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inna [77]
2 years ago
5

Vertex A in quadrilateral ABCD lies at (-3, 2). If you rotate ABCD 180° clockwise about the origin, what will be the coordinates

of A′ of the rotated quadrilateral A′B′C′D′?
A. (3, -2)
B. (-3, 2)
C. (-2, 3)
D. (-3, -2)
Mathematics
2 answers:
Olegator [25]2 years ago
7 0

Correct answer. A: (3, -2)

Marysya12 [62]2 years ago
3 0
I think it might be B hope I’m right
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Solve 7x- c= k for x. <br>A. X = 7(k+C) <br>B. x = 7(K-C) <br>C. X = k+c/7 <br>D. X= k-c/7​
antoniya [11.8K]

Answer:

C

Step-by-step explanation:

you add c to both side which will then make it 7x=k+c then you would divide 7 from both sides leaving you with x=k+c/7. Except 7 will be under the k and c.

7 0
1 year ago
Excel:In cell B13, create a formula using the VLOOKUP function that looks up the value from cell A11 in the range A5:B7, returns
sp2606 [1]

Answer:

=Vlookup'B13' A11' 7'false

Press enter.

Step-by-step explanation:

Vlookup is a technique in excel which enables users to search for criterion values. It is vertical lookup function in excel which return a value from a different column. The formula for Vlookup function is:

=Vlookup'select cell you want to look up in' select cell you want to lookup from' select column index number' true/false.

where true is approximate match and false is exact match.

8 0
2 years ago
Page:<br>If A+B+C=180 , then prove that:cosA+cosB+cosC= 1+4SINA/2SINB/2SINC/2​
lakkis [162]

Answer:

A + B + C = π ...... (1)

...........................................................................................................

L.H.S.

= ( cos A + cos B ) + cos C

= { 2 · cos[ ( A+B) / 2 ] · cos [ ( A-B) / 2 ] } + cos C

= { 2 · cos [ (π/2) - (C/2) ] · cos [ (A-B) / 2 ] } + cos C

= { 2 · sin( C/2 ) · cos [ (A-B) / 2 ] } + { 1 - 2 · sin² ( C/2 ) }

= 1 + 2 sin ( C/2 )· { cos [ (A -B) / 2 ] - sin ( C/2 ) }

= 1 + 2 sin ( C/2 )· { cos [ (A-B) / 2 ] - sin [ (π/2) - ( (A+B)/2 ) ] }

= 1 + 2 sin ( C/2 )· { cos [ (A-B) / 2 ] - cos [ (A+B)/ 2 ] }

= 1 + 2 sin ( C/2 )· 2 sin ( A/2 )· sin( B/2 ) ... ... ... (2)

= 1 + 4 sin(A/2) sin(B/2) sin(C/2)

= R.H.S. ............................. Q.E.D.

...........................................................................................................

In step (2), we used the Factorization formula

cos x - cos y = 2 sin [ (x+y)/2 ] · sin [ (y-x)/2 ]

Step-by-step explanation:

6 0
1 year ago
Solve the following multiplication and division problems. a. 8 T. 1,398 lb. 14 oz. × 6 b. 349 lb. 6 oz. ÷ 130 c. 6 T. 294 lb. ÷
tester [92]

Answer:  a) 278382 oz

b) 43 oz

c) 64.1 oz

Step-by-step explanation:

a) 8 T. 1,398 lb. 14 oz. × 6

First we need to change it in 'oz'.

As we know that

1\ ton=32000\ oz\\\\1\ lb=16\ oz

so, it becomes,

8\times 32000+1398\times 16+14\ oz\\\\=278382\ oz

8 T. 1,398 lb. 14 oz. × 6 becomes

278382\times 6\\\\=1670292\ oz

b) 349 lb. 6 oz. ÷ 130

It becomes,

349\times 16+6\ oz\\\\=5590\ oz\\\\\text{ at last it becomes}\\\\=\frac{5590}{130}\\\\=43\ oz

c) 6 T. 294 lb. ÷ 3,071

First it becomes,

6\times 32000+294\times 16\ oz\\\\=196704\ oz

At last it becomes,

\frac{196704}{3071}\\\\=64.1\ oz

Hence, a) 278382 oz

b) 43 oz

c) 64.1 oz

6 0
2 years ago
Read 2 more answers
25. To produce the graph of the function y=0.5cos(0.5x), what transformations should be applied to the graph of the parent funct
soldi70 [24.7K]

Answer:

C. A horizontal stretch to produce a period of 2\pi and a vertical compression.

Step-by-step explanation:

We are given the parent function as y= \cot x

It is given that, transformations are applied to the parent function in order to obtain the function y=0.5\cot (0.5x) i.e. y=\frac{1}{2}\cot (\frac{x}{2})

That is, we see that,

The parent function y= \cot x is stretched horizontally by the factor of \frac{1}{2} which gives the function y=\cot (\frac{x}{2}).

Further, the function is compressed vertically by the factor of \frac{1}{2} which gives the function y=\frac{1}{2}\cot (\frac{x}{2}).

Now, we know,

If a function f(x) has period P, then the function cf(bx) will have period \frac{P}{|b|}.

Since, the period of y= \cot x is \pi, so the period of y=\frac{1}{2}\cot (\frac{x}{2}) is \frac{\pi}{1/2} = 2\pi

Hence, we get option C is correct.

3 0
2 years ago
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