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Alex73 [517]
4 years ago
9

Plato math algebra 2

Mathematics
2 answers:
Bas_tet [7]4 years ago
8 0

Answer:

do you have algebra 2 for plato as well

Step-by-step explanation:

wel4 years ago
3 0
Is there a specific question that you want to ask?
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what's the width of the base of a rectangle if the length is 70ft less than twice the width and the perimeter is 760
schepotkina [342]

Answer:

Let width be x ft

Length=2x-70 ft

Perimeter=760ft

Perimeter=2(x+2x-70)

760 ft=2(3x-70)

380ft=3x-70 ft

450ft=3x

x=450/3

x=150ft.

6 0
3 years ago
21 points !!!! .For each of the following, the given angle is the base angle of an
Ilia_Sergeevich [38]

Answer:

150°

Step-by-step explanation:

180 - 2(15)

180 - 30

150

8 0
3 years ago
Can someone please help me with the answer
pychu [463]

Answer:

x =  \frac{46}{3}

Step-by-step explanation:

A straight angle is equal to 180° degrees. Since both 9x +4 and 3x -18 is supplementary then I can make this claim below

(9x + 4 ) + (3x - 8) = 180

All we have to do now is to solve for x

(9x + 4) + (3x - 8) = 180 \\ 9x + 4+ 3x - 8 = 180 \\ 12x - 4 = 180 \\ 12x = 184 \\ x = \frac{184}{12}  \\ x =  \frac{46}{3}

6 0
3 years ago
What is the area, in square units, of the parallelogram shown below? A parallelogram ABCD is shown with height 8 units and base
GenaCL600 [577]

Answer:

Step-by-step explanation:

Area of parallelogram = base * height

                                      = 6 * 8

                                      = 48 square units

5 0
3 years ago
Read 2 more answers
Use Theorem 2.1.1 to verify the logical equivalence. Give a reason for each step. -(pv –q) v(-p^q) = ~p
sertanlavr [38]

Answer:

The statement \lnot(p\lor\lnot q)\lor(\lnot p \land \lnot q) is equivalent to \lnot p, \lnot(p\lor\lnot q)\lor(\lnot p \land \lnot q) \equiv \lnot p

Step-by-step explanation:

We need to prove that the following statement \lnot(p\lor\lnot q)\lor(\lnot p \land \lnot q) is equivalent to \lnot p with the use of Theorem 2.1.1.

So

\lnot(p\lor\lnot q)\lor(\lnot p \land \lnot q) \equiv

\equiv (\lnot p \land \lnot(\lnot q))\lor(\lnot p \land \lnot q) by De Morgan's law.

\equiv (\lnot p \land q)\lor(\lnot p \land \lnot q) by the Double negative law

\equiv \lnot p \land (q \lor \lnot q) by the Distributive law

\equiv \lnot p \land t by the Negation law

\equiv \lnot p by Universal bound law

Therefore \lnot(p\lor\lnot q)\lor(\lnot p \land \lnot q) \equiv \lnot p

4 0
3 years ago
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