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3 years ago

Which costs the most: 12 cat toys, 4 bags of cat food, or 3 litter boxes? Explain.

Mathematics

1 answer:

lukranit [14]3 years ago

7 0

Answer:

3 litter boxes

Step-by-step explanation:

It depends what type it is

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What is the measurement of x in the equation. 8x+5x+2x-11+6=180

34kurt

Answer:

x = 37/3 = 12.333

Step-by-step explanation:

Step 1 :

Pulling out like terms :

1.1 Pull out like factors :

15x - 185 = 5 • (3x - 37)

Equation at the end of step 1 :

Step 2 :

Equations which are never true :

2.1 Solve : 5 = 0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation :

2.2 Solve : 3x-37 = 0

Add 37 to both sides of the equation :

3x = 37

Divide both sides of the equation by 3:

x = 37/3 = 12.333

4 0

3 years ago

In ΔVWX, the measure of ∠X=90°, VW = 93 feet, and XV = 57 feet. Find the measure of ∠W to the nearest tenth of a degree.

Maksim231197 [3]

Answer:

37.8°

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

What is the answer to this

weeeeeb [17]

If all the cheese slices have same mass, say m grams

and mass of 3 cheese slices is 27

so,

m+m+m=27

3m=27

or m=27/3

m=9

thus

mass of one cheese slice is 9 grams

3 0

4 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

Help pls I will give 20 points

dalvyx [7]

Answer:

neon-0

oxide- -2

copper-2

tin-0

4 0

3 years ago