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puteri [66]
3 years ago
5

What are the 4 parts of the Missouri comprise

Mathematics
1 answer:
boyakko [2]3 years ago
8 0

Answer:

Missouri admitted as a slave state.

Maine admitted as a free state.

Slavery disallowed in future territories north of 36°30' except within Missouri itself.

Step-by-step explanation:

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It is >5. put a black dot on 5 with a around pointing to one
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PLS HURRY!!<br> Simplify.<br> 7/3+3(2/3−1/3)^2 Enter your answer in the box.
AveGali [126]

Do distributed property   7/3+3(2/3-1/3)^2 7/3+3×(2/3-1/3)^2 7/3+3×(1/3)^2 7/3×3×1/9 7/3×1/3 7+1/3 8/3 or 2 and 2/3 or 2.66667

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4 years ago
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In the diagram below, TUVW is a rectangle. Prove triangle TZW is congruent to triangle UZV.
otez555 [7]

Answer:

By S.S.S. congruence property;  ΔTZW ≅ ΔVZU

Step-by-step explanation:

Given:

TUVW is a rectangle.

To Prove : TZW ≅ UZV

Proof:

Since TUVW is a rectangle, and we know that opposite side of a rectangle is equal.

So,

TW = UV \ \ .\ .\ .\ .\ 1

And also TV and WU are the diagonals of the rectangle.

And the diagonals of rectangle bisects each other.

Therefore;

TZ=ZV.........\ 2\\\\WZ=ZU..........\ 3

Now In ΔTZW and ΔVZU

TW = UV (from 1)

TZ = ZV (from 2)

WZ = ZU (from 3)

So, by S.S.S. congruence property;

ΔTZW ≅ ΔVZU

Hence proved.

8 0
4 years ago
Solve the system of equations using the substitution method.
Firdavs [7]

y = -1 + 3/8x

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Substitute the first equation into the second equation, since y is already by itself.

2x - 5(-1 + 3/8x) = 6

2x + 5 - 15/8x = 6

2x - 15/8x = 1

16/8x - 15/8x = 1

1/8x = 1 Multiply 8 on both sides to get x by itself

x = 8


Plug x into either of the equations.

y = -1 + 3/8(8)

y = -1 + 3

y = 2


2(8) - 5y = 6

16 - 5y = 6

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(8,2)

8 0
3 years ago
Find the equation of a parabola with a vertical axis and its vertex at the origin and passing through the point (-2, 3)
vredina [299]

a vertical axis, I assume it means a vertical axis of symmetry, thus it'd be a vertical parabola, like the one in the picture below.

\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} y=a(x- h)^2+ k\qquad \qquad \leftarrow vertical\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=0\\ k=0 \end{cases}\implies y=a(x-0)^2+0 \\\\\\ \textit{we also know that } \begin{cases} x=-2\\ y=3 \end{cases}\implies 3=a(-2-0)^2+0\implies 3=4a \\\\\\ \cfrac{3}{4}=a~\hspace{10em}y=\cfrac{3}{4}(x-0)^2+0\implies \boxed{y=\cfrac{3}{4}x^2}

8 0
3 years ago
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