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3 years ago

Solve this inequality: j/4 – 8 < 4. A. j < –48 B. j < 48 C. j > –12 D. j < 12

Mathematics

2 answers:

postnew [5]3 years ago

4 0

J/4 - 8 < 4

j/4 < 4 + 8
j4 < 12
j < 12 X 4
j < 48

Kruka [31]3 years ago

3 0

$\frac{j}{4} - 8 \ \textless \ 4 \\ \\ \frac{j}{4} \ \textless \ 4 + 8 \\ \\ \frac{j}{4} \ \textless \ 12 \\ \\ j \ \textless \ 12 \times 4 \\ \\ j \ \textless \ 48 \\ \\$

The final result is: j < 48.

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Answer:

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3 years ago

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What is the domain and range ?

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Answer: domain are the ones that go side to side or the x range is the ones that go up and down or the y

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2 years ago

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)]

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3 years ago

sarah used 3/4 pound of blueberries to make 1/2 cup of jam. how many pounds of blueberries would she need to make a cup of jam?

Ksenya-84 [330]

Answer: She needed $1\frac{1}{2}\text{ pounds of blueberries}$ to make a cup of jam .

Step-by-step explanation:

Since we have given that

Sarah used $\frac{3}{4}$ pound of blueberries to make $\frac{1}{2}$ cup of jam.

As we need to find the number of pounds of blueberries she would need to make a cup of jam.

So, we will use "Unitary Method":

$\text{For }\frac{1}{2}\text{ cup of jam},\\\\\text{ she need }\frac{3}{4}\text{ pound of blueberries }\\\\\text{ For 1 cup of jam}.\\\\\text{She need }=\frac{3}{4}\times 2=\frac{3}{2}=1\frac{1}{2}\text{ pounds of blueberries }$

Hence, She needed $1\frac{1}{2}\text{ pounds of blueberries}$ to make a cup of jam .

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3 years ago

Point A lies outside of plane P. How many lines can be drawn perpendicular to plane P through point A?

Ivanshal [37]

<em>Answer</em>

D) 1

<em>Explanation</em>

A perpendicular line makes a angle of 90° with a line or a plane.

If a line is to be drawn from a point to a line or a plain it can only be one. In this case a line is to be drawn through a point A to a plane P. If the line is to be perpendicular, then is is only one.

From the choices, the appropriate answer is D) 1

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3 years ago

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