All categories

2 years ago

Correct answer plz. 15 points. Reported for wrong answer. Thx

Mathematics

1 answer:

34kurt2 years ago

7 0

Third option is the right answer:

−2x + 4y ≥ 0

You might be interested in

What equation matches this situation.mai biked x miles han biked 2/3 more than that

Olin [163]

Answer:

i need more information like how much mai biked

Step-by-step explanation:

3 0

2 years ago

Solve the simultaneous equation y-2x+1=0 and 4x^2+3y^2-2xy=7

Lapatulllka [165]

The first equation is $y-2x+1=0$

$y=2x-1$ (Equation 1)

The second equation is $4x^{2}+3y^{2}-2xy=7$ (Equation 2)

Putting the value of x from equation 1 in equation 2.

we get,

$4x^{2}+3(2x-1)^{2}-2x(2x-1)=7$

$4x^{2}+3(4x^{2}+1-4x)-4x^{2}+2x=7$

by simplifying the given equation,

$12x^{2}-10x-4=0$

$6x^{2}-5x-2=0$

Using discriminant formula,

$D=b^{2}-4ac$

$D=25-4 \times 6 \times -2 = 73$

Now the formula for solution 'x' of quadratic equation is given by:

$x=\frac{-b+\sqrt{D}}{2a} and x=\frac{-b-\sqrt{D}}{2a}$

$x=\frac{5+\sqrt{73}}{12} and x=\frac{5-\sqrt{73}}{12}$

Hence, these are the required solutions.

8 0

3 years ago

What is the 10th term in the pattern with the formula 9n + 9?

Studentka2010 [4]

A(10) = <span>9(10) + 9= 90+9 =99

---------------------------------------------</span>

6 0

2 years ago

Read 2 more answers

Hi!! Please help me with this question. I got an answer of 63.25 but it was incorrect and need to redo it. Please also explain h

matrenka [14]

Answer:

x ≈ 63.27

Step-by-step explanation:

Using the tangent ratio in the right triangle

tan49° = $\frac{opposite}{adjacent}$ = $\frac{EF}{DE}$ = $\frac{x}{55}$ ( multiply both sides by 55 )

55 × tan49° = x , then

x ≈ 63.27 ( to the nearest hundredth )

4 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