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3 years ago

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HELP I NEED HELP ASAP HELP I NEED HELP ASAP HELP I NEED HELP ASAP HELP I NEED HELP ASAP

1 answer:

TiliK225 [7]3 years ago

8 0

Answer:

I believe the correct answer is option A.

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PLEASE HELP ILL MARK BRAINLIEST !!!

Alexus [3.1K]

Answer:

a) the common difference is 20

b) $x_8=115 , x_{12}=195$

c) the common difference is -13

d) $a_{12}=52, a_{15}=13$

Step-by-step explanation:

a) what is the common difference of the sequence xn

Looking at the table, we get x_3=16, x_4=36 and x_5= 56

Deterring the common difference by subtracting x_4 from x_3 we get

36-16 =20

So, the common difference is 20

b) what is x_8? what is x_12

The formula used is: $x_n=x_1+(n-1)d$

We know common difference d= 20, we need to find $x_1$

Using $x_3=16$ we can find $x_1$

$x_n=x_1+(n-1)d\\x_3=x_1+(3-1)d\\15=x_1+2(20)\\15=x_1+40\\x_1=15-40\\x_1=-25$

So, We have $x_1 = -25$

Now finding $x_8$

$x_n=x_1+(n-1)d\\x_8=x_1+(8-1)d\\x_8=-25+7(20)\\x_8=-25+140\\x_8=115$

So, $\mathbf{x_8=115}$

Now finding $x_{12}$

$x_n=x_1+(n-1)d\\x_{12}=x_1+(12-1)d\\x_{12}=-25+11(20)\\x_{12}=-25+220\\x_{12}=195$

So, $\mathbf{x_{12}=195}$

c) what is the common difference of the sequence $a_m$

Looking at the table, we get a_7=104, a_8=91 and a_9= 78

Deterring the common difference by subtracting a_7 from a_8 we get

91-104 =-13

So, the common difference is -13

d) what is a_12? what is a_15?

The formula used is: $a_n=a_1+(n-1)d$

We know common difference d= -13, we need to find $a_1$

Using $a_7=104$ we can find $x_1$

$a_n=a_1+(n-1)d\\a_7=a_1+(7-1)d\\104=a_1+7(-13)\\104=a_1-91\\a_1=104+91\\a_1=195$

So, We have $a_1 = 195$

Now finding $a_{12}$ , put n=12

$a_n=a_1+(n-1)d\\a_{12}=a_1+(12-1)d\\a_{12}=195+11(-13)\\a_{12}=195-143\\a_{12}=52$

So, $\mathbf{a_{12}=52}$

Now finding $a_{15}$ , put n=15

$a_n=a_1+(n-1)d\\a_{15}=a_1+(15-1)d\\a_{15}=195+14(-13)\\a_{15}=195-182\\a_{15}=13$

So, $\mathbf{a_{15}=13}$

5 0

3 years ago

During a snowstorm, the temperature drops from 0°F to -12°F in 3 hours. On

Anastasy [175]

Answer:D.

-4°F

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

If f(x) = 4x + k and f (2) = f (1), then calculate the value<br>of K.

andrezito [222]

Answer:

-7

Step-by-step explanation:

given f(x)=4x+k

therfore, f(2)=4*2+k

we know that f(2) =1

so, 4*2+k=1

8+k=1

k=1-8

k=-7

8 0

3 years ago

Regina receive a score of 39 out of 52 on a quiz. What percentage did she earn?

Verdich [7]

Answer:

78%

Convert fraction (ratio) 39 / 50 Answer: 78%

8 0

3 years ago

Read 2 more answers

Consider the linear transformation T from V = P2 to W = P2 given by T(a0 + a1t + a2t2) = (2a0 + 3a1 + 3a2) + (6a0 + 4a1 + 4a2)t

Svet_ta [14]

Answer:

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

Step-by-step explanation:

First we start by finding the dimension of the matrix [T]EE

The dimension is : Dim (W) x Dim (V) = 3 x 3

Because the dimension of P2 is the number of vectors in any basis of P2 and that number is 3

Then, we are looking for a 3 x 3 matrix.

To find [T]EE we must transform the vectors of the basis E and then that result express it in terms of basis E using coordinates and putting them into columns. The order in which we transform the vectors of basis E is very important.

The first vector of basis E is e1(t) = 1

We calculate T[e1(t)] = T(1)

In the equation : 1 = a0

$T(1)=(2.1+3.0+3.0)+(6.1+4.0+4.0)t+(-2.1+3.0+4.0)t^{2}=2+6t-2t^{2}$

$[T(e1)]E=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

And that is the first column of [T]EE

The second vector of basis E is e2(t) = t

We calculate T[e2(t)] = T(t)

in the equation : 1 = a1

$T(t)=(2.0+3.1+3.0)+(6.0+4.1+4.0)t+(-2.0+3.1+4.0)t^{2}=3+4t+3t^{2}$

$[T(e2)]E=\left[\begin{array}{c}3&4&3\\\end{array}\right]$

Finally, the third vector of basis E is $e3(t)=t^{2}$

$T[e3(t)]=T(t^{2})$

in the equation : a2 = 1

$T(t^{2})=(2.0+3.0+3.1)+(6.0+4.0+4.1)t+(-2.0+3.0+4.1)t^{2}=3+4t+4t^{2}$

Then

$[T(t^{2})]E=\left[\begin{array}{c}3&4&4\\\end{array}\right]$

And that is the third column of [T]EE

Let's write our matrix

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

T(X) = AX

Where T(X) is to apply the transformation T to a vector of P2,A is the matrix [T]EE and X is the vector of coordinates in basis E of a vector from P2

For example, if X is the vector of coordinates from e1(t) = 1

$X=\left[\begin{array}{c}1&0&0\\\end{array}\right]$

$AX=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]\left[\begin{array}{c}1&0&0\\\end{array}\right]=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

Applying the coordinates 2,6 and -2 to the basis E we obtain

$2+6t-2t^{2}$

That was the original result of T[e1(t)]

8 0

3 years ago