All categories

3 years ago

PLEASE HELPPPPPPPP....?.......

Mathematics

2 answers:

m_a_m_a [10]3 years ago

8 0

Answer:

1) 117

2) 9

3) 45 or -45

Step-by-step explanation:

a=9

b=6

1) a²+b²=9²+6²=81+36=117

2) (a-b)²=(9-6)²=3²=9

3) a²-b²=81-36=45

It’s also possible that a=6 and b=9, in which case

1) (same)

2) (same)

3) 36-81=-45

Kobotan [32]3 years ago

4 0

Answer:

a = 9

b = 6

I hope you can do the rest yoyrself, as I coulsnt find how to type it out ;-;

You might be interested in

X + 16 = 9z Help please

diamong [38]

Answer:x=9z + 16

z= x/9 + 16/9

Step-by-step explanation: dont know if i solve for x or z so i did both subtract 16 from both sides for x

Isolate the variable by dividing each side by factors that don't contain the variable. for z let me know if u need help

3 0

3 years ago

Read 2 more answers

Please help me fill in the blanks

jeka57 [31]

E is an exponential function with an approximate value of 1.

6 0

3 years ago

Read 2 more answers

The sum of z and 7? What is it plz

wolverine [178]

Answer:

z + 7

Step-by-step explanation:

Step 1: Convert words into an expression

The sum of z and 7

z + 7

Answer: z + 7

If needed solve:

z + 7 - 7 = 0 - 7

z = -7

5 0

3 years ago

Read 2 more answers

Find the? inverse, if it? exists, for the given matrix. [4 3] [3 6]

True [87]

Answer:

Therefore, the inverse of given matrix is

$=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}$

Step-by-step explanation:

The inverse of a square matrix $A$ is $A^{-1}$ such that

$A A^{-1}=I$ where I is the identity matrix.

Consider, $A = \left[\begin{array}{ccc}4&3\\3&6\end{array}\right]$

$\mathrm{Matrix\:can\:only\:be\:inverted\:if\:it\:is\:non-singular,\:that\:is:}$

$\det \begin{pmatrix}4&3 \\3&6\end{pmatrix}\ne 0$

$\mathrm{Find\:2x2\:matrix\:inverse\:according\:to\:the\:formula}:\quad \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}^{-1}=\frac{1}{\det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}}\begin{pmatrix}d\:&\:-b\:\\ -c\:&\:a\:\end{pmatrix}$

$=\frac{1}{\det \begin{pmatrix}4&3\\ 3&6\end{pmatrix}}\begin{pmatrix}6&-3\\ -3&4\end{pmatrix}$

$\mathrm{Find\:the\:matrix\:determinant\:according\:to\:formula}:\quad \det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}\:=\:ad-bc$

$4\cdot \:6-3\cdot \:3=15$

$=\frac{1}{15}\begin{pmatrix}6&-3\\ -3&4\end{pmatrix}$

$=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}$

Therefore, the inverse of given matrix is

$=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}$

4 0

4 years ago

Question 18

kobusy [5.1K]

Answer:

Undefined

Step-by-step explanation:

$y=-7$ has a slope of $0$ , since it is a horizontal line. The only types of lines that are perpendicular to horizontal lines are vertical lines, which are usually in the form $x=a$ (where $a$ is any real number). Vertical lines have an undefined slope, so the answer is undefined. Hope this helps!

6 0

3 years ago