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irga5000 [103]
3 years ago
8

Someone please help me ASAP

Mathematics
1 answer:
givi [52]3 years ago
4 0
It’s 212.00 why does it have to be 20 words ugh
You might be interested in
Order the numbers from least to greatest -1/3, -0.3, 4/3, 1.2, 3/2
Nikolay [14]

Answer:

1/3,-0.3,1.2,4/3,3/2

Step-by-step explanation:

-1/3,-0.3,1.2,4/3,3/2

7 0
3 years ago
Determine all real values of p such that the set of all linear combination of u = (3, p) and v = (1, 2) is all of R2. Justify yo
Rama09 [41]

Answer:

p ∈ IR - {6}

Step-by-step explanation:

The set of all linear combination of two vectors ''u'' and ''v'' that belong to R2

is all R2 ⇔

u\neq 0_{R2}      

v\neq 0_{R2}

And also u and v must be linearly independent.

In order to achieve the final condition, we can make a matrix that belongs to R^{2x2} using the vectors ''u'' and ''v'' to form its columns, and next calculate the determinant. Finally, we will need that this determinant must be different to zero.

Let's make the matrix :

A=\left[\begin{array}{cc}3&1&p&2\end{array}\right]

We used the first vector ''u'' as the first column of the matrix A

We used the  second vector ''v'' as the second column of the matrix A

The determinant of the matrix ''A'' is

Det(A)=6-p

We need this determinant to be different to zero

6-p\neq 0

p\neq 6

The only restriction in order to the set of all linear combination of ''u'' and ''v'' to be R2 is that p\neq 6

We can write : p ∈ IR - {6}

Notice that is p=6 ⇒

u=(3,6)

v=(1,2)

If we write 3v=3(1,2)=(3,6)=u , the vectors ''u'' and ''v'' wouldn't be linearly independent and therefore the set of all linear combination of ''u'' and ''b'' wouldn't be R2.

7 0
3 years ago
What is the image point of (1, -3) after a translation right 5 units and up 2 units?
Natali [406]

Answer:(6, -1)

Step-by-step explanation:

8 0
3 years ago
Bob wants to create two pens, as shown in the figure. One pen is for a garden and it needs a heavy duty fence to keep out the cr
sattari [20]
\bf \textit{the area of either pen is 2744, thus}\\\\
A=xy\implies 2744=xy\implies \cfrac{2744}{x}=y

\bf \begin{array}{llll}
\textit{perimeter of the garden pen}\\\\
P=2(x+y)\\\\
P=2\left( x+ \cfrac{2744}{x}\right)\\\\
P=2x+\cfrac{5488}{x}\\
----------\\

\textit{cost of it}\\\\
C=15\left( 2x+\cfrac{5488}{x} \right)\\\\
C=30x+\cfrac{82320}{x}
\end{array}\qquad 
\begin{array}{llll}
\textit{perimeter of the dog pen}\\\\
P=2x+y\\\\
P=2x+\cfrac{2744}{x}\\
----------\\
\textit{cost of it}\\\\
C=5\left( 2x+\cfrac{2744}{x} \right)\\\\
C=10x+\cfrac{27440}{x}
\end{array}

notice... the dog's pen perimeter, does not include the side that's bordering the garden's, since that side will use the heavy duty fence, instead of the light one

so, the sum of both of those costs, will be the C(x)

\bf C(x)=\left( 30x+\cfrac{82320}{x} \right)+\left( 10x+\cfrac{27440}{x} \right)
\\\\\\
C(x)=40x+\cfrac{109760}{x}

so, just take the derivative of it, and set it to 0 to find the extremas, and do a first-derivative test for any minimum

8 0
3 years ago
Dennis wants to buy a card for his wife. Dennis calculates the amount of the card as $4.50. The actual price of a card is $4. Wh
Alex17521 [72]

Answer:

12.5%

Step-by-step explanation:

$4.50-$4/$4 x 100%

0.5/4x100%

50/4= 25/2 = 12.5%

8 0
3 years ago
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