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3 years ago

15

We would like to cut exactly 160 meters of wire. Instead of measuring the length, we measure the weight. It is known that 3 mete

rs of wire weighs 75 grams.
If x meters of wire weighs y grams, express y in terms of x.

What is the weight of 160 meters of wire?

1 answer:

ivanzaharov [21]3 years ago

7 0

The weight is 4000 grams, or 4 kg

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Consider the triangle below. Determine the angle measure for the bottom two angles in order for the triangle to be classified as

FinnZ [79.3K]

Answer:

The angle measure for the bottom two angles in order for the triangle to be classified as an isosceles triangle are 30 degrees each

Explanation:

An isosceles triangle has two of its angles equal.

Given that one angle is 120 degrees, assuming the remaining angles are equal, since the sum of the angles in a triangle is 180 degrees, we have:

120 + x + x = 180

120 + 2x = 180

2x = 180 - 120 = 60

x = 60/2 = 30

Each of the remaining angles measure 30 degrees

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9 months ago

La ecuación de la recta que pasa por el punto P(1,3) y es paralela a la recta

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Answer: No puedo responder a esto sin que me muestres el problema.

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2 years ago

The perimeter of a triangular field is 120m. Two of the sides are 21m and 40m.Calculate the largest angle of the field.

Mama L [17]

Answer:

the largest angle of the field is 149⁰

Step-by-step explanation:

Given;

perimeter of the triangular filed, P = 120 m

length of two known sides, a and b = 21 m and 40 m respectively

The length of the third side is calculated as follows;

a + b + c = P

21 m + 40 m + c = 120 m

61 m + c = 120 m

c = 120 m - 61 m

c = 59 m

B

↓ ↓

↓ ↓

↓ ↓

A → → → → → → → → → → → C

Consider ABC as the triangular field;

Angle A is calculated by applying cosine rule;

$a^2 = b^2 + c^2 - 2bc \ Cos A\\\\Cos \ A = \frac{b^2 + c^2 - a^2}{2bc} \\\\Cos \ A = \frac{40^2 + 59^2 - 21^2}{2 \times 40 \times 59} \\\\Cos \ A = 0.983\\\\A = Cos ^{-1} (0.983)\\\\A = 10.6 \ ^0$

Angle B is calculated as follows;

$Cos \ B = \frac{a^2 + c^2 - b^2}{2ac} \\\\Cos \ B = \frac{21^2 + 59^2 - 40^2}{2 \times 21 \times 59} \\\\Cos \ B = 0.937\\\\B= Cos ^{-1} (0.937)\\\\B = 20.5 \ ^0$

Angle C is calculated as follows;

$Cos \ C = \frac{a^2 + b^2 - c^2}{2ab} \\\\Cos \ C = \frac{21^2 + 40^2 - 59^2}{2 \times 21 \times 40} \\\\Cos \ C = -0.857\\\\C = Cos ^{-1} (-0.857)\\\\C = 149\ ^0$

Therefore, the largest angle of the field is 149⁰.

8 0

3 years ago

What number could replace kkk below? \dfrac{k}{100} = \dfrac{1}{10} 100 k = 10 1 start fraction, k, divided by, 100, end fra

Ivanshal [37]

Answer:

<h3>k = 10</h3>

Step-by-step explanation:

Given the equation;

$\dfrac{k}{100} = \dfrac{1}{10}$

We are to find the value of k.

Cross multiply;

$\dfrac{k}{100} = \dfrac{1}{10} \\10 \times k = 100 \times 1\\10k = 100\\$

Divide both sides by 10

$10k/10 = 100/10\\k = 10$

Hence the value of k is 10

6 0

2 years ago

Please help me out on this one

vivado [14]

it is definitely the answer B

8 0

2 years ago