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4 years ago

7

What is the length of leg s of the triangle below?

2 answers:

allsm [11]4 years ago

8 0

B because in a 45-45-90 triangle, the leg lengths are the same (isosceles).

OverLord2011 [107]4 years ago

6 0

Answer:

Length of leg s is 3 units.

Step-by-step explanation:

Given the 45-45-90 triangle, in which length of hypotenuse is $\sqrt{18}$ and one leg is 3 units.

we have to find the length of other leg i.e leg s

As two angles of triangle are equal implies given triangle is isosceles.

As the two sides of isosceles triangle are equal.

Hence, the length of leg s must be equal to the length of other leg which is given 3 units.

Hence, length of leg s is 3 units.

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3 years ago

Read 2 more answers

if an airplane travels 660 mph how long will it take an airplane to travel to a distance of 2970 miles?

vazorg [7]

Answer:

4 hours and 30 minutes.

Step-by-step explanation:

660 miles = 1 hour
1320 miles = 2 hours
1980 miles = 3 hours
2640 miles = 4 hours

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So we know that it will be taking 4 hours, you would then add the 30 minutes which gives you 4 hours and 30 minutes.

Hope this helps!

4 0

2 years ago

The product of a number and 9 is -450

Softa [21]

Answer:

-50

Well to get the answer of the product of a number and 9 is -450. You must divide 9 by -450 using long divison.

6 0

3 years ago

Experimentally verify sum of three angles of an triangle is 180⁰.

svetlana [45]

Draw line a through points A and B. Draw line b through point C and parallel to line a.

Since lines a and b are parallel, <)BAC = <)B'CA and <)ABC = <)BCA'.

It is obvious that <)B'CA + <)ACB + <)BCA' = 180 degrees.

Thus <)ABC + <)BCA + <)CAB = 180 degrees.

Lemma

If ABCD is a quadrilateral and <)CAB = <)DCA then AB and DC are parallel.

Proof

Assume to the contrary that AB and DC are not parallel.

Draw a line trough A and B and draw a line trough D and C.

These lines are not parallel so they cross at one point. Call this point E.

Notice that <)AEC is greater than 0.

Since <)CAB = <)DCA, <)CAE + <)ACE = 180 degrees.

Hence <)AEC + <)CAE + <)ACE is greater than 180 degrees.

Contradiction. This completes the proof.

Definition

Two Triangles ABC and A'B'C' are congruent if and only if

|AB| = |A'B'|, |AC| = |A'C'|, |BC| = |B'C'| and,

<)ABC = <)A'B'C', <)BCA = <)B'C'A', <)CAB = <)C'A'B'.

4 0

3 years ago

15 points for someone who can answer & explain this to me thanks

iris [78.8K]

Answer:

see explanation

Step-by-step explanation:

Given

$\frac{9}{a^2+9a+20}$ , $\frac{7a}{a^2+11a+28}$

Factor the denominators of both fractions

$\frac{9}{(a+4)(a+5)}$ , $\frac{7a}{(a+4)(a+7)}$

The lowest common denominator of both fractions is (a + 4)(a + 5)(a + 7)

Multiply numerator/denominator of first fraction by (a + 7)

Multiply numerator/denominator of second fraction by (a + 5 )

$\frac{9(a+7)}{(a+4)(a+5)(a+7)}$ , $\frac{7a(a+5)}{(a+4)(a+5)(a+7)}$ , then

$\frac{9a+63}{(a+4)(a+5)(a+7)}$ , $\frac{7a^2+35a}{(a+4)(a+5)(a+7)}$

8 0

3 years ago