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denpristay [2]
1 year ago
12

If point B, shown on the coordinate plane below, is reflected over the y-axis to create B’, what will be the coordinates of B’?(

-5, 2)(5, 2)(-5, -2)(5, -2)

Mathematics
1 answer:
Reil [10]1 year ago
5 0

Solution

- The transformation for reflection over the y-axis is given below:

(x,y)\to(-x,y)

- We have been given the coordinate of B to be (-5, -2) as shown below:

- Thus, applying the transformation formula given above, we have:

\begin{gathered} (x,y)\to(-x,y) \\ (-5,-2)\to(-(-5),-2)=(5,-2) \end{gathered}

- Thus, the reflected point B' is

(5,-2)

- This is shown below:

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dangina [55]

Answer: Option 'D' is correct.

Step-by-step explanation:

Since we have given that

First term = 13

Increase each time = 20

Number of clients = 25

So, it forms an arithmetic progression:

So, 25 th term would be

a_{25}=a+(n-1)d\\\\a_{25}=13+(25-1)\times 20\\\\a_{25}=13+24\times 20\\\\a_{25}=13+480\\\\a_{25}=493

Hence, list would look like 13,33,............493.

Therefore, Option 'D' is correct.

3 0
3 years ago
4.62x10<br> 1.2x 10<br> What’s the scientific notation
MArishka [77]
46200000000.
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5 0
3 years ago
Solve. 6 Everyone at a park was either hiking or riding a bike. There were 24 more hikers than bike riders. If there were a tota
nydimaria [60]

b r = 178 + h;

b r + h = 676;

then, 178 + 2h = 676 => h = 498/2 = 249 => b r = 178 + 249 = 427

Answer:

676

Step-by-step explanation:

h=b+178=427

b=b=249

-------------

(b+178)+b=676

2b+178=676

2b=498

b=249

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249+427=676

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5 0
2 years ago
Given m&lt;12= 121 and m&lt;6= 75, find the measure of each missing angle.
slava [35]

Answer/Step-by-step explanation:

Given:

m<12 = 121°

m<6 = 75°

a. m<1 = m<6 (vertical angles)

m<1 = 75° (substitution)

b. m<12 = m<1 + m2 (alternate exterior angles)

121° = 75° + m<2 (substitution)

121° - 75° = m<2 (subtraction property of equality)

46° = m<2

m<2 = 46°

c. m<1 + m<2 + m<3 = 180° (angles on a straight line)

75° + 46° + m<3 = 180° (substitution)

121° + m<3 = 180°

m<3 = 180° - 121° (subtraction property of equality)

m<3 = 59°

d. m<4 = m<3 (vertical angles)

m<4 = 59° (substitution)

e. m<5 + m<4 + m<6 = 180° (angles on a straight line)

m<5 + 59° + 75° = 180° (substitution)

m<5 + 134° = 180°

m<5 = 180° - 134° (Subtraction property of equality)

m<5 = 46°

f. m<7 = m<12 (vertical angles)

m<7 = 121° (substitution)

g. m<8 = m<4 (vertical angles)

m<8 = 59° (substitution)

h. m<9 = m<6 (Alternate Interior Angles)

m<9 = 75° (substitution)

i. m<10 + m<9 = 180° (Linear Pair)

m<10 + 75° = 180° (substitution)

m<10 = 180° - 75° (Subtraction property of equality)

m<10 = 105°

j. m<11 = m<8 (vertical angles)

m<11 = 59° (substitution)

k. m<13 = m<10 (vertical angles)

m<13 = 105° (substitution)

l. m<14 = m<9 (vertical angles)

m<14 = 75° (substitution)

5 0
3 years ago
Consider the function ​f(x)equalscosine left parenthesis x squared right parenthesis. a. Differentiate the Taylor series about 0
dybincka [34]

I suppose you mean

f(x)=\cos(x^2)

Recall that

\cos x=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}

which converges everywhere. Then by substitution,

\cos(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{(x^2)^{2n}}{(2n)!}=\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n)!}

which also converges everywhere (and we can confirm this via the ratio test, for instance).

a. Differentiating the Taylor series gives

f'(x)=\displaystyle4\sum_{n=1}^\infty(-1)^n\frac{nx^{4n-1}}{(2n)!}

(starting at n=1 because the summand is 0 when n=0)

b. Naturally, the differentiated series represents

f'(x)=-2x\sin(x^2)

To see this, recalling the series for \sin x, we know

\sin(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^{n-1}\frac{x^{4n+2}}{(2n+1)!}

Multiplying by -2x gives

-x\sin(x^2)=\displaystyle2x\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n+1)!}

and from here,

-2x\sin(x^2)=\displaystyle 2x\sum_{n=0}^\infty(-1)^n\frac{2nx^{4n}}{(2n)(2n+1)!}

-2x\sin(x^2)=\displaystyle 4x\sum_{n=0}^\infty(-1)^n\frac{nx^{4n}}{(2n)!}=f'(x)

c. This series also converges everywhere. By the ratio test, the series converges if

\displaystyle\lim_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{(n+1)x^{4(n+1)}}{(2(n+1))!}}{(-1)^n\frac{nx^{4n}}{(2n)!}}\right|=|x|\lim_{n\to\infty}\frac{\frac{n+1}{(2n+2)!}}{\frac n{(2n)!}}=|x|\lim_{n\to\infty}\frac{n+1}{n(2n+2)(2n+1)}

The limit is 0, so any choice of x satisfies the convergence condition.

3 0
4 years ago
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