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dangina [55]
3 years ago
11

HELP PLEASE!!! Maia walks 12 miles. Then she walks 1/4 of that distance. select ALL of the expressions that represent the total

distance she walks.

Mathematics
1 answer:
posledela3 years ago
5 0

Answer:

1, 2, and 3

Step-by-step explanation:

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Given f(x) = -4x - 2, solve for x when f(x) = 6.
Andrej [43]

Answer:

use

Step-by-step explanation:

calculator

6 0
3 years ago
What’s the answer I need The answer ASAP
ser-zykov [4K]

Answer:

The only one that applies is g = 6 because 102 divided by 6 is 17 which is less that 33 and all the other answers are greater that 33.

Hope this helps!

Step-by-step explanation:

3 0
3 years ago
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Formulate the situation as a system of two linear equations in two variables. Be sure to state clearly the meaning of your x- an
slega [8]

Answer:

1) There were 33 $4,000 investors and 27 $8,000 investors.

2) The solution in x = 4, y = 9.

3) There were 24 nickels and 56 dimes.

Step-by-step explanation:

1) A lawyer has found 60 investors for a limited partnership to purchase an inner-city apartment building, with each contributing either $4,000 or $8,000. If the partnership raised $348,000, then how many investors contributed $4,000 and how many contributed $8,000?

I am going to say that:

x is the number of investors that contributed 4,000.

y is the number of investors that contributed 8,000.

Building the system:

There are 60 investors. So:

x + y = 60

In all, the partnership raised $348,000. So:

4000x + 8000y = 348000

I am going to simplify by 4000. So:

x + 2y = 87

Solving the system:

The elimination method is a method in which we can transform the system such that one variable can be canceled by addition. So:

1)x + y = 60

2)x + 2y = 87

I am going to multiply 1) by -1. So we have

1)-x - y = -60

2)x + 2y = 87

By addition, the x are going to cancel each other

-x + x - y + 2y = -60 + 87

y = 27

For x:

x + y = 60

x = 60-y = 60-27 = 33

There were 33 $4,000 investors and 27 $8,000 investors.

2) Solve the system by row-reducing the corresponding augmented matrix.

2x + y = 17

x + y = 13

This system has the following augmented matrix:

\left[\begin{array}{ccc}2&1&17\\1&1&13\end{array}\right]

To help the row reducing, i am going to swap the first with the second line:

L1  L2

So we have:

\left[\begin{array}{ccc}1&1&13\\2&1&17\end{array}\right]

Now, reducing the first column.

L2 = L2 - 2L1

So we have:

\left[\begin{array}{ccc}1&1&13\\0&-1&-9\end{array}\right]

Now we do:

L2 = -L2

And the matrix is:

\left[\begin{array}{ccc}1&1&13\\0&1&9\end{array}\right]

Now to reduce the second column, we do:

L1 = L1 - L2

\left[\begin{array}{ccc}1&0&4\\0&1&9\end{array}\right].

So the solution is:

x = 4, y = 9.

3) A jar contains 80 nickels and dimes worth $6.80. How many of each kind of coin are in the jar?

I am going to say that x is the number of nickels and y is the number of dimes.

Each nickel is worth 5 cents and each dime is worth 10 cents.

Building the system:

There are 80 coins in all:

x + y = 80

They are worth $6.80. So:

0.05x + 0.10y = 6.80

Solving the system:

1)x + y = 80

2)0.05x + 0.10y = 6.80

I am going to divide 1) by -10, so we can cancel y. So:

1)-0.10x - 0.10y = -8

2)0.05x + 0.10y = 6.80

Adding:

-0.10x + 0.05x - 0.10y + 0.10y = -8 + 6.80

-0.05x = -1.2 *(-100)

5x = 120

x = \frac{120}{5}

x = 24

Also

x + y = 80

y = 80-x = 80-24 = 56

There were 24 nickels and 56 dimes.

8 0
3 years ago
13<br> 5<br> 11<br> True/False...the given triangle is a right triangle.<br> O True<br> O False
Gala2k [10]

Answer:

False

Step-by-step explanation:

11 times 11 is 121

5 times 5 is 25

13 times 13 is 169

121+25 doesnt equal 169

this means it is not a right angled triangle as it does not follow pythag

7 0
3 years ago
Evaluate exactly the value of the integral from -1 to 0 (2x^4+8x)^3 (4x^3+4),dx. Include the use of substitution and the antider
ddd [48]
Hello,

Let's be

 z=2x^4+8x \\&#10;dz=(8x^3+8)\ dx=8(x^3+1)\ dx \\&#10;&#10; \int\limits^0_{-1} {(2x^4+8x)^3*(4x^3+4)} \, dx =4* \int\limits^0_{-6} { \dfrac{z^3}{2}} \, dz \\&#10;= 2* \left[\begin{array}{ccc} \dfrac{z^4}{4}\end{array}\right] ^0_{-6}=&#10;\left[\begin{array}{ccc} \dfrac{z^4}{2}\end{array}\right] ^0_{-6}=0-648=-648

3 0
3 years ago
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