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Anastaziya [24]
2 years ago
7

Plz help me with this

Mathematics
1 answer:
gayaneshka [121]2 years ago
7 0

Answer:

The Placebo Effect.

Step-by-step explanation:

The sugar pills are called Placebos. If a sick person gets better taking them, it is known as the Placebo Effect.

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the lenght of a rectangle is 2 cm less than 7 times the width. the perimeter is 60 cm . find the width and lenght
vampirchik [111]
First, you want to establish your equations.

L=7W-2
P=60

This is what we already know. To find the width, we have to plug in what we know into P=2(L+W), our equation to find perimeter.

60=2(7W-2+W)

Now that we only have 1 variable, we can solve.

First, distribute the 2.
60=14W-4+2W

Next, combine like terms.

60=16W-4

Then, add four to both sides.

64=16W

Lastly, divide both sides by 16


W=4

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7W-2

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3 years ago
Which problem has the same answer as 48 6 ?
pshichka [43]
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Points M, N, and P are respectively the midpoints of sides AC , BC , and AB of △ABC. Prove that the area of △MNP is on fourth of
Hunter-Best [27]

Answer:

The area of △MNP is one fourth of the area of △ABC.

Step-by-step explanation:

It is given that the points M, N, and P are the midpoints of sides AC, BC and AB respectively. It means AC, BC and AB are median of the triangle ABC.

Median divides the area of a triangle in two equal parts.

Since the points M, N, and P are the midpoints of sides AC, BC and AB respectively, therefore MN, NP and MP are midsegments of the triangle.

Midsegments are the line segment which are connecting the midpoints of tro sides and parallel to third side. According to midpoint theorem the length of midsegment is half of length of third side.

Since MN, NP and MP are midsegments of the triangle, therefore the length of these sides are half of AB, AC and BC respectively. In triangle ABC and MNP corresponding side are proportional.

\triangle ABC \sim \triangle NMP

MP\parallel BC

MP=\frac{BC}{2}

By the property of similar triangles,

\frac{\text{Area of }\triangle MNP}{\text{Area of }\triangle ABC}=\frac{PM^2}{BC^2}

\frac{\text{Area of }\triangle MNP}{\text{Area of }\triangle ABC}=\frac{(\frac{BC}{2})^2}{BC^2}

\frac{\text{Area of }\triangle MNP}{\text{Area of }\triangle ABC}=\frac{1}{4}

Hence proved.

5 0
3 years ago
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