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4 years ago

Can you prove the two triangles congruent? By which postulate?

Mathematics

1 answer:

SpyIntel [72]4 years ago

6 0

Given 2 equal sides and a non-included equal angle as in the diagram are not enough to prove congruency

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How do I find his out

Setler [38]

You would divide 1/16 by 3/8

3 0

4 years ago

Find an equation for y in terms of x

snow_tiger [21]

Answer:

y = $\frac{9}{x^2}$

Step-by-step explanation:

Given y is inversely proportional to x² then the equation relating them is

y = $\frac{k}{x^2}$ ← k is the constant of proportion

To find k use an ordered pair from the table

Using x = 3 when y = 1 , then

1 = $\frac{k}{3^2}$ = $\frac{k}{9}$ ( multiply both sides by 9 )

9 = k

y = $\frac{9}{x^2}$ ← equation of proportion

3 0

3 years ago

33 is 15less than k.

RoseWind [281]

Answer:

k = 48

Step-by-step explanation:

Given expression:

33 is 15 less than k

Convert this into equation and solve for k:

33 = k - 15
k = 33 + 15
k = 48

So the <u>value of k is 48</u>

4 0

2 years ago

Read 2 more answers

Express the form n:1 give n as a decimal 14:8

Alexus [3.1K]

I’m pretty sure it’s 1. 8:1

6 0

3 years ago

What is the quotient of (x3 - 3x2 + 5x – 3) = (x - 1)?

IRINA_888 [86]

Note: Consider "÷" sign instead of "=".

Given:

$(x^3-3x^2+5x-3)\div (x-1)$

To find:

The quotient.

Solution:

We have,

$(x^3-3x^2+5x-3)\div (x-1)$

It can be written as

$=\dfrac{x^3-3x^2+5x-3}{x-1}$

Splitting the middle terms, we get

$=\dfrac{x^3-x^2-2x^2+2x+3x-3}{x-1}$

$=\dfrac{x^2(x-1)-2x(x-1)+3(x-1)}{x-1}$

Taking out the common factor (x-1), we get

$=\dfrac{(x-1)(x^2-2x+3)}{x-1}$

Cancel the common factors.

$=x^2-2x+3$

The quotient is $x^2-2x+3$ .

Therefore, the correct option is D.

7 0

3 years ago

Read 2 more answers