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3 years ago

Help with question 2 please.

Mathematics

1 answer:

Firdavs [7]3 years ago

3 0

Words would be "x is smaller than 1"

Algebra would be x < 1

Possible solutions are 0, -1, -2

Hope this helps.

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3 years ago

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What is the first step to solve this equation 4m=44

son4ous [18]

Answer: H

The first step is dividing 4 from both sides to find what m equals. By the way, its the <u>only</u> step :D

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Please help me out with this!!<br> BRAINLIEST AVAILABLE!!

lapo4ka [179]

Answer:

xy = 1

k = 79

Step-by-step explanation:

Question One

The first and third frames look to me to be the same. I'll treat them that way.

y = x^2 Equate y = x^2 to the result of 2y + 6 = 2x + 6

2y + 6 = 2(x + 3) Remove the brackets

2y + 6 = 2x + 6 Subtract 6 from both sides

2y = 2x Divide by 2

y = x

Now solve these two equations.

so x^2 = x

x > 0

1 solution is x = 0 from which y = 0. This won't work. x must be greater than 0. So the other is

x(x) = x Divide both sides by x

x = 1

y = x^2 Put x = 1 into x^2

y = 1^2 Solve

y = 1

The second solution is

(1,1)

xy = 1*1

xy = 1

Answer: A

Question Two

square root(k + 2) - x = 0

Subtract x from both sides

sqrt(k + 2) = x Square both sides

k + 2 = x^2 Let x = 9

k + 2 = 9^2 Square 9

k + 2 = 81

k = 81 - 2

k = 79

4 0

3 years ago

que condiciones debe cumplir las medidas de un triangulo para que puedas asegurar que es un triangulo rectangulo

nirvana33 [79]

Answer:

ENGLISH

One of the angles must equal 90 degrees and the other two angles must add up to 90 degrees.

SPANISH TRANSLATION

Uno de los ángulos debe ser igual a 90 grados y los otros dos ángulos deben sumar 90 grados.

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3 years ago

The following describes a sample. The information given includes the five number summary, the sample size, and the largest and s

lys-0071 [83]

Answer:

$L= Q_1 - 1.5*IQR = 210 -1.5*40 =150$

$U= Q_3 + 1.5*IQR = 250 +1.5*40 =310$

And if we analyze the info provided we have 500 values

Tails: 160,165,167,171,175,...,268,269,269,269,270,270

So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

Step-by-step explanation:

For this case we have the 5 number summary

(160,210,220,250,270)

So then we have:

minimum = 160 , Q1 = 210, Q2= Median=220, Q3 = 250, Max=270

If we find the interquartile range we got:

$IQR = Q_3 -Q_1 = 250-210 =40$

For this case we need to find the lower and upper limit with the following formulas:

$L= Q_1 - 1.5*IQR = 210 -1.5*40 =150$

$U= Q_3 + 1.5*IQR = 250 +1.5*40 =310$

And if we analyze the info provided we have 500 values

Tails: 160,165,167,171,175,...,268,269,269,269,270,270

So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

4 0

4 years ago