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3 years ago

PLZZZ HELP ME OUT THIS DUE RIGHT NOW AND I FREAKING OUT PLEASE HELP ME P.S YOU HAVE WRITE ANSWERS FOR ALL OF THEM

Mathematics

2 answers:

Levart [38]3 years ago

6 0

Answer:

a then b each have a value work out whether its numbered or distributed on a number line -2 , -1 0 , 1, 2, 3, 4, 5 etc if a is -4 then if 1-a = 1--4 the answer is -5 then when the question ab +1 comes up if b = 3 then ab-1 will be written out again as -4 x 3 -1 = -12-1=-13 etc. etc.

ab+1 is different if a = -4 and b = 3 then ab+1 = -4 x 3 +1 = -12 +1 = -11

its count back for( minus then times) count back for( minus then subtract ) and count back for any subtract and forward for positive add minus

Hope that helps branliest please =)

AnnZ [28]3 years ago

5 0

Answer:

Step-by-step explanation:

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Consider the two data sets below:

alina1380 [7]

Answer:

Option c) The data sets will have the same values of their interquartile range.

Explanation:

1. The values are in order: they are in increasing oder, from lowest to highest value.

2. Calculate the interquartile range.

Interquartile range, IQR, is the third quartile, Q3, less the first quartile Q1:

IQR = Q3 - Q1

To find the first and the third quartile, first find the median:

Data Set 1: 19, 25, 35, 38, 41, 49, 50, 52, 59

[19, 25, 35, 38], 41, [49, 50, 52, 59]

↑

median = 41

Data Set 2: 19, 25, 35, 38, 41, 49, 50, 52, 99

[19, 25, 35, 38] , 41, [49, 50, 52, 99]

↑

median = 41

Now find the median of each subset: the values below the median and the values above the median.

Data set 1: First quartile

[19, 25, 35, 38],

↑

Q1 = [25 + 35] / 2 = 30

Third quartile

[49, 50, 52, 59]

↑

Q3 = [50 + 52] / 2 = 51

IQR = Q3 - Q1 = 51 - 30 = 21

Data set 2: First quartile

[19, 25, 35, 38]

↑

Q1 = [25 + 35] / 2= 30

Third quartile

[49, 50, 52, 99]

↑

Q3 = [52 + 50]/2 = 51

IQR = 51 - 30 = 21

Thus, it is shown that the data sets have will have the same values for the interquartile range: IQR = 21. (option c)

This happens because replacing one extreme value (in this case the maximum value) by other extreme value does not affect the median.

An outlier will change the range because the range is the maximum value less the minimum value.

5 0

2 years ago

What is the perimeter of triangle ABC? Round each step to the nearest tenth. Please help me

Ad libitum [116K]

I would use the pythagorean theorem to find the lengths of each side. a² + b² = c².

Side AB is one we're looking for. If you make other right triangle with that same side you can see that one length is 4 and the other is 3. So, 4² + 3² = c² → 25 = c² → 5 = c. Side AB is length 5.

Side AC is another. Do the same with that side and you get that one length is 4 and the other is 3. (This is the same one as above) so side AC is length 5.

Side BC is the last one. One of the lengths is 1 and the other is 1 → 1² + 1² = c² → 2 = c² → 1.414213562 = c so side BC is approximately length 1.41.

Add each length up and you get a perimeter of roughly 11.4

8 0

3 years ago

Simplify 12^3/12^7 A,B,C or D....???

Lesechka [4]

Answer:

1/12^4

Step-by-step explanation:

I'm pretty sure this is the answer sorry if I am wrong

3 0

3 years ago

The table shows some values of a function of the form y = ax2 + bx + c.

nadezda [96]

Answer:

c = 5

Step-by-step explanation:

We can find a, b, c by filling in values from the table into the equation:

17 = a(2²) +b(2) +c

32 = a(3²) +b(3) +c

53 = a(4²) +b(4) +c

There are numerous ways to solve 3 linear equations in 3 unknowns. We can use elimination.

Subtracting the first equation from each of the other two, we get ...

15 = 5a +b . . . . . . . . . note that c has been eliminated from the equations

36 = 12a +2b

Subtracting twice the first from the second gives ...

(12a +2b) -2(5a +b) = 36 -2(15)

2a = 6

a = 3

Now that we have a value for "a", we can "back substitute" into the equations to find "b" and "c".

Substituting this into 15 = ..., we get ...

15 = 5(3) +b

0 = b

And substituting for "a" and "b" in the first original equation gives ...

17 = 4(3) +c

5 = c

The value of c, the constant of the function, is 5.

8 0

3 years ago

Does anyone know how to do this? I’m confused

nikklg [1K]

Answer:

cos(θ)

Step-by-step explanation:

Para una función f(x), la derivada es el límite de

f(x+h)−f(x)

, ya que h va a 0, si ese límite existe.

dθ

(sin(θ))=(

h→0

lim

sin(θ+h)−sin(θ)

)

Usa la fórmula de suma para el seno.

h→0

lim

sin(h+θ)−sin(θ)

Simplifica sin(θ).

h→0

lim

sin(θ)(cos(h)−1)+cos(θ)sin(h)

Reescribe el límite.

(

h→0

lim

sin(θ))(

h→0

lim

cos(h)−1

)+(

h→0

lim

cos(θ))(

h→0

lim

sin(h)

)

Usa el hecho de que θ es una constante al calcular límites, ya que h va a 0.

sin(θ)(

h→0

lim

cos(h)−1

)+cos(θ)(

h→0

lim

sin(h)

)

El límite lim

θ→0

sin(θ)

es 1.

sin(θ)(

h→0

lim

cos(h)−1

)+cos(θ)

Para calcular el límite lim

h→0

cos(h)−1

, primero multiplique el numerador y denominador por cos(h)+1.

(

h→0

lim

cos(h)−1

)=(

h→0

lim

h(cos(h)+1)

(cos(h)−1)(cos(h)+1)

)

Multiplica cos(h)+1 por cos(h)−1.

h→0

lim

h(cos(h)+1)

(cos(h))

−1

Usa la identidad pitagórica.

h→0

lim

−

h(cos(h)+1)

(sin(h))

Reescribe el límite.

(

h→0

lim

−

sin(h)

)(

h→0

lim

cos(h)+1

sin(h)

)

El límite lim

θ→0

sin(θ)

es 1.

−(

h→0

lim

cos(h)+1

sin(h)

)

Usa el hecho de que

cos(h)+1

sin(h)

es un valor continuo en 0.

(

h→0

lim

cos(h)+1

sin(h)

)=0

Sustituye el valor 0 en la expresión sin(θ)(lim

h→0

cos(h)−1

)+cos(θ).

cos(θ)

5 0

3 years ago

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