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Vadim26 [7]
3 years ago
15

jen uses 8.2 of blue paint and white paint to paint her bedroom walls. 4/5ths of this amount is blue paint, and the rest is whit

e paint. how many pints of white paint did she use to paint her bedroom walls?

Mathematics
1 answer:
lara31 [8.8K]3 years ago
5 0

Answer:

The amount of white paint Jen used to paint the walls in her room is:

  • <u>1.64 pints.</u>

Step-by-step explanation:

To solve the exercise you only have to pay attention to the statement, in the section that says that 4/5 parts of the total painting is blue, therefore, if 1 is the total painting, you must do a subtraction:

  • 1 - 4/5 = 1/5

Since the remaining white paint is 1/5 of the total, you have two ways to solve the exercise: multiply the total paint (8.2 pints) by 1/5 or divide the number by 5, as shown below:

  • <u>Amount of white paint = 8.2 pints * (1/5) = 1.64 pints. </u>
  • <u>Amount of white paint = 8.2 pints / 5 = 1.64 pints. </u>

As you can see, the two methods provide a <u>value of 1.64 pints, which corresponds to 1/5 of the total paint and is the amount of white paint used</u>.

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1) Anna and Jason have summer jobs stuffing envelopes for two different companies. Anna earns $20 for every 400 envelopes she fi
AleksAgata [21]

Answer:

The answer is below

Step-by-step explanation:

A) i)

For Anna initially, she has $0 from making 0 envelopes. After making 400 envelopes she has $20. Let x represent the number of envelopes and y the earnings. Hence this can be represented by the points (0, 0) and (400, 20). Using the equation of a line:

y-y_1=\frac{y_2-y_1}{x_2-x_1} (x-x_1)\\\\y-0=\frac{20-0}{400-0}(x-0)\\\\y=\frac{1}{20} x

The table is:

x:   200     400       600     800     1000

y:    10        20          30        40       50

ii)

For Jason initially, he has $0 from making 0 envelopes. For every 250 envelopes he has $10. Let x represent the number of envelopes and y the earnings. Hence this can be represented by the points (0, 0) and (250, 10). Using the equation of a line:

y-y_1=\frac{y_2-y_1}{x_2-x_1} (x-x_1)\\\\y-0=\frac{10-0}{250-0}(x-0)\\\\y=\frac{1}{25} x

The table is:

x:   200     400       600     800     1000

y:    8         16           24        32       40

The graph is plotted using geogebra online graphing

b) From the table above we can see that Anna makes more stuffing than Jason.

c) Anna has a savings of $100. Hence this can be represented by the points (0, 100) and (250, 10). Using the equation of a line:

y-y_1=\frac{y_2-y_1}{x_2-x_1} (x-x_1)\\\\y-100=\frac{20-0}{400-100}(x-0)\\\\y=\frac{1}{15} x+100

We can see from the graph that there is a y intercept at 100. That is the earnings starts from 100.

The equation of a line is given as y = mx + b, where m is the slope and b is the y intercept (initial value)

For the first graph, the slope is 1/20 and the initial value is 0 while for the second graph the slope is 1/15 and the initial value is 100

D) The line pass through (10, 10) and (100, 40), hence:

y-y_1=\frac{y_2-y_1}{x_2-x_1} (x-x_1)\\\\y-10=\frac{40-10}{100-10}(x-10)\\\\y-10=\frac{1}{3} (x-10)\\\\y=\frac{1}{3}x+\frac{20}{3}

3 0
3 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)]

8 0
3 years ago
if it takes 10 men 10 days to dig a hole how long will it take 5 men to dig a hole that's half as big
Nataly [62]

Answer:

10 days

Step-by-step explanation:

It will take 10 men 5 days to dig a hole half as big.

It will take 5 men 10 days to dig a hole half as big.

( Remember, the less people, the more time it will take. )

8 0
2 years ago
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How can I solve 900-382?
BlackZzzverrR [31]

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<h3>   517</h3><h3>Hope this helps</h3>
6 0
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Which function listed below represents a greater rate of change, and what is the slope of the function?
Flauer [41]

Answer:

B

Step-by-step explanation:

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