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3 years ago

19g - 4h + 4 - 20(g - 1) Can you plss help me

Mathematics

1 answer:

alukav5142 [94]3 years ago

8 0

Answer:

I think its A

Step-by-step explanation:

correct me if im wrong

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Which other angles could be in that triangle?

CaHeK987 [17]

The other angles in the isosceles triangle with an angle of 100° will be 40°

Step-by-step explanation:

Lets define an isosceles triangle first.

"An isosceles triangle is a triangle with two equal sides and two equal angles.

Given that an angle of the triangle is 100°

We know that the sum of internal angles of a triangle is 180°

The sum of remaining two angles is:

=180°-100°

=80°

As the triangle is an isosceles triangle, the two angles will be equal.

So the angles will be:

$=\frac{80}{2}\\=40$

The other angles in the isosceles triangle with an angle of 100° will be 40°

Keywords: Triangle, isosceles triangle

Learn more about isosceles triangle at:

#LearnwithBrainly

7 0

3 years ago

Which expression does not represent "the sum of n and 6"

nignag [31]

The expression that does not represent the sum of n and 6 is 6n

6 0

3 years ago

Read 2 more answers

What is the measure of the central angle of a circle with radius 60 cm the intercepts a 15 pi centimeters arc

alekssr [168]

The measure of the central angle of a circle is 45⁰

<h3>What is Angle?</h3>

A figure which is formed by two rays or lines that shares a common endpoint is called an angle.

Here,

Theta (angle) θ = arc length / radius

θ = 15 π / 60 cm

θ = 0.79 radians

Convert to degrees

θ = 0.79 radians = 0.79 X 180/ π

= 0.79 X 180/ 3.14

= 45⁰

Thus, the measure of the central angle of a circle is 45⁰.

Learn more about Angle from:

brainly.com/question/7116550

#SPJ1

6 0

1 year ago

The bases of a trapezoid are 11 meters and 14 meters. Its height is 10 meters. What is the area of the trapezoid?

kenny6666 [7]

Area of Trapezoid = (1/2) * (Sum of bases) * height

Area = (1/2) * (11 + 14) * 10

= (1/2) * 25 * 10 = 5 *25 = 125

Area = 125 m²

4 0

3 years ago

Rationalise the denominator of:<br>1/(√3 + √5 - √2)

Paul [167]

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} + \sqrt{5} - \sqrt{2} }$

can be re-arranged as

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} - \sqrt{2} + \sqrt{5} }$

$\rm \: = \: \dfrac{1}{( \sqrt{3} - \sqrt{2} ) + \sqrt{5} }$

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{1}{( \sqrt{3} - \sqrt{2} ) + \sqrt{5} } \times \dfrac{( \sqrt{3} - \sqrt{2} ) - \sqrt{5} }{( \sqrt{3} - \sqrt{2} ) - \sqrt{5} }$

We know,

$\rm :\longmapsto\:\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}$

So, using this, we get

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{ {( \sqrt{3} - \sqrt{2} )}^{2} - {( \sqrt{5}) }^{2} }$

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{3 + 2 - 2 \sqrt{6} - 5}$

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{5 - 2 \sqrt{6} - 5}$

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{ - 2 \sqrt{6}}$

$\rm \: = \: \dfrac{ - ( - \sqrt{3} + \sqrt{2} + \sqrt{5}) }{ - 2 \sqrt{6}}$

$\rm \: = \: \dfrac{- \sqrt{3} + \sqrt{2} + \sqrt{5}}{2 \sqrt{6}}$

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{- \sqrt{3} + \sqrt{2} + \sqrt{5}}{2 \sqrt{6}} \times \dfrac{ \sqrt{6} }{ \sqrt{6} }$

$\rm \: = \: \dfrac{- \sqrt{18} + \sqrt{12} + \sqrt{30}}{2 \times 6}$

$\rm \: = \: \dfrac{- \sqrt{3 \times 3 \times 2} + \sqrt{2 \times 2 \times 3} + \sqrt{30}}{12}$

$\rm \: = \: \dfrac{- 3\sqrt{2} + 2 \sqrt{3} + \sqrt{30}}{12}$

Hence,

$\boxed{\tt{ \rm \dfrac{1}{ \sqrt{3} + \sqrt{5} - \sqrt{2} } =\dfrac{- \sqrt{3 \times 3 \times 2} + \sqrt{2 \times 2 \times 3} + \sqrt{30}}{12}}}$

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<h3><u>More Identities to </u><u>know:</u></h3>

$\purple{\boxed{\tt{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2}}}}$

$\purple{\boxed{\tt{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}}}}$

$\purple{\boxed{\tt{ {(x + y)}^{3} = {x}^{3} + 3xy(x + y) + {y}^{3}}}}$

$\purple{\boxed{\tt{ {(x - y)}^{3} = {x}^{3} - 3xy(x - y) - {y}^{3}}}}$

$\pink{\boxed{\tt{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}}}$

$\pink{\boxed{\tt{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}}}$

6 0

3 years ago