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3 years ago

14

Is it possible for a pair of angles to be both corresponding angles and alternate interior angles

1 answer:

marin [14]3 years ago

3 0

Answer:

When the two lines intersected by the transversal are parallel, corresponding angles are congruent, alternate interior angles are congruent, alternate exterior angles are congruent, and consecutive interior angles become supplementary, which means they have a sum of 180 degrees.

Step-by-step explanation:

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Please help I will reward brainly! Need done ASAP

hoa [83]

Fort Sumter, Battle of Bull Run, <span>Battle of Shiloh, Battle of Vicksburg and Battle of </span>Gettysburg

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4 years ago

Giving like 62 points PLZ giving brainliest Which expressions are equivalent to 9x(2+y) ? Select each correct answer.

grigory [225]

<span>9x(2+y)
= 18x + 9xy

answer

</span><span>18x+(9x⋅y)
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3 years ago

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Let f(x) = - 4x + 5. Find and simplify f(x + 2).

aniked [119]

Answer:

-4x - 3.

Step-by-step explanation:

f(x) = -4x + 5.

f(x + 2) = -4(x + 2) + 5

= -4x - 8 + 5

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Hope this helps!

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3 years ago

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Can somebody explain please?

Scilla [17]

Answer:

34 degrees

Step-by-step explanation:

considering the two lines are parallel:

(3x + 7) + (2x + 3) = 180 degrees

3x + 2x + 7 + 3 = 180 degrees

5x + 10 = 180

5x = 180 - 10

5x = 170

x = 170/5 = 34

the value of x = 34 degrees

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4 years ago

Read 2 more answers

Surface-finish defects in a small electric appliance occur at random with a mean rate of 0.3 defects per unit. find the probabil

mario62 [17]

Answer: Probability that a randomly selected unit will contain at least two surface- finish defect is 0.04.

Step-by-step explanation:

Since we have given that

Mean rate defects per unit = 0.3

Since we will use "Poisson distribution":

$P(X=K)=\frac{e^{-\lambda}\lambda^k}{k!}$

But we need to find the probability that a randomly selected unit will contain at least two surface-finish defect.

$P(X\geq 2)=1+P(X=0)+P(X=1)$

So,

$P(X=0)=\frac{e^{-0.3}0.3^0}{0!}=e^{-0.3}=0.74\\\\P(X=1)=\frac{e^{-0.3}\times 0.3}{1}=0.22$

so, it becomes,

$P(X>2)=1-(0.74+0.22)=1-0.96=0.04$

Hence, probability that a randomly selected unit will contain at least two surface- finish defect is 0.04.

4 0

4 years ago