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Elena-2011 [213]
3 years ago
5

There are 0.97 grams of salt in a container and nayan uses 0.56 grams

Mathematics
2 answers:
gogolik [260]3 years ago
7 0

Answer:

Hey, what is the question for this problem. Is it asking how much salt is left after nayan uses some of it?

if so, the answer is .41 grams of salt

Step-by-step explanation:

.97 grams (total) minus .56 grams( what is used)

is equal to .41 grams

belka [17]3 years ago
7 0

Answer:

0.41

Step-by-step explanation:

0.97-0.56=0.41 because she used 0.56 so subtract it from the total number

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Which expression is equivalent to 30 (1/2x - 2) +40 (3/4y-4)? (will add pic)
bixtya [17]

Answer:

C

Step-by-step explanation:

Given

30( \frac{1}{2} x - 2) + 40( \frac{3}{4} y - 4)

Multiply each of the terms in the first parenthesis by 30 and multiply each of the terms in the second parenthesis by 40.

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3 years ago
A. Dos números enteros positivos cuya diferencia sea 42
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4 0
3 years ago
In triangle $ABC$, let angle bisectors $BD$ and $CE$ intersect at $I$. The line through $I$ parallel to $BC$ intersects $AB$ and
Umnica [9.8K]

Answer:

41

Step-by-step explanation:

If you work through a series of obscure calculations involving area and the radius of the incircle, they boil down to a simple fact:

... For MN║BC, perimeter ΔAMN = perimeter ΔABC - BC = AB+AC

.. = 17+24 = 41

_____

Wow! Thank you for an interesting question with a not-so-obvious answer.

_____

<em>A little more detail</em>

The point I that you have defined is the incenter—the center of an inscribed circle in the triangle. Its radius is the distance from I to any side, such as BC, for example.

If we use "Δ" to represent the area of the triangle and "s" to represent the semi-perimeter, (AB+BC+AC)/2, then the incircle has radius Δ/s. The area Δ can be computed from Heron's formula by ...

... Δ = √(s(s-a)(s-b)(s-c)) . . . . where a, b, c are the side lengths

For this triangle, the area is Δ = √38480 ≈ 196.1632 units². That turns out to be irrelevant.

The altitude to BC will be 2Δ/(BC), so the altitude of ΔAMN = (2Δ/(BC) -Δ/s). Dividing this by the altitude to BC gives the ratio of the perimeter of ΔAMN to the perimeter of ΔABC, which is 2s.

Putting these ratios and perimeters together, we get ...

... perimeter ΔAMN = (2Δ/(BC) -Δ/s)/(2Δ/(BC)) × 2s

... = (2/(BC) -1/s) × BC × s = 2s -BC

... perimeter ΔAMN = AB +AC

8 0
3 years ago
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