part is finding the length of the curve at the top. To find arc length, multiply the circumference by the arc angle divided by full angle (360°) in a circle. Or:
$s = (2\pi r)( \frac{x}{360})$
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3 years ago

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aivan3 [116]

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4 years ago

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What is used in place of a word or words??

Makovka662 [10]

Answer:

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Step-by-step explanation:

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6 0

3 years ago

Plz help with full process

BabaBlast [244]

Answer:

$\displaystyle \frac{(a - b)^{2} - {c}^{2} }{{a}^{2} - {(b+ c)}^{2} } + \frac{(b - c)^{2} - {a}^{2} }{{b}^{2} - {(c+ a)}^{2} } + \frac{(c - a)^{2} - {b}^{2} }{{c}^{2} - {(a+ b)}^{2} }=1$

Step-by-step explanation:

<u>Algebra Simplifying</u>

We are given the expression:

$\displaystyle T=\frac{(a - b)^{2} - {c}^{2} }{{a}^{2} - {(b+ c)}^{2} } + \frac{(b - c)^{2} - {a}^{2} }{{b}^{2} - {(c+ a)}^{2} } + \frac{(c - a)^{2} - {b}^{2} }{{c}^{2} - {(a+ b)}^{2} }$

We need to repeatedly use the following identity:

$(a^2-b^2)=(a-b)(a+b)$

For example, the first numerator has a difference of squares thus we factor as:

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

The first denominator can be factored also:

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

Applying the same procedure to all the expressions:

$\displaystyle T=\frac{(a - b- c)(a-b+c) }{a-b-c)(a+b+c) } + \frac{(b - c - a)(b-c+a) }{(b-c-a)(b+c+a) } + \frac{(c - a-b)(c-a+b) }{(c-a-b)(c+a+b)}$

Simplifying all the fractions:

$\displaystyle T=\frac{a - b- c}{a+b+c} + \frac{b-c+a}{b+c+a } + \frac{c-a+b }{c+a+b}$

Since all the denominators are equal:

$\displaystyle T=\frac{a - b+ c+b-c+a+c-a+b}{a+b+c}$

Simplifying:

$\displaystyle T=\frac{a +c+b}{a+b+c}$

Simplifying again:

T = 1

Thus:

$\boxed{\displaystyle \frac{(a - b)^{2} - {c}^{2} }{{a}^{2} - {(b+ c)}^{2} } + \frac{(b - c)^{2} - {a}^{2} }{{b}^{2} - {(c+ a)}^{2} } + \frac{(c - a)^{2} - {b}^{2} }{{c}^{2} - {(a+ b)}^{2} }=1}$

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3 years ago

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