Do these equations have an identity or no solution?

Find the Value of each variable that makes the equation true.

alexdok [17]

1.) 0
2.) -2
3.) -3

Please comment for concerns please likeb

8 0

2 years ago

Tina was asked to determine the possible dimensions of a given rectangle whose area is 12a4b3-18a6b3-72a7b3. Tina stated that th

Rufina [12.5K]

Answer:

Tina is correct

Step-by-step explanation:

Given

$Area = 12a^4b^3 - 18a^6b^3 - 72a^7b^3$

Required

State if $3a^4b^3(4-6a^2-24a^3)$ is a possible dimension

To do this, we simply expand $3a^4b^3(4-6a^2-24a^3)$

$3a^4b^3(4-6a^2-24a^3)$

$3a^4b^3 * 4-3a^4b^3 * 6a^2-3a^4b^3 * 24a^3$

$12a^4b^3 - 18a^{4+2}b^3 * -72a^{4+3}b^3$

$12a^4b^3 - 18a^{6}b^3 * -72a^{7}b^3$

By comparison, the result of the expansion

$12a^4b^3 - 18a^{6}b^3 * -72a^{7}b^3$

and the given expression

$Area = 12a^4b^3 - 18a^6b^3 - 72a^7b^3$

are the same.

Hence, Tina is correct

7 0

2 years ago

How many triangles does a=6 b=10 A=33° create?

GaryK [48]

Answer:

2 triangles are possible.

Step-by-step explanation:

Given

a=6

b=10

$\angle$ A=33°

To find:

Number of triangles possible ?

Solution:

First of all, let us use the sine rule:

As per Sine Rule:

$\dfrac{a}{sinA}=\dfrac{b}{sinB}$

And let us find the angle B.

$\dfrac{6}{sin33}=\dfrac{10}{sinB}\\sinB = \dfrac{10}{6}\times sin33\\B =sin^{-1}(1.67 \times 0.545)\\B =sin^{-1}(0.9095) =65.44^\circ$

This value is in the 1st quadrant i.e. acute angle.

One more value for B is possible in the 2nd quadrant i.e. obtuse angle which is: 180 - 65.44 = $114.56^\circ$

For the value of $\angle B = 65.44^\circ$ , let us find $\angle C$ :

$\angle A+\angle B+\angle C = 180^\circ\\\Rightarrow 33+65.44+\angle C = 180\\\Rightarrow \angle C = 180-98.44 = 81.56^\circ$

Let us find side c using sine rule again:

$\dfrac{6}{sin33}=\dfrac{c}{sin81.56^\circ}\\\Rightarrow c = 11.02 \times sin81.56^\circ = 10.89$

So, one possible triangle is:

a = 6, b = 10, c = 10.89

$\angle$ A=33°, $\angle$ A=65.44°, $\angle$ C=81.56°

For the value of $\angle B =$ $114.56^\circ$ , let us find $\angle C$ :

$\angle A+\angle B+\angle C = 180^\circ\\\Rightarrow 33+114.56+\angle C = 180\\\Rightarrow \angle C = 180-147.56 = 32.44^\circ$

Let us find side c using sine rule again:

$\dfrac{6}{sin33}=\dfrac{c}{sin32.44^\circ}\\\Rightarrow c = 11.02 \times sin32.44^\circ = 5.91$

So, second possible triangle is:

a = 6, b = 10, c = 5.91

$\angle$ A=33°, $\angle$ A=114.56°, $\angle$ C=32.44°

So, answer is : 2 triangles are possible.

8 0

3 years ago

Simply the expressions 12(x-6)+6(7+x)

andrew11 [14]

Answer:

18x-30

Step-by-step explanation:

7 0

3 years ago

PLEASE HELP OMG PLEASE IM LITERALLY BEGGING ANYONE WHAT DO I DO

Veseljchak [2.6K]

Answer:

Bro every school in every country in every place in the world does different things for their first semester, not every school does the exact same thing. I recommend just learning everything you can when it comes to things like maths, science, english, etc. That way you might be ahead of everyone, which is always better than being behind.

Step-by-step explanation:

5 0

2 years ago

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