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2 years ago

Write an equation in slope intercept form for the line with slope -4 and y intercept 1

Mathematics

1 answer:

aalyn [17]2 years ago

8 0

Answer:

y = -4x + 1

Step-by-step explanation:

you said, slope of -4, therefore it's -4x and you also said y intercept of 1, therefore it's a positive 1 : +1

combine them together: y = -4x +1

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Henry can write 5 pages of his novel in 3 hours. How many pages can he read in 8 hours

Neporo4naja [7]

Answer:40 pages in 8 hours

Step-by-step explanation:

5 times 8 is 40 3 times five is 15

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3 years ago

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Is FGH ~ JKL? If so, identify the similarity postulate or theorem that applies.

Leokris [45]

Yes, for triangles to similar two angles have to be the same, in this case, they both have 55 and 30 degrees), this makes the third angle the same. Triangles angles add up to 180 degrees and if you subtract 30 and 55 from 180..
180-30=150-55=95
95+30+55=180

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3 years ago

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Let X be the temperature in at which a certain chemical reaction takes place, and let Y be the temperature in (so Y = 1.8X + 32)

Black_prince [1.1K]

Answer:

See explanation

Step-by-step explanation:

Solution:-

The random variable, Y be the temperature of chemical reaction in degree fahrenheit be a linear expression of a random variable X : The temperature in at which a certain chemical reaction takes place.

Y = 1.8*X + 32

- The median of the random variate "X" is given to be equal to "η". We can mathematically express it as:

P ( X ≤ η ) = 0.5

- Then the median of "Y" distribution can be expressed with the help of the relation given:

P ( Y ≤ 1.8*η + 32 )

- The left hand side of the inequality can be replaced by the linear relation:

P ( 1.8*X + 32 ≤ 1.8*η + 32 )

P ( 1.8*X ≤ 1.8*η ) ..... Cancel "1.8" on both sides.

P ( X ≤ η ) = 0.5 ...... Proven

Hence,

- Through conjecture we proved that: (1.8*η + 32) has to be the median of distribution "Y".

- Recall that the definition of proportion (p) of distribution that lie within the 90th percentile. It can be mathematically expressed as the probability of random variate "X" at 90th percentile :

P ( X ≤ p_.9 ) = 0.9 ..... 90th percentile

- Now use the conjecture given as a linear expression random variate "Y",

P ( Y ≤ 1.8*p_0.9 + 32 ) = P ( 1.8*X + 32 ≤ 1.8*p_0.9 + 32 )

= P ( 1.8*X ≤ 1.8*p_0.9 )

= P ( X ≤ p_0.9 )

= 0.9

- So from conjecture we saw that the 90th percentile of "X" distribution is also the 90th percentile of "Y" distribution.

- The more general relation between two random variate "Y" and "X" is given:

Y = aX + b

Where, a : is either a positive or negative constant.

- Denote, (np) as the 100th percentile of the X distribution, so the corresponding 100th percentile of the Y distribution would be : (a*np + b).

- When a is positive,

P ( Y ≤ a*p_% + b ) = P ( a*X + b ≤ a*p_% + b )

= P ( a*X ≤ a*p_% )

= P ( X ≤ p_% )

= np_%

- When a is negative,

P ( Y ≤ a*p_% + b ) = P ( a*X + b ≤ a*p_% + b )

= P ( a*X ≤ a*p_% )

= P ( X ≥ p_% )

= 1 - np_%

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2 years ago

Using the triangle theorem solve this triangle.

Blizzard [7]

Answer:

8 and 8√3

Step-by-step explanation:

Sin90 = b/8

1 x 8 =b

b=8

Tan 30 = 8/a

1/√3 = 8/a

a=8√3

3 0

2 years ago

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If ABCD is an A4 sheet and BCPO is the square, prove that △OCD is an isosceles triangle. And find the angles marked as 1 to 8 wi

Dmitry [639]

Answer:

The diagram for the question is missing, but I found an appropriate diagram fo the question:

Proof:

since OC = CD = 297mm Therefore, Δ OCD is an isoscless triangle

∠BCO = 45°

∠BOC = 45°

∠PCO = 45°

∠POC = 45°

∠DOP = 22.5°

∠PDO = 67.5°

∠ADO = 22.5°

∠AOD = 67.5°

Step-by-step explanation:

Given:

AB = CD = 297 mm

AD = BC = 210 mm

BCPO is a square

∴ BC = OP = CP = OB = 210mm

Solving for OC

OCB is a right anlgled triangle

using Pythagoras theorem

(Hypotenuse)² = Sum of square of the other two sides

(OC)² = (OB)² + (BC)²

(OC)² = 210² + 210²

(OC)² = 44100 + 44100

OC = √(88200

OC = 296.98 = 297

OC = 297mm

An isosceless tringle is a triangle that has two equal sides

Therefore for △OCD

CD = OC = 297mm; Hence, △OCD is an isosceless triangle.

The marked angles are not given in the diagram, but I am assuming it is all the angles other than the 90° angles

Since BC = OB = 210mm

∠BCO = ∠BOC

since sum of angles in a triangle = 180°

∠BCO + ∠BOC + 90 = 180

(∠BCO + ∠BOC) = 180 - 90

(∠BCO + ∠BOC) = 90°

since ∠BCO = ∠BOC

∴ ∠BCO = ∠BOC = 90/2 = 45

∴ ∠BCO = 45°

∠BOC = 45°

∠PCO = 45°

∠POC = 45°

For ΔOPD

$Tan\ \theta = \frac{opposite}{adjacent}\\ Tan\ (\angle DOP) = \frac{87}{210} \\(\angle DOP) = Tan^-1(0.414)\\(\angle DOP) = 22.5 ^{\circ}$

Note that DP = 297 - 210 = 87mm

∠PDO + ∠DOP + 90 = 180

∠PDO + 22.5 + 90 = 180

∠PDO = 180 - 90 - 22.5

∠PDO = 67.5°

∠ADO = 22.5° (alternate to ∠DOP)

∠AOD = 67.5° (Alternate to ∠PDO)

3 0

3 years ago