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2 years ago

8

The Spanish club sold hot pretzels as a fundraiser. The pretzels normally sold for $2.00, but near the end of the sale the price

was reduced by 25%. What was the new price for a hot pretzel?

1 answer:

Maurinko [17]2 years ago

5 0

The new price is 1.50$

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Triangle DEF has vertices D(−4, 1), E(2, 3), and F(2, 1) and is dilated by a factor of 3 using the point (1, 1) as the point of

CaHeK987 [17]

Answer: $D'(-14,1);\ E'(4,7);\ F'(4,1)$

Step-by-step explanation:

Since the center of dilation is not at the origin, we can use the following formula in order to find the coordinates of the vertices of the triangle D'E'F':

$D_{O,k}(x,y)=(k(x-a)+a, k(y-b)+b)$

Where "O" is the center of dilation at (a,b) and "k" is the scale factor.

In this case you can identify that:

$(a,b)=(1,1)\\k=3$

Therefore, susbtituting values into the formula shown above, you get that the coordinates ot the resulting triangle D'E'F, are the following:

Vertex D' → $(3(-4-1)+1,\ 3(1-1)+1)=(-14,1)$

Vertex E' → $(3(2-1)+1,\ 3(3-1)+1)=(4,7)$

Vertex F' → $(3(2-1)+1,\ 3(1-1)+1)=(4,1)$

6 0

3 years ago

The diagram shows corresponding lengths in two similar figures. Find the area of the smaller figure. 240cm

7nadin3 [17]

Hey! My last name's Alvarez too! Anyway, the answer came out to 180cm squared. Hope I helped! Since you can barely get more than one or two answers on this site

5 0

3 years ago

How to covert feet into miles

Naya [18.7K]

1 mile is equal to 5280 feet

5 0

2 years ago

Find an equation of the plane through the point (−5,−1,2) with normal vector ????=⟨−1,−5,2⟩.

Ganezh [65]

Answer:

$x + 5y - 2z + 14 = 0$

Step-by-step explanation:

A plane that passes through the point $(x_{0}, y_{0}, z_{0})$ with a normal vector of $(a,b,c)$ has the following equation, initially:

$a(x - x_{0}) + b(y - y_{0}) + c(z - z_{0}) = 0$

After we solve this, we have

$ax + by + cz + d = 0$

In this problem, we have that:

$(x_{0}, y_{0}, z_{0}) = (-5,-1,2)$

$(a,b,c) = (−1,−5,2)$

So

$a(x - x_{0}) + b(y - y_{0}) + c(z - z_{0}) = 0$

$-1(x - (-5)) - 5(y - (-1)) + 2(z - 2) = 0$

$-(x + 5) - 5(y + 1) + 2(z - 2) = 0$

$-x - 5 - 5y - 5 + 2z - 4 = 0$

$-x - 5y + 2z - 14 = 0$

Multplying everything by -1

$x + 5y - 2z + 14 = 0$

3 0

3 years ago

Match the cofactors to their corresponding entries in the matrix.

Masja [62]

Answer:

The co-factors and their values are shown in the table below.

Step-by-step explanation:

We are given the matrix, $\begin{bmatrix}3&7&1\\7&1&-3\\8&5&1\end{bmatrix}$ .

It is required to match the co-factors with the corresponding values.

As, the co-factors are given by,

$A_{c_{ij}}$ = $(-1)^{i+j}(d)$ , where the d= determinant of the matrix after removing the i- row and j- column.

So, we have,

1. $A_{c_{11}}$ .

So, $A_{c_{11}}=(-1)^{1+1}(1\times 1-5\times (-3))$

i.e. $A_{c_{11}}=(-1)^{2}(1+15)=16$

2. $A_{c_{13}}$ .

So, $A_{c_{13}}=(-1)^{1+3}(7\times 5-1\times 8)$

i.e. $A_{c_{13}}=(-1)^{4}(35-8)=27$

3. $A_{c_{21}}$ .

So, $A_{c_{21}}=(-1)^{2+1}(7\times 1-5\times 1)$

i.e. $A_{c_{21}}=(-1)^{3}(7-5)=-2$

4. $A_{c_{22}}$ .

So, $A_{c_{22}}=(-1)^{2+2}(3\times 1-8\times 1)$

i.e. $A_{c_{22}}=(-1)^{4}(3-8)=-5$

5. $A_{c_{31}}$ .

So, $A_{c_{31}}=(-1)^{3+1}(7\times (-3)-1\times 1)$

i.e. $A_{c_{31}}=(-1)^{4}(-21-1)=-22$

Thus, we get,

Co-factor Value

$A_{c_{11}}$ 16

$A_{c_{13}}$ 27

$A_{c_{21}}$ -2

$A_{c_{22}}$ -5

$A_{c_{31}}$ -22

4 0

3 years ago