All categories

3 years ago

Help please 2+2-1-1+6=x

Mathematics

2 answers:

natita [175]3 years ago

8 0

Answer:

$2+2-1-1+6=x : x = 8$

Step-by-step explanation:

$2+2-1-1+6=x\\$

Switch sides:

$x=2+2-1-1+6$

Add/Subtract the numbers: $2+2-1-1+6=8$

$x = 8$

Hope I helped, if so may I get brainliest and a thanks?

Thank you, have a good day! =)

Anestetic [448]3 years ago

6 0

Answer:

Step-by-step explanation:

You might be interested in

Will give crown thanks

Tatiana [17]

Answer:

A is right

B is acute

C is obtuse

D is right

Step-by-step explanation:

90 is right

anything under 90 is acute

over 90 obtuse

8 0

2 years ago

Read 2 more answers

If m∠F = 90° and m∠G = 70°, then ∠F and ∠G are not supplementary.

miss Akunina [59]

a) Proof by contradiction is different from traditional proof as it accepts a single example showing that a statement is false, instead of having the need to derive a general relationship for all input values.

b) The statement is true by contradiction as the sum of the measures is of 160º, and not 180º.

<h3>What are supplementary angles?</h3>

Two angles are called supplementary angles if the sum of their measures has a value of 180º.

The measures of the angles in this problem are given as follows:

m∠F = 90°.

m∠G = 70°.

Then the sum of the measures of this angles is given as follows:

90 + 70 = 160º.

Which is a different sum of 160º, confirming the statement that the angles are not supplementary by contradiction.

A similar problem, involving proof by contradiction and supplementary angles, is presented at brainly.com/question/28889480

#SPJ1

5 0

8 months ago

4=32b<br> solve for b WILL GIVE BRAINLIEST

Makovka662 [10]

Answer:

1/8

Step-by-step explanation:

$32b \div 32 = 4 \div 32 \\$

5 0

2 years ago

ILL MAKE BRAINLY??!!!?!!!!!

Leno4ka [110]

Answer:

Mizuki here to answer! B is the correct answer!

Step-by-step explanation:

1 - 0.4 = 0.6

2.1 x 0.6 = 1.26

6 0

2 years ago

Which of the following functions are homomorphisms?

Vikentia [17]

Part A:

Given $f:Z \rightarrow Z,$ defined by $f(x)=-x$

$f(x+y)=-(x+y)=-x-y \\ \\ f(x)+f(y)=-x+(-y)=-x-y$

but

$f(xy)=-xy \\ \\ f(x)\cdot f(y)=-x\cdot-y=xy$

Since, f(xy) ≠ f(x)f(y)

Therefore, the function is not a homomorphism.

Part B:

Given $f:Z_2 \rightarrow Z_2,$ defined by $f(x)=-x$

Note that in $Z_2$ , -1 = 1 and f(0) = 0 and f(1) = -1 = 1, so we can also use the formular $f(x)=x$

$f(x+y)=x+y \\ \\ f(x)+f(y)=x+y$

and

$f(xy)=xy \\ \\ f(x)\cdot f(y)=xy$

Therefore, the function is a homomorphism.

Part C:

Given $g:Q\rightarrow Q$ , defined by $g(x)= \frac{1}{x^2+1}$

$g(x+y)= \frac{1}{(x+y)^2+1} = \frac{1}{x^2+2xy+y^2+1} \\ \\ g(x)+g(y)= \frac{1}{x^2+1} + \frac{1}{y^2+1} = \frac{y^2+1+x^2+1}{(x^2+1)(y^2+1)} = \frac{x^2+y^2+2}{x^2y^2+x^2+y^2+1}$

Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.

Part D:

Given $h:R\rightarrow M(R)$ , defined by $h(a)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)$

$h(a+b)= \left(\begin{array}{cc}-(a+b)&0\\a+b&0\end{array}\right)= \left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right) \\ \\ h(a)+h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)+ \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)=\left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right)$

but

$h(ab)= \left(\begin{array}{cc}-ab&0\\ab&0\end{array}\right) \\ \\ h(a)\cdot h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)\cdot \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)= \left(\begin{array}{cc}ab&0\\-ab&0\end{array}\right)$

Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.

Part E:

Given $f:Z_{12}\rightarrow Z_4$ , defined by $\left([x_{12}]\right)=[x_4]$ , where $[u_n]$ denotes the lass of the integer $u$ in $Z_n$ .

Then, for any $[a_{12}],[b_{12}]\in Z_{12}$ , we have

$f\left([a_{12}]+[b_{12}]\right)=f\left([a+b]_{12}\right) \\ \\ =[a+b]_4=[a]_4+[b]_4=f\left([a]_{12}\right)+f\left([b]_{12}\right)$

and

$f\left([a_{12}][b_{12}]\right)=f\left([ab]_{12}\right) \\ \\ =[ab]_4=[a]_4[b]_4=f\left([a]_{12}\right)f\left([b]_{12}\right)$

Therefore, the function is a homomorphism.

7 0

3 years ago