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2 years ago

PLEASE HELP WILL MARK BRAINLIEST

Mathematics

1 answer:

prisoha [69]2 years ago

8 0

Answer:

Step-by-step explanation:

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-5(x+7)<10. What does x equal?

gulaghasi [49]

Answer:

X>-9

Step-by-step explanation:

Distribute

-5x-35<10

add 35 on both sides

-5x<45

Divide -5 on both sides and flip the sign since your dividing by a NEGATIVE number

x>-9

4 0

2 years ago

Read 2 more answers

Last year 80 students signed up for a summer trip to Washington, D.C. This summer 50 students have signed up to go. What is the

Fynjy0 [20]

Given:

Last year 80 students signed up for a summer trip to Washington, D.C.

This summer 50 students have signed up to go.

To find:

The percent decrease in the number of students.

Solution:

We have,

Students in last year = 80

Students in this year = 50

Now,

$\text{Decrease}\%=\dfrac{\text{Students in last year - Students in this year }}{\text{Students in last year }}\times 100$

$\text{Decrease}\%=\dfrac{80-50}{80}\times 100$

$\text{Decrease}\%=\dfrac{30}{80}\times 100$

$\text{Decrease}\%=\dfrac{3}{8}\times 100$

$\text{Decrease}\%=37.5$

Therefore, the correct option is B.

7 0

2 years ago

How many of the scores from the group that studied in question eight are in one mean absolute deviation of the mean? Which score

n200080 [17]

Answer:

Step-by-step explanation:

Add all of the numbers and divide them by the number of numbers there are.

Hope this helped ·ω·

4 0

2 years ago

Money Flow The rate of a continuous money flow starts at $1000 and increases exponentially at 5% per year for 4 years. Find the

77julia77 [94]

Answer:

Present value = $4,122.4

Accumulated amount = $4,742

Step-by-step explanation:

Data provided in the question:

Amount at the Start of money flow = $1,000

Increase in amount is exponentially at the rate of 5% per year

Time = 4 years

Interest rate = 3.5% compounded continuously

Now,

Accumulated Value of the money flow = $1000e^{0.05t}$

The present value of the money flow = $\int\limits^4_0 {1000e^{0.05t}(e^{-0.035t})} \, dt$

= $1000\int\limits^4_0 {e^{0.015t}} \, dt$

= $1000\left [\frac{e^{0.015t}}{0.015} \right ]_0^4$

= $1000\times\left [\frac{e^{0.015(4)}}{0.015} -\frac{e^{0.015(0)}}{0.015} \right]$

= 1000 × [70.7891 - 66.6667]

= $4,122.4

Accumulated interest = $e^{rt}\int\limits^4_0 {1000e^{0.05t}(e^{-0.035t}} \, dt$

= $e^{0.035\times4}\times4,122.4$

= $4,742

8 0

2 years ago

What is the 5th term of an arithmetic sequence if t2 = -5 and t6 = 7

RUDIKE [14]

An=Asub1+d(n-1)
Asub5= -5+½(4)
=-5+7
=2

5 0

3 years ago