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3 years ago

Solve the inequality 2x + 9 < 4x + 3

Mathematics

2 answers:

Maslowich3 years ago

6 0

Answer:

x>3

Step-by-step explanation:

2x+9<4x+3

2x-4x<3-9 (collect the like terms first)

-2x<-6 (divide both sides by -2 to get the value of x)

$\frac{-2x}{-2}$ ( the < sign changes because both sides are divided by -2)

x>3.

{x:x>3}= {x=4,5,6,7...}

Check

2x+9<4x+3

2(4)+9<4(4)+3

8+9<16+3

17<19

Diano4ka-milaya [45]3 years ago

5 0

Answer:

x > 3

Step-by-step explanation:

2x+9<4x+3

or, 9-3<4x-2x

or, 6<2x

or, (6/2)<x

or, 3<x

or, x >3

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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

Omas is making trail mix using granola and walnuts. He can spend a total of $12 on the ingredients. He buys 3 pounds of granola

Serga [27]

Answer:

Omas can buy 1 pound of walnuts.

Step-by-step explanation:

From the information given, you can say that the total cost would be equal to the result of multiplying the price per pound of granola for the number of pounds of granola plus the result of multiplying the price per pound of walnuts for the number of pounds of walnuts, which would be:

Total cost=2x+6y, where

x is the number of pounds of granola

y is the number of pounds of walnuts

Now, you can replace the values on the formula and solve for y:

12=(2*3)+6y

12=6+6y

6=6y

y=1

According to this, the answer is that Omas can buy 1 pound of walnuts.

4 0

2 years ago

Read 2 more answers

20% of 55 is what number

Vitek1552 [10]

20% of 55 is 11

Workingout: 20%/100% = 0.2
0.2 x 55 = 11

:)

7 0

3 years ago

The total length of pencils A, B and C is 29 cm. Pencil a is 11 cm shorter then pencil B, and pencil B is twice as long a pencil

satela [25.4K]

The length of pencil A is 5 cm

Solution:

Let the length of pencil A be "x"

Let the length of pencil B be "y"

Let the length of pencil C be "z"

The total length of pencils A, B and C is 29 cm

Therefore,

length of pencil A + length of pencil B + length of pencil C = 29

x + y + z = 29 ------------ eqn 1

Pencil A is 11 cm shorter then pencil B

x = y - 11 ------- eqn 2

Pencil B is twice as long a pencil C

y = 2z

$z = \frac{y}{2}$ ------ eqn 3

Substitute eqn 2 and eqn 3 in eqn 1

$y - 11 + y + \frac{y}{2} = 29\\\\2y + \frac{y}{2} = 29 + 11\\\\\frac{4y+y}{2} = 40\\\\5y = 80\\\\y = 16$

Substitute y = 16 in eqn 2

x = 16 - 11

x = 5

Thus length of pencil A is 5 cm

5 0

3 years ago

2. 3m - 9n = 24 solve for m

PIT_PIT [208]

Answer:

m=8+3n

Step-by-step explanation:

Move all terms that don't contain m to the right side and solve.

Hope this helps <3

5 0

2 years ago