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3 years ago

Need help :(................................

Mathematics

1 answer:

Zanzabum3 years ago

7 0

It is the transitive property. C It's actually a very good example of the transitive property.

The associative property would be (a + b) + c = a + (b + c) If you add a and b together and then add on c, that's the same thing as adding b + c together and then add a to it.

The Commutitive property would be a + b = b + a. Again it does not matter which entry you put into your calculator first.

The symmetry property is a = b or b = a. They are interchangeable. That's not what you are given.

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Given: △ABC, m∠A=60° m∠C=45°, AB=8 Find: Perimeter of △ABC, Area of △ABC . FIRST CORRECT ANSWER GETS POINTS AND BRAINLIEST!!!! T

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Answer:

Given : In △ABC, m∠A=60°, m∠C=45°,and AB=8 unit

Firstly, find the angles B

Sum of measures of the three angles of any triangle equal to the straight angle, and also expressed as 180 degree

∴m∠A+ m∠B+m∠C=180 ......[1]

Substitute the values of m∠A=60° and m∠C=45° in [1]

$60^{\circ}+ m\angle B+45^{\circ}=180^{\circ}$

$105^{\circ}+ m\angle B=180^{\circ}$

Simplify:

$m\angle B=75^{\circ}$

Now, find the sides of BC

For this, we can use law of sines,

Law of sine rule is an equation relating the lengths of the sides of a triangle to the sines of its angles.

$\frac{\sin A}{BC} = \frac{\sin C}{AB}$

Substitute the values of ∠A=60°, ∠C=45°,and AB=8 unit to find BC.

$\frac{\sin 60^{\circ}}{BC} =\frac{\sin 45^{\circ}}{8}$

then,

$BC = 8 \cdot \frac{\sin 60^{\circ}}{\sin 45^{\circ}}$

$BC=8 \cdot \frac{0.866025405}{0.707106781} =9.798$ unit

Similarly for AC:

$\frac{\sin B}{AC} = \frac{\sin C}{AB}$

Substitute the values of ∠B=75°, ∠C=45°,and AB=8 unit to find AC.

$\frac{\sin 75^{\circ}}{AC} =\frac{\sin 45^{\circ}}{8}$

then,

$AC = 8 \cdot \frac{\sin 75^{\circ}}{\sin 45^{\circ}}$

$AC=8 \cdot \frac{0.96592582628}{0.707106781} =10.9283$ unit

To find the perimeter of triangle ABC;

Perimeter = Sum of the sides of a triangle

i,e

Perimeter of △ABC = AB+BC+AC = 8 +9.798+10.9283 = 28.726 unit.

To find the area(A) of triangle ABC ;

Use the formula:

$A = \frac{1}{2} \times AB \times AC \times \sin A$

Substitute the values in above formula to get area;

$A=\frac{1}{2} \times 8 \times 10.9283 \times \sin 60^{\circ}$

$A = 4 \times 10.9283 \times 0.86602540378$

Simplify:

Area of triangle ABC = 37.856 (approx) square unit

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Assume {v1, . . . , vn} is a basis of a vector space V , and T : V ------> W is an isomorphism where W is another vector spac

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Answer:

Step-by-step explanation:

To prove that $w_1,\dots w_n$ form a basis for W, we must check that this set is a set of linearly independent vector and it generates the whole space W. We are given that T is an isomorphism. That is, T is injective and surjective. A linear transformation is injective if and only if it maps the zero of the domain vector space to the codomain's zero and that is the only vector that is mapped to 0. Also, a linear transformation is surjective if for every vector w in W there exists v in V such that T(v) =w

Recall that the set $w_1,\dots w_n$ is linearly independent if and only if the equation

$\lambda_1w_1+\dots \lambda_n w_n=0$ implies that

$\lambda_1 = \cdots = \lambda_n$ .

Recall that $w_i = T(v_i)$ for i=1,...,n. Consider $T^{-1}$ to be the inverse transformation of T. Consider the equation

$\lambda_1w_1+\dots \lambda_n w_n=0$

If we apply $T^{-1}$ to this equation, then, we get

$T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) =T^{-1}(0) = 0$

Since T is linear, its inverse is also linear, hence

$T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) = \lambda_1T^{-1}(w_1)+\dots + \lambda_nT^{-1}(w_n)=0$

which is equivalent to the equation

$\lambda_1v_1+\dots + \lambda_nv_n =0$

Since $v_1,\dots,v_n$ are linearly independt, this implies that $\lambda_1=\dots \lambda_n =0$ , so the set $\{w_1, \dots, w_n\}$ is linearly independent.

Now, we will prove that this set generates W. To do so, let w be a vector in W. We must prove that there exist $a_1, \dots a_n$ such that

$w = a_1w_1+\dots+a_nw_n$

Since T is surjective, there exists a vector v in V such that T(v) = w. Since $v_1,\dots, v_n$ is a basis of v, there exist $a_1,\dots a_n$ , such that

$a_1v_1+\dots a_nv_n=v$

Then, applying T on both sides, we have that

$T(a_1v_1+\dots a_nv_n)=a_1T(v_1)+\dots a_n T(v_n) = a_1w_1+\dots a_n w_n= T(v) =w$

which proves that $w_1,\dots w_n$ generate the whole space W. Hence, the set $\{w_1, \dots, w_n\}$ is a basis of W.

Consider the linear transformation $T:\mathbb{R}^2\to \mathbb{R}^2$ , given by T(x,y) = T(x,0). This transformations fails to be injective, since T(1,2) = T(1,3) = (1,0). Consider the base of $\mathbb{R}^2$ given by (1,0), (0,1). We have that T(1,0) = (1,0), T(0,1) = (0,0). This set is not linearly independent, and hence cannot be a base of $\mathbb{R}^2$