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3 years ago

14

A box of popsicles contains 4 of each flavor: cherry, lemon-lime, grape, and orange. Find the probability of one person randomly

picking a cherry popsicle then another person selecting a grape popsicle from the box.
Group of answer choices

1/8

1/4

1/15

1/20

1 answer:

sashaice [31]3 years ago

8 0

because if u pick a grape that was 25% chance and then pick another grape that is 1/8

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What’s the midpoint of the line segment joining A and B<br><br> A(2,-5);B(6,1)

IRISSAK [1]

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{2}~,~\stackrel{y_1}{-5})\qquad B(\stackrel{x_2}{6}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{6+2}{2}~~,~~\cfrac{1-5}{2} \right)\implies \left( \cfrac{8}{4}~,~\cfrac{-4}{2} \right)\implies (4,-2)$

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2 years ago

NEED HELP EMERGENCY ‼️

cestrela7 [59]

Answer:

A. Reflect across the y-axis

Step-by-step explanation:

if the negative sign was outside the radical then it would've been a reflection on the x-axis.

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3 years ago

Read 2 more answers

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 $\mathbb{R}^2$

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3 years ago

Find the perimeter of the following triangle.

pashok25 [27]

Hello!

To find the perimeter of the triangle, we need to find the length of all the sides using the <u>distance formula</u>.

The distance formula is: $d =\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ .

First, we can find the distance between the points (-3, -1) and (2, -1). The point (-3, -1) can be assigned to $(x_{1},y_{1})$ , and (2, -1) is assigned to $(x_{2},y_{2})$ . Then, substitute the values into the formula.

$d =\sqrt{(2 - (-3))^{2}+ (-1 - (-1))^{2}}$

$d =\sqrt{5^{2}+0^{2}}$

$d =\sqrt{25} = 5$

The distance between the points (-3, -1) and (2, -1) is 5 units.

Secondly, we need to find the distance between the points (2, 3) and (2, -1). Assign those points to $(x_{1},y_{1})$ and $(x_{2},y_{2})$ , then substitute it into the formula.

$d =\sqrt{(2 - 2)^{2}+ (-1 - 3)^{2}}$

$d =\sqrt{0^{2}+(-4)^{2}}$

$d =\sqrt{16} = 4$

The distance between the two points (2, 3) and (2, -1) is 4 units.

Finally, we use the distance formula again to find the distance between the points (-3, -1) and (2, 3). Remember the assign the ordered pairs to $(x_{1},y_{1})$ and $(x_{2},y_{2})$ and substitute!

$d =\sqrt{(2 -(-3))^{2}+ (3 - (-1))^{2}}$

$d =\sqrt{5^{2}+4^{2}}$

$d =\sqrt{25 + 16}$

$d =\sqrt{41}$ This is equal to approximately 6.40 units.

The last step is to find the perimeter. To find the perimeter, add of the three sides of the triangle together.

P = 5 units + 4 units + 6.4 units

P = 15.4 units

Therefore, the perimeter of this triangle is choice A, 15.4.

6 0

3 years ago

Write an absolute value equation that represents these heights 10 and 15

djverab [1.8K]

Answer:

An equation for the absolute value that represents the midpoint of 10 and 15 would be (10 + 15)/2 = 12.5

6 0

3 years ago