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Novosadov [1.4K]
3 years ago
12

LATS SET THIS TIME graphs GEOEMTRY. SPEEDDEMON ANSWER PLEASE

Mathematics
1 answer:
zaharov [31]3 years ago
4 0
3) idk how to answer that... If its a true or false question true
12) um... That one's kind of confusing but my best guess is somewhere in the 3rd quadrant

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statuscvo [17]

It's true. That's how parallelograms are defined. Here's a diagram to show you what the question means. The opposite sides are parallel and there are 2 sets of parallel lines.

3 0
3 years ago
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The decimal equivalent of 10^-2 is
Vesna [10]

Answer:

Step-by-step explanation:

10^(-2) = 1/10^2 =1/100 = 0.01

8 0
3 years ago
According to the Joy Cone Company, their waffle cones have a diameter of inches 2 3\8 and a height of 6 inches.
svetlana [45]

Answer:

so idc\sqrt[n]{x} \sqrt{x} \alpha \pi x^{2} \\ \left \{ {{y=2} \atop {x=2}} \right. x_{123} \int\limits^a_b {x} \, dx  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]

443

Step-by-step explanation: its 2 6\7

7 0
2 years ago
What is x. PLEASE HELP NEED ANSWERS NOW. M=​
Agata [3.3K]
2(6x+20) +2(8x-16)=180
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4 0
3 years ago
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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
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