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Ray Of Light [21]
3 years ago
6

Scarborough High School ordered several replacement books for the mathematics department. When the box of books arrived at the h

igh school, it contained Algebra I, Geometry, and Algebra II textbooks. A label on the box reads: “Contents: 23 books, Weight: 93 lbs.” An Algebra I book weighs 4 pounds, a Geometry book weighs 3 pounds, and an Algebra II book weighs 5 pounds. The number of Geometry books and Algebra II books combined is one less than the number of Algebra I books.
Formulate a system of equations to represent the problem situation.
Solve the system of equations to determine the number of Algebra I textbooks, the number of Geometry textbooks, and the number of Algebra II textbooks that are in the box.
Mathematics
1 answer:
Alekssandra [29.7K]3 years ago
6 0

Answer:

Step-by-step explanation:

Let us represent the number of books as:

Algebra 1 = A1

Algebra 2 = A2

Geometry = G

A1 + A2 + G = 23........Equation 1

An Algebra I book weighs 4 pounds, a Geometry book weighs 3 pounds, and an Algebra II book weighs 5 pounds.

4A1 + 5A2 + 3G = 93

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Number A is a function
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Write a subtraction fact with the same difference as 16 -7
hoa [83]
14-5=9, 12-3=9, 15-6=9
7 0
3 years ago
Read 2 more answers
Si un triángulo tiene una hipotenusa de 15 cm y un cateto que mide 12 cm, ¿cual es la longitud del otro cateto?
BartSMP [9]

Answer:

   

9cm

Explicación:

El teorema de pitagoras nos dice que la hipotenusa al cuadrado es igual a la suma de los catetos al cuadrado:

c^{2} =a^{2} + b^{2}

Donde c es la hipotenusa y a  y b son los catetos.

Por lo cual, podemos reemplazar c por 15cm y a por 12cm :

15^{2} =12^{2} +b^{2}

Finalmente , debemos resolver la ecuación para b, así que b es igual a :

225=144+b^{2} \\225-144=144+b^{2} -144\\81=b^{2} \\\sqrt{81} =b\\9=b

Por lo tanto, la longitud del otro cateto es 9 cm

3 0
3 years ago
Does anyone have the answers for all?
Shkiper50 [21]

Answer and Step-by-step explanation:

1-6

a.

What you can do is plug in 6 and start by doing the first machine first, then the second. If this doesn't work, do the second machine, then the first machine.

<em><u>The order that works is second machine, then first machine.</u></em>

<u>Second Machine:</u>

y = (6)^{2} -6 = 36 - 6 = 30

Plug into first machine now.

<u>First Machine:</u>

y = \sqrt{(30) - 5}  = \sqrt{25} = 5

We see that the result is 5.

b. <em><u>If we were to do the order of second machine, then first machine, we will never get a negative result (or -5 for that matter), due to the fact that you can't algebraically square root a negative number.</u></em>

<em><u>However, if we were to do the order of first machine, second machine, we can get a negative number. To get -5, we plug in 6 for x in the first machine, getting an answer of 1. We plug this into the second machine, getting 1^2 - 6, which results in -5.</u></em>

1-7. The absolute value of a value is the positive version of that value.

a. |54| = <em><u>54</u></em>

b. -|-7\frac{3}{5}| = -7\frac{3}{5}

c. |3| - |-1| = 3 - 1 = <em><u>2</u></em>

d. |2.2 - 5.13| = |-2.93| = <em><u>2.93</u></em>

1-8.

a.

_

<u>|  |</u>

<u>|  |</u>

<u>|  |</u>

<u>|  |</u>

<u>|  |</u> <u>  </u> <u>   </u> <u>  </u> <u>   </u>

<u>|  ||  ||  ||  ||  |</u>

<u>(5 boxes on each side)</u>

_

<u>|  |</u>

<u>|  |</u>

<u>|  |</u>

<u>|  |</u>

<u>|  |</u> <u>  </u> <u>   </u> <u>  </u> <u>   </u> <u>   </u>

<u>|  ||  ||  ||  ||  ||  |</u>

<u>(6 boxes on each side)</u>

b.<em><u> The amount of squares on each side of the figure increases by 1.</u></em>

c. <em><u>There would be 1 square, because we know that the pattern is there is 2 squares added to the next figure, 1 at each end. If we go back a figure, we take away 2 squares.</u></em>

1-9.

a. -42 - 17 = <em><u>-59</u></em>

b. 8 - (-9) = 8 + 9 = <em><u>17</u></em>

c. 8(-9) = <em><u>-72</u></em>

d. -42 ÷ (-7) = <em><u>6</u></em>

e. -2(-3)(-4) = 6(-4) = <em><u>24</u></em>

f. -18 - 7 = <em><u>-25</u></em>

g. (-5)^{2} = <em><u>25</u></em>

h. -5^{2}  = -(5)^2 = -(25) = <em><u>-25</u></em>

i. \sqrt{49} = <em><u>7</u></em>

1-10.  x = 2 Plug 2 in for x, then solve.

a. y = 7 - 3(2) = 7 - 6 = <em><u>1</u></em>

b. y = (2)^2 -1 = 4 - 1 = <em><u>3</u></em>

c. y = \frac{18}{2} = <em><u>9</u></em>

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

<em><u>I hope this helps!</u></em>

8 0
3 years ago
The price of a four star dinner was $100 two decades ago. A report says that a dinner is 250% of what it was 20 years ago, how m
leva [86]

For this case we can solve the problem by means of the following rule of three:

100 $ --------------> 100%

x -------------------> 250%

From here, we clear the value of x.

We have then:

x = (\frac{250}{100}) (100)

Rewriting we have:

x = 250

Answer:

The price of a four star dinner today is $ 250.

5 0
3 years ago
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