I will give brainliest. Help Me, please

5(5t + 1) = 25t-7 Can someone help me with this solution??

masha68 [24]

Firstly you multiply 5 with everything in the brackets on the left side of = . and than you put t on the left side and change sign (if it was + then its going to be -) and numbers on right side and do the same with signs. then just calculate the rest

5(5t +1) = 25t - 7
25t + 5 = 25t - 7
25t - 25t = -7 + 5
0=-2
the statement is false

7 0

3 years ago

2.8, 3.4, 4.0, 4.6, . . . Write an equation for the nth term of the arithmetic sequence. Then find a50.

vova2212 [387]

Answer:

a_n = 2.2 + 0.6 n

a_50 = 32.2

Step-by-step explanation:

What's the common difference of this series?

$a_1 = 2.8$

$a_2 = 3.4$

Common difference = $a_2 - a_1 = 3.4 - 2.8 = 0.6$ .

Expression for the nth term:

$a_n = a_1 + (n - 1) \cdot\text{Common Difference} \\\phantom{a_n } = 2.8 + 0.6 \; (n-1) \\\phantom{a_n} = 2.8 + 0.6 \; n - 0.6\\\phantom{a_n} = 2.2 + 0.6\; n$

n = 50 for the fiftieth term. Therefore

$a_{50} = 2.2 + 0.6 \times 50 = 32.2$ .

6 0

3 years ago

Please could someone explain this to me?

qwelly [4]

Ok so A is the same as the square root of 4b+c but all of the numbers and the answer have to be prime numbers

4 0

2 years ago

Find k so that the following function is continuous: f(x)={kx8x2if0≤x<5if5≤x.

tankabanditka [31]

Check the one-sided limits:

$\displaystyle \lim_{x\to5^-}f(x) = \lim_{x\to5}kx = 5k$

$\displaystyle \lim_{x\to5^+}f(x) = \lim_{x\to5}8x^2 = 200$

If f(x) is to be continuous at x = 5, then these two limits should have the same value, which means

5k = 200

k = 200/5

k = 40

3 0

2 years ago

The number of tickets purchased by an individual for Beckham College's holiday music festival is a uniformly distributed random

Zanzabum

1) Mean: 6.5

2) Standard deviation: 3.2

Step-by-step explanation:

1)

A uniform random distribution is the distribution of a variable that can assume n different values, and each of these values is equally likely to occur.

This means that the probability of occurring of each of these values is the same:

$p(x_1)=p(x_2)=...=p(x_n)$

For a uniform random distribution, the mean of the distribution is simply equal to the median value of the distribution, which is exactly at the center of the distribution itself. It can be calculated as

$\bar x = \frac{1}{2}(a+b)$