All categories

4 years ago

Please help thanks a bunch

Mathematics

2 answers:

mash [69]4 years ago

6 0

Hypotenuse ² =(leg1)² +(leg2)²
h²=6²+8²=36+64=100
√(h²)=√(100)
h=10

Minchanka [31]4 years ago

5 0

Pythagorean Theorem: a² + b² = c²

a and b = legs
c = hypotenuse

a = 6

6² = 6 · 6 = 36

b = 8

8² = 8 · 8 = 64

Add:

64 + 36 = 100

√100 = 10

Hypotenuse = 10

Hope this helped :)

You might be interested in

What is a11. A12. A21. A22

Ierofanga [76]

Answer:

$a_{11} =-1\\a_{12} =3\\a_{21} =5\\a_{22} =3/2\\$

Step-by-step explanation:

$2X-\left[\begin{array}{ccc}-4&8\\2&-2\end{array}\right] =\left[\begin{array}{ccc}2&-2\\8&5\end{array}\right]\\X=\frac{1}{2} (\left[\begin{array}{ccc}-4&8\\2&-2\end{array}\right] +\left[\begin{array}{ccc}2&-2\\8&5\end{array}\right])\\X=\frac{1}{2} (\left[\begin{array}{ccc}-4+2&8-2\\2+8&-2+5\end{array}\right] )\\X=\frac{1}{2} (\left[\begin{array}{ccc}-2&6\\10&3\end{array}\right] )\\X= (\left[\begin{array}{ccc}\frac{1}{2}(-2)&\frac{1}{2}(6)\\\frac{1}{2}(10)&\frac{1}{2}(3)\end{array}\right] )\\$

$X= (\left[\begin{array}{ccc}-1&3\\5&3/2\end{array}\right] )$

$a_{11} =-1\\a_{12} =3\\a_{21} =5\\a_{22} =3/2\\$

8 0

3 years ago

Please help me fast

Tju [1.3M]

Answer:

face value=6

roman numeral 68=LXVIII

All even numbers are divisible by 2

13,26, 39,52,65,,78

4 0

3 years ago

Alvin is running at a rate of 3 meters per second. The meters he runs, m, in s seconds is given by the equation m=3s. Identify t

dem82 [27]

It is 6 because I got the awnser multiple times

4 0

3 years ago

Read 2 more answers

What is the decimal multiplier to increase by 93%?

Tom [10]

Answer:

1.93

Step-by-step explanation:

1 = base multiplier

0.93 + 1 = 1.93 new multiplier

8 0

4 years ago

30 POINTS! Consider the system of linear equations and the matrix equations below.

fiasKO [112]

Answer:

Option C is correct.

Step-by-step explanation:

-x+3y=2

4x-2y=22

In matrix form is represented as:

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

AX=B

$X = A^{-1}B$

$A^{-1} = |A|/Adj A$

|A| = (-1)(-2)-(3)(4)

|A| = 2-12

|A| = -10

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

A^-1 = -1/10 $\left[\begin{array}{cc}-2&-3\\-4&-1\end{array}\right]$

A^-1 = 1/10 $\left[\begin{array}{cc}2&3\\4&1\end{array}\right]$

X= A^-1 B

X = 1/10 $\left[\begin{array}{cc}2&3\\4&1\end{array}\right]$ $\left[\begin{array}{c}2&22\end{array}\right]$

X=1/10 $X=1/10\left[\begin{array}{c}2*2+3*22\\4*2+1*22\end{array}\right]\\X=1/10\left[\begin{array}{c}4+66\\8+22\end{array}\right]\\X=1/10\left[\begin{array}{c}70\\30\end{array}\right]\\X=\left[\begin{array}{c}70/10\\30/10\end{array}\right]\\X=\left[\begin{array}{c}7\\3\end{array}\right]$

So, x = 7 and y =3

Hence Option C is correct.

4 0

3 years ago

Read 2 more answers