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3 years ago

Need some friends on ma friends list...

Mathematics

2 answers:

Stels [109]3 years ago

6 0

Answer:

i'll be your friend lol

Step-by-step explanation:

tekilochka [14]3 years ago

4 0

Answer:

I will be ur friend

Step-by-step explanation:

I will add you right now

You might be interested in

What is the length of the segment that joins points (4,-1) and (7,5)?

GrogVix [38]

Answer:

3√5

Step-by-step explanation:

d = √(difference of x)² + (difference of y)²

d = √(7 - 4)² + (5 - (-1))² = √9 + 36 = √45 = 3√5

4 0

3 years ago

ASAP!!!!!

In-s [12.5K]

Think of the given points as endpoints of vectors.

The point (3, 1) is the terminus of the vector 3 i + j.

The point (-4, 2) is the terminus of -4 i + 2 j.

By the "triangle rule" for vector summation, the vector starting at (3, 1) and ending at (-4, 2) is the difference between these two vectors.

(-4 i + 2j) - (3i + j) = -7 i + j

4 0

3 years ago

In a 45-45-90 triangle, what is the length of the hypotenuse when the length of one of the legs is 11 in.?

lana [24]

Answer:

= 15.556 in

Step-by-step explanation:

Since this is a 45-45-90 triangle, we can tell that;

This is a right triangle
This is an isosceles triangle

One of the theorems of geometry, the Isosceles Right Triangle Theorem, says that the hypotenuse is √2 times the length of a leg.

Therefore;

hypotenuse = leg × √2

If we take hypotenuse as x

Since the leg is 11;

Then; x = 11√2

x = 11/√2

x = 15.556 in

3 0

4 years ago

Jacob solves the system of equations by forming a matrix equation.

Lyrx [107]

Simultaneous equations can be solved using inverse matrix operation.

The complete steps of Jacob's solution are:

$\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]^{-1} \cdot \left[\begin{array}{cc}4&1\\-2&3\end{array}\right]\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14}\left[\begin{array}{cc}3&-1\\2&4\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{cc}4&1\\-2&3\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14} \left[\begin{array}{c}28&-84\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}2&-6\end{array}\right]$

We have:

$4x + y = 2$

$-2x + 4y = -22$

Calculate the determinant of $\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]$

$|A| = 4 \times 3 -1 \times -2$

$|A| = 12 +2$

$|A| = 14$

So, the inverse matrix becomes

$A = \frac{1}{14}\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]$

Replace the first column with $\left[\begin{array}{c}2&-22\end{array}\right]$ to calculate the value of x

$x = \frac{1}{14}\left[\begin{array}{cc}2&1\\-22&3\end{array}\right]$

So, we have:

$x = \frac{1}{14}(2 \times 3 - 1 \times -22)$

$x = \frac{1}{14}(6 +22)$

$x = \frac{1}{14}(28)$

$x = 2$

Replace the second column with $\left[\begin{array}{c}2&-22\end{array}\right]$ to calculate the value of y

$y = \frac{1}{14}\left[\begin{array}{cc}4&2\\-2&-22\end{array}\right]$

So, we have:

$y = \frac{1}{14}(4 \times -22 - 2 \times -2)$

$y = \frac{1}{14}(-88 +4)$

$y = \frac{1}{14}(-84)$

$y = -6$

Hence, the complete process is:

$\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{cc}4&1\\-2&3\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14} \left[\begin{array}{c}28&-84\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}2&-6\end{array}\right]$