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3 years ago

ADDITION OF INTEGERS 21) 2 +6+-6 A) 0 B) 8 C) 1 D) 2

Mathematics

1 answer:

MrMuchimi3 years ago

5 0

Answer:

Step-by-step explanation:

add up the positives then subtract the negative

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Pls help me with trig :)

trapecia [35]

Answer:

Step-by-step explanation:

let ∠GEF=x

then ∠GFE=90-x

$\frac{GH}{EH}=tan x\\GH=EH~ tan x\\or~ GH=a ~tan x=6 tan x\\\frac{GH}{FH}=tan (90-x)=cot x\\GH=FH*cot x\\GH=b*cot x=27 cot x\\6 tan x=27 cotx\\\frac{tan x}{cot x}=\frac{27}{6}\\tan ^2 x=\frac{27}{6}=\frac{9}{2}\\sec^2x-tan^2x=1\\sec^2 x=1+tan ^2x=1+\frac{9}{2}=\frac{11}{2}\\\frac{EG}{EH}=sec x\\ EG=EH sec x=a sec x= 6 sec x=6\sqrt{\frac{11}{2} } =3\sqrt{22} \approx 14.07$

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3 years ago

Helpppppp poooorrrfavooor plsssssss

nikklg [1K]

Answer:

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Write the equation of the line parallel to Y equals 2/3X +1 through the point (0,-4) use slope intercept form

Andrew [12]

Writing the slope-intercept form of a linear equation, we have:

$y=mx+b$

Where m is the slope and b is the y-intercept.

Since parallel lines have the same slope, we can see that the slope of the line y = 2/3x + 1 is equal m = 2/3, so for our equation we also have m = 2/3.

Now, using the point (0, -4), we have:

$\begin{gathered} y=\frac{2}{3}x+b \\ (0,-4)\colon \\ -4=\frac{2}{3}\cdot0+b \\ b+0=-4 \\ b=-4 \end{gathered}$

So our equation is:

$y=\frac{2}{3}x-4$

y = 2/3x - 4

7 0

1 year ago

Using the weighted average approach to process costing, floridyne would use what number of equivalent units in 2019 to calculate

timofeeve [1]

Floridyne would use an equivalent unit labor of 166,340 unit to calculate the cost per equivalent unit for direct labor.

<h3>How do we get the equivalent unit labor?</h3>

Under weighted average, we do not make distinction between started and finished and just finished. Thus we work with finished and ending WIP only:

Finished 162,000

Ending - WIP 6,200 ending at 70% complete

Equivalent units for labor = Finished + Percentage of completion ending units

= 162,000 + 6,200 x 70%

= 166,340

Therefore, he would use an equivalent unit labor of 166,340 unit to calculate the cost per equivalent unit for direct labor.

Missing word "Floridyne, Inc. manufactures mouthwash. They had no finished goods inventory at the beginning of 2019. They have only one processing department for this product. A review of the company’s inventory records shows the following: At the beginning of January 2019, Floridyne has 4,500 gallons of mouthwash in process. (costs $8,410 for materials, 1,663 for labor and 4,990 for overhead) During 2019, Floridyne finishes/transfers 162,000 gallons of mouthwash. On December 31, 2019, Floridyne has 6,200 gallons of mouthwash that is 70% complete. Direct materials are added half at the beginning of the process and half after the process is 60% complete. During 2019 $349,000 of direct materials and $92,500 of direct labor were added. Using the weighted average approach to process costing,"

Read more about equivalent unit labor

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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