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3 years ago

Charles has $1600 dollars in his account and makes automatic $400

Mathematics

1 answer:

kipiarov [429]3 years ago

3 0

1600(-400)= f is the y axis and -400 because its taking money from his account

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Which of the following are solutions to the equation below (5x+1)^2=7?<br><br> Check all that apply.

stiv31 [10]

(5x+1)² = 7; expand: 25x² + 10x +1 = 7 or 25x² + 10x -6 = 0

Solve this quadratic for x:

x = [-b + √(b² - 4ac)]/2a and x = [-b - √(b² - 4ac)]/2a
Plug the values and you'll find :
x = (- √7 - 1)/5 (answer C)
x = (+√7 -1)/5 (answer E)

4 0

3 years ago

Read 2 more answers

Radio signals travel at a rate of 3 x 10^8 meters per second. How many seconds would it take for a radio signal to travel from a

wel

Time = Distance / Speed [ Already mentioned ]

Then, substitute the known values,

t = (9.6 × 10⁶) / (3 × 10⁸)

t = 3.2 × 10⁻² s

In short, Your Answer would be Option B

Hope this helps!

4 0

3 years ago

Joe ate twice as many pieces of Halloween candy as Ed. Together they ate 48 pieces.

gulaghasi [49]

Answer: 16

Step-by-step explanation:

Let the number of candy Ed took be denoted as "x"

Since Joe ate twice the amount of candy Ed ate, then the number of candies joe ate will be denoted as "2x"

Since all together they ate a total of 48 candies, then the sum of the candies eaten by Joe and Ed is what gives a total of 48

This means:

2x + x = 48

3x = 48

X = 16.

8 0

3 years ago

Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

Bezzdna [24]

Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

$a_2 = a_1 + 2$

$a_3 = a_2 + 2 = a_1 + 2\cdot2$

$a_4 = a_3+2 = a_1+3\cdot2$

$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$