Giving 25 points please hurry

Please help 4 math pages with questions. 1 has all the circle graphs, and all the other 3 have each circle graphs question. Plea

oksano4ka [1.4K]

Answer:

Favorite chip Answers v

Step-by-step explanation:

1. How many were surveyed? 82+8+18+62=100

2. How many said Jalepeno was their favorite? 8/100 0r 2/25 or 8%

3. How many said BBQ was their favorite?62/100 or 31/50 or 62%

4. How would it effect your final answer if the percentages did not add up to 100%? This means that your percentages are wrong. In a circle graph the percentages always equal 100%!

5. Is it possible to determine a missing percentage in a circle graph? Yes, we can always assume the total percentage is 100 so you add your given percentages together and then subtract that total from 100 to find the missing percentage.

8 0

2 years ago

I need help with this question

den301095 [7]

Answer:

m

Step-by-step explanation:

6 0

3 years ago

A 14-foot ladder is leaning against a building as shown. It touches a point 11 feet up on a building. How far away from the base

Luden [163]

We can assume that the point the ladder creates with the ground and building is a triangle. You can use the Pythagorean theorem to solve this.

A^2 + B^2 = C^2

The ladder is C, and the building can act as A or B, so for the purpose of this explanation, I’ll make it A.

11^2 + B^2 = 14^2
Figure out the squares

121 + B^2 = 196
Subtract 121 from both sides

B^2 = 75
Square root B^2 and 75

B = 5 root3

6 0

3 years ago

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

Degger [83]

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 $\mathbb{R}^2$

8 0

3 years ago

If m∠A = m∠B and m∠A + m∠C = m∠D, then<br> m∠B + m∠C = m∠D.

just olya [345]

Answer:

True. The statement is also equivalent to: If m<B + m<C does not equal m<D, then m<A does not equal m<B and m<A + m<C is not equal to m<D.

8 0

3 years ago