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quester [9]
3 years ago
8

Kerry has a net spendable income of $1,800 per month. She also has a significant amount of debt and payments of $150 per month.

She is looking for an apartment to rent. Which of the following apartments would best fit her budget? A. Rent: $450 Electricity: $60 Water: $40 Gas: $80 B. Rent: $500 Electricity: $50 Water: $40 Gas: $ 70 C. Rent: $640 Electricity: included in rent Water: included in rent Gas: included in rent D. Rent: $475 Electricity: $50 Water: $40 Gas: $35
Mathematics
1 answer:
Firdavs [7]3 years ago
7 0
A the rent all together is
450 + 60 + 40 + 80 = 630
b the rent alltogether would be
500 + 50 + 40 + 70 = 660
the rent for c would cost

640 including everything
and the rent for d would be
475 + 50 + 40 + 35 = 60
so the answer would be D
hope this helped sorry for late answer
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The population at Bishop High School is 2000 students in 2011. Every year the population decreases by 50 students
Artemon [7]

Answer:

  • Let p be the population at t be the number of years since 2011.  Then, p=2000-50t
  • The projected population of the high school in 2015=1800
  • In <u>2019</u> the population be 1600 students

Step-by-step explanation:

Given: The population at Bishop High School students in 2011 =2000

Also,  Every year the population decreases by 50 students which implies the rate of decrease in population is constant.

So, the function is a linear function.

Let p be the population at t be the number of years since 2011.

Then, p=2000-50t

So at t=0, p=2000

In year 2015, t=4, substitute t=4 in the above equation ,we get

p=2000-50(4)\\\Rightarrow\ p=2000-200\\\Rightarrow\ p=1800

Hence, the projected population of the high school in 2015=1800

Now, put p=1600 in the function , we get

1600=2000-50t\\\Rightarrow50t=2000-1600\\\Rightarrow50t=400\\\Rightarrow\ t=\frac{400}{50}\\\Rightarrow\ t=8

Now, 2011+8=2019

Hence, in <u>2019</u> the population be 1600 students




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Nancy’s 50-meter freestyle swim time was 28.004 seconds. Betty’s 50-meter freestyle swim time was 27.949 seconds. What was the d
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If you subtract Betty's time for the swim, 27.948 seconds, from Nancy's time, 28.004 seconds, you will find that Betty was faster than Nancy by 0.056 seconds. She would be considered the winner of the 50-meter freestyle swim.
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jonny [76]

Answer:

∠13 ≅ ∠16   - Vertical Angles Theorem

∠10 ≅ ∠14  - corresponding angles for parallel line p and q cut by the transversal s

∠5 ≅ ∠13  - corresponding angles for

parallel lines r and s cut by

the transversal q

∠1 ≅ ∠5 - corresponding angles for

parallel lines r and s cut by

the transversal q

Step-by-step explanation:

Linear Pair Theorem won't be used. When you look at the lines on the image you see that 13 and 16 are vertical from each other making there answer the vertical angles theorem. When you look at 10 and 14 you see that they lie on p and q with s going in the center of them. When you look at 5 and 13 they lie on s and r with q going down the middle of them. With 1 and 5  they also lie on p  and q but r goes down the center of them instead of s.

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In Exercises 5-7, find all the exact t-values for which the given statement is true,
bearhunter [10]

Answer:  See Below

<u>Step-by-step explanation:</u>

NOTE: You need the Unit Circle to answer these (attached)

5) cos (t) = 1

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In radians:     t = 0π + 2πn

In degrees:   t = 0° + 360n

******************************************************************************

6)\quad sin (t) = \dfrac{1}{2}

Where on the Unit Circle does   sin = \dfrac{1}{2}

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\text{Answer: at}\  \dfrac{\pi}{6}\ (30^o)\ \text{and at}\ \dfrac{5\pi}{6}\ (150^o)\ \text{and all rotations of}\ 2\pi \ (360^o)

\text{In radians:}\ t = \dfrac{\pi}{6} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{6} + 2\pi n

In degrees:    t = 30° + 360n  and  150° + 360n

******************************************************************************

7)\quad tan (t) = -\sqrt3

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<em>Hint: sin and cos are only opposite signs in Quadrants II and IV</em>

\text{Answer: at}\  \dfrac{2\pi}{3}\ (120^o)\ \text{and at}\ \dfrac{5\pi}{3}\ (300^o)\ \text{and all rotations of}\ 2\pi \ (360^o)

\text{In radians:}\ t = \dfrac{2\pi}{3} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{3} + 2\pi n

In degrees:    t = 120° + 360n  and  300° + 360n

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