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Ira Lisetskai [31]
3 years ago
11

Help Me Please~

Mathematics
1 answer:
Mkey [24]3 years ago
3 0

Answer:

45^0.999

Step-by-step explanation:

The bases are equal so we wont divide them, we shall only divide the exponents.

<em>2799/2800=0.999</em>

<em>hope it helps</em>

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Two Parnell lines (lines, L, and L2) are intersected by a transversal (L,3). Fill in the missing angle measure for each diagram
almond37 [142]

Apply the : congruent (corresponding alternate exterior and alternate interior angles) and supplementary (consecutive interior angles) theorems

step-by-step explanation:

The property to apply here is : congruent (corresponding alternate exterior and alternate interior angles) and supplementary (consecutive interior angles)

1.h = 180°-120°=60°=e=d=a

  f=120° = c=b

2.b=180°-105°=75°=c=f=g

   a=105°=e=h

3. f=180°-36°=144°=g=c=b

   a=36°=d=h

4. a=180°-161°=19°=e=h=d

   c=161°=f=g

5. f=180°-145°=35°=g=c=b

   a=145°=d=h

6. b=180°-25°=155°=c=f=g

    25°=d=e=h

Learn More

Angles :brainly.com/question/13209798

Keywords: parallel lines, intersect, transversal, angle

#LearnwithBrainly

7 0
4 years ago
Identify the expression with nonnegative limit values. More info on the pic. PLEASE HELP.
marshall27 [118]

Answer:

\lim _{x\to 2}\:\frac{x-2}{x^2-2}\\\\  \lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}\\\\ \lim _{x\to \frac{5}{2}}\left\frac{2x^2+x-15}{2x-5}\right

Step-by-step explanation:

a) \lim _{x\to 3}\:\frac{x^2-10x+21}{x^2+4x-21}=\lim \:_{x\to \:3}\:\frac{\left(x-7\right)\left(x-3\right)}{\left(x+7\right)\left(x-3\right)}=\lim \:_{x\to \:3}\:\frac{x-7}{x+7}=\frac{3-7}{3+7}=-\frac{4}{10}=-\frac{2}{5}

b) \lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=\lim \:_{x\to -\frac{3}{2}}\:\frac{\left(2x+3\right)\left(x-4\right)}{\left(2x+3\right)}=\lim \:\:_{x\to \:-\frac{3}{2}}\:\left(x-4\right)=-\frac{3}{2}-4\\ \\ \lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=-\frac{11}{2}

c) \lim _{x\to 2}\:\frac{x-2}{x^2-2}=\frac{2-2}{\left(2\right)^2-2}=\frac{0}{4-2}=0

d) \lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}=\lim _{x\to 11}\:\frac{\left(x-11\right)\left(x+17\right)}{\left(x-11\right)\left(x+14\right)}=\lim _{x\to 11}\:\frac{\left(x+17\right)}{\left(x+14\right)}=\frac{11+17}{11+14}=\frac{28}{25}

e) \lim _{x\to 3}\:\frac{x^2-8x+15}{x-3}=\lim \:_{x\to \:3}\:\frac{\left(x-3\right)\left(x-5\right)}{x-3}=\lim _{x\to 3}\left(x-5\right)=3-5=-2

f) \lim _{x\to \frac{5}{2}}\left(\frac{2x^2+x-15}{2x-5}\right)=\lim \:_{x\to \:\frac{5}{2}}\frac{\left(2x-5\right)\left(x+3\right)}{2x-5}=\lim \:\:_{x\to \:\:\frac{5}{2}}\left(x+3\right)=\frac{5}{2}+3=\frac{11}{2}

4 0
3 years ago
Please Help, and show work..
azamat
Hi there! So the the new price is $93, and went up from $55. Let's subtract the numbers. 93 - 55 is 38. The change is 38. To find the percent markup, we can write and solve a proportion. Set it up like this:

38/55 = x/100

This is because the amount of change goes above the original (55), and x represents the percent out of 100. We are looking for the markup. Let's cross multiply the values. 100 * 38 is 3,800. 55 * x is 55x. 3,800 = 55x. Now, divide each side by 55 to isolate the x. 55x/55 cancels out. 3,800/55 is 69.0909 or 69 when rounded to the nearest whole number. There. The percent of markup is approx. 69%.
3 0
3 years ago
If Jay ran 1 mile in 10 minutes on Monday, 2 miles in 20 minutes on Tuesday,
earnstyle [38]

Answer:

4 miles

Step-by-step explanation:

5 0
4 years ago
What is the probability of getting either a sum of 66 or at least one 66 in the roll of a pair of​ dice?
Sindrei [870]
So I'm hoping you mean 6, or else the probability is 0. So the possible ways of getting a sum of 6 is 1&5, 2&4, 3&3, 4&2, 5&1. There are 36 different ways of rolling the dice, so divide the 5 possible ways by 36. Then, we need to add the possibility of getting at least one 6 in the roll. It doesn't matter what the other die is, so for each die, the probability is 1/6, and for both, the probability is 2/6. Add this to the probability of getting a sum of 6, and you get your answer.
7 0
3 years ago
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