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2 years ago

Choose all of the functions that are linear.

Mathematics

1 answer:

xxTIMURxx [149]2 years ago

5 0

Answers:answer is b because if you multiple y by the x you will get

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X^2 - 16 = 0 by extracting the square root

scoray [572]

Answer:

{-4, 4}

Step-by-step explanation:

x² - 16 = 0
x² = 16
x = √16
x = ± 4

6 0

2 years ago

NEED ANSWER ASAP <br> MARKING BRAINLIEST

Digiron [165]

Area of a sector is x over 360 * pi r squares so in this case x would be 210 so it would be 210 over 360 and the pi r squared would be 24 pi and when we times that our answer is 14 pi

Answer:14 pi

8 0

3 years ago

SEAGULLS A biologist studying seagulls estimates that 5% of the seagulls in a flock have been banded. If 22 of the seagulls have

mina [271]

Answer:

440 seagulls

Step-by-step explanation:

Total number of seagulls in flock = Number of seagulls have banded / Percentage of seagulls have banded

Total number of seagulls in flock = 22 / 5%

Total number of seagulls in flock = 22 / 0.05

Total number of seagulls in flock = 440 seagulls

6 0

2 years ago

Choose one of the factors of 500x3 + 108y18

sammy [17]

Answer:

Option C

Therefore, one of the factors of $500x^3 +108y^\left (18\right )$ is $(5x+3y^6)$ .

Step-by-step explanation:

Given: $500x^3 +108y^\left (18\right )$

the common factor from $500x^3$ and $108y^\left (18\right )$ is 4.

therefore, $4\cdot \left ( 125x^3+27y^\left (18\right ) \right )$

$4\cdot \left ( (5x)^3+(3y^6)^3 \right))$

Now, use the formula for above expression: $(a^3+b^3)=(a+b)(a^2-ab+b^2)$

here, $a=5x$ and $b=3y^6$

$( (5x)^3+(3y^6)^3 \right))$ $=(5x+3y^6)(25x^2-15xy^6+9y^12)$

Therefore, we have

$500x^3 +108y^\left (18\right )$ $=4\cdot (5x+3y^6)\left (25x^2-15xy^6+9y^\left ( 12 \right ) \right )$

Therefore, one of the factors of $500x^3 +108y^\left (18\right )$ is $(5x+3y^6)$ .

7 0

3 years ago

Read 2 more answers

Exhibit 6-2 the weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 poun

Evgen [1.6K]

The question is asking for the lower bound of the 95% two tailed Confidence interval of the normally distributed population.

95% C.I. is given by 200 + or - 1.96(25) = 200 + or - 49 = (151, 249)
Therefore, the minimum weight of the middle 95% of players is 151 pounds.

5 0

3 years ago