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vodka [1.7K]
3 years ago
15

Please help quickly !

Mathematics
2 answers:
bixtya [17]3 years ago
8 0

Answer:

A

Step-by-step explanation:

xxMikexx [17]3 years ago
4 0

Answer:

x- 41.5 = 13.5

Step-by-step explanation:

The normal rate is x and the sick rate is 41.5

The difference is 13.5

x- 41.5 = 13.5

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Determine which postulate or theorem can be used to prove that ABC=DCB
lianna [129]

Answer with explanation:

In Δ ABC and ΔD BC

∠A=∠D-------Given

∠ABC=∠DCB-------Each being 90° given in the Diagram.

Side, BC is Common.

⇒⇒Δ ABC ≅ ΔD BC-------[AAS]

When two triangles are congruent , their areas are equal.

So, Area(ABC)=Area (DCB)

Option B :⇒ A AS

3 0
3 years ago
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Some1 help please :) dont answer if u are not 100% sure thank you
Natalija [7]

Answer:

Step-by-step explanation:

It's never negative.

5 0
3 years ago
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-(y-7)=14 solve the following equation
enyata [817]

First, distribute the negative sign and get -y + 7 = 14

then subtract 7 from both sides and get -y = 7

Divide both sides by -1 and get y = -7

Hope this helps :)

3 0
3 years ago
On a coordinate plane, triangle A B C is shown. Point A is at (0, 0), point B is at (3, 4), and point C is at (3, 2). What is th
Elena L [17]

Answer:

The area of triangle for the given coordinates is  1.5\sqrt{4.6}

Step-by-step explanation:

Given coordinates of triangles as

A = (0,0)

B = (3,4)

C = (3,2)

So, The measure of length AB = a = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}

Or, a = \sqrt{(3-0)^{2}+(4-0)^{2}}

Or, a =  \sqrt{9+16}

Or, a =   \sqrt{25}

∴ a = 5 unit

Similarly

The measure of length BC = b = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}

Or, b = \sqrt{(3-3)^{2}+(2-4)^{2}}

Or, a =  \sqrt{0+4}

Or, b =   \sqrt{4}

∴ b = 2 unit

And

So, The measure of length CA = c = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}

Or, c = \sqrt{(3-0)^{2}+(2-0)^{2}}

Or, c =  \sqrt{9+4}

Or, c =   \sqrt{13}

∴ c = \sqrt{13} unit

Now, area of Triangle written as , from Heron's formula

A = \sqrt{s\times (s-a)\times (s-b)\times (s-c)}

and s = \frac{a+b+c}{2}

I.e  s = \frac{5+2+\sqrt{13}}{2}

Or. s =  \frac{7+\sqrt{13}}{2}

So, A = \sqrt{(\frac{(7+\sqrt{13})}{2})\times ((\frac{(7+\sqrt{13})}{2})-5)\times (\frac{7+\sqrt{13}}{2}-2)\times (\frac{7+\sqrt{13}}{2}-\sqrt{13})}

Or, A = \sqrt{(\frac{(7+\sqrt{13})}{2})\times (\frac{(\sqrt{13}-3)}{2})\times (\frac{4+\sqrt{13}}{2})\times (\frac{7-\sqrt{13}}{2})}

Or, A = \frac{3}{2} × \sqrt{1+\sqrt{13} }

∴  Area of triangle = 1.5\sqrt{4.6}

Hence The area of triangle for the given coordinates is  1.5\sqrt{4.6}  Answer

7 0
3 years ago
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Can someone help me? I'll reward points and brainalist
UNO [17]

Answer:

135 degrees

Step-by-step explanation:

45 is for half of the square so u would add 45+45+45 or 45x3. there is a half and 1 whole square.

7 0
3 years ago
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