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2 years ago

Kaelyn has some yarn that she wants to use to make hats and scarves. Each hat uses 0.2 kilograms of yarn and

Mathematics

2 answers:

ioda2 years ago

4 0

Answer:

s=3h

0.2h+0.1s=5

Step-by-step explanation:

Khan Acadamy

slavikrds [6]2 years ago

3 0

using 5 kilograms of yarn Kaelyn can make 16 (or) 17 hats and 16 (or) 17 scarfs and

5.55 times available to fulfill her wish of making 3 times as many scarves as hats using 5 kilograms of yarn.

Step-by-step explanation

Let

h be the number of hats uses 0.2 kilograms of yarn

s be the number of scarves uses 0.1 kilograms of yarn

The equation will be = 3h + 3s ≤ 5 kilograms

= 3(0.2) + 3(0.1) ≤ 5

= 5 ÷ 0.6 + 0.3

$= 3h + 3s ≤ 5 kilograms = 3(0.2) + 3(0.1) ≤ 5 = 5 ÷ 0.6 + 0.3$

= 5 ÷ 0.9 = 5.55 times when Kaelyn wants to make 3 times as many scarves as hats

Number of hats = 5.55×3 = 16.65 ≤17 hats

$Number of hats = 5.55×3 = 16.65 ≤17 hats$

Number of scarfs = 5.55×3 = 16.55 ≤ 17 scarfs

$Number of scarfs = 5.55×3 = 16.55 ≤ 17 scarfs$

using 5 kilograms of yarn Kaelyn can make 16 (or) 17 hats and 16 (or) 17 scarfs and

5.55 times available to fulfill her wish of making 3 times as many scarves as hats using 5 kilograms of yarn.

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Weight of an object=J
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2 years ago

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How many solutions are there to the system of equations?

hichkok12 [17]

Answer: The system of equations has NO SOLUTION.

Step-by-step explanation:

The equation of the line in Slope-Intercept form is:

$y=mx+b$

Where "m" is the slope and "b" is the y-intercept.

Given the following system of equations:

$\left \{ {{4x-5y=5} \atop {-0.08x+0.10y=0.10}} \right.$

Write the first equation and solve for "y" in order to express it in Slope-Intercept form:

$4x-5y=5\\\\-5y=-4x+5\\\\y=\frac{-4x}{-5}+\frac{5}{-5}\\\\y=0.8x-1$

You can identify that:

$m=0.8\\b=-1$

Apply the same procedure with the second equation. Then:

$-0.08x+0.10y=0.10\\\\0.10y=0.08x+0.10\\\\y=\frac{0.08x}{0.10} +\frac{0.10}{0.10}\\\\y=0.8x+1$

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2 years ago

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Which statement is not used to prove that ΔABC is similar to ΔADE? triangles ABC and ADE in which point E is between points A an

Levart [38]

Answer:

The correct option is;

The sum of angles A and B are supplementary to angle C

Step-by-step explanation:

The statements are analysed as follows

1. Angle A is congruent to itself reflective property

Which shows that ΔABC and ΔADE have a common and equal angle

2. Segment ED and CB are parallel

From the transversal line passing EB and CB which shows that the angles ∠ADE and ∠ABC are equal and also ∠AED and ∠ACB are equal

The statement is used to prove similarity between the ΔABC and ΔADE

3. The sum of angles A and B are supplementary to angle C

The above statements relates to only ΔABC and i does not show similarity between ΔABC and ΔADE.

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2 years ago

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One of the tallest buildings in a country is topped by a high antenna. The angle of elevation from the position of a surveyor on

irina1246 [14]

Answer:

a. distance of the surveyor to the base of the building = 2051.90 ft

b. height of the building = 1384 ft

c. Angle of elevation from the surveyor to the top of the antenna = 38.31°

d. Height of antenna = 237.08 ft

Step-by-step explanation:

The picture above is a illustration of the described event.

a = the height of the flag

b = the height of the building

c = distance of the surveyor from the base of the building

the angle of elevation from the position of the surveyor on the ground to the top of the building = 34°

distance from her position to the top of the building = 2475 ft

distance from her position to the top of the flag = 2615 ft

(a) How far away from the base of the building is the surveyor located?

using the SOHCAHTOA principle

cos 34° = c/2475

c = 0.8290375726 × 2475

c = 2051.8679921

c = 2051.90 ft

(b) How tall is the building

The height of the building = b

sin 34° = opposite /hypotenuse

0.5591929035 = b/2475

b = 0.5591929035 × 2475

b = 1384.0024361

b = 1384.00 ft

(c) What is the angle of elevation from the surveyor to the top of the antenna?

let the angle = ∅

cos ∅ = adjacent/hypotenuse

cos ∅ = 2051.90/2615

cos ∅ = 0.784665392

∅ = cos-1 0.784665392

∅ = 38.310258303

∅ = 38.31°

(d) How tall is the antenna?

height of the antenna = a

sin 38.31° = opposite/hypotenuse

sin 38.31° = (a + b)/2615

sin 38.31° × 2615 = (a + b)

(a + b) = 0.6199159917 × 2615

(a + b) = 1621.0803182

(a + b) = 1621. 08 ft

Height of antenna = 1621. 08 - 1384.00 = 237.08031822 ft

Height of antenna = 237.08 ft

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2 years ago

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