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3 years ago

(2x-1) (4x+1) solve equation

2 answers:

8 0

(2x-1)(4x+1) just multiply the terms in the parentesis

Do the following:

2x*4x
2x*1
-1*4x
-1*1

You get:

8x^2+2x-4x-1

Final result:

8x^2-2x-1

Ymorist [56]3 years ago

4 0

(6x-1) is the answer to (2x-1) (4x+1)

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tom always keeps his garden clean. he takes 20 minutes to clean a single garden. how much time will it take to clean all four ga

neonofarm [45]

80 minutes to clean all four gardens

8 0

3 years ago

Read 2 more answers

Over the last 3 evenings, Keisha received a total of 81 phone calls at the call center. The first evening, she received 6 few

tensa zangetsu [6.8K]

Answer:

First: 15

Second: 21

Third: 45

Step-by-step explanation:

In order to create the equation we need to represent the evenings with an alebraic term, in thsi case we are going to represent the second evening with an X

Second evening:x

The first night she got 6 fewer calls than the second: Second-6=x-6

The third night she received 3 times the first: 3(first night)=3(x-6)

The equation is First Night plus second night plus third night equals 81.

$First+second+third=81\\x+(x-7)+3(x-6)=81\\x+x-7+3x-18=81\\5x=81+21+6\\5x=105\\x=\frac{105}{5} \\x=21$

So the first evening he received 15 calls, the second he received 21 and the third one he received 45.

8 0

3 years ago

Find the sum of the positive integers less than 200 which are not multiples of 4 and 7

taurus [48]

Answer:

$12942$ is the sum of positive integers between $1$ (inclusive) and $199$ (inclusive) that are not multiples of $4$ and not multiples $7$ .

Step-by-step explanation:

For an arithmetic series with:

$a_1$ as the first term,
$a_n$ as the last term, and
$d$ as the common difference,

there would be $\displaystyle \left(\frac{a_n - a_1}{d} + 1\right)$ terms, where as the sum would be $\displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n)$ .

Positive integers between $1$ (inclusive) and $199$ (inclusive) include:

$1,\, 2,\, \dots,\, 199$ .

The common difference of this arithmetic series is $1$ . There would be $(199 - 1) + 1 = 199$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}$ .

Similarly, positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $4$ include:

$4,\, 8,\, \dots,\, 196$ .

The common difference of this arithmetic series is $4$ . There would be $(196 - 4) / 4 + 1 = 49$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $7$ include:

$7,\, 14,\, \dots,\, 196$ .

The common difference of this arithmetic series is $7$ . There would be $(196 - 7) / 7 + 1 = 28$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $28$ (integers that are both multiples of $4$ and multiples of $7$ ) include:

$28,\, 56,\, \dots,\, 196$ .

The common difference of this arithmetic series is $28$ . There would be $(196 - 28) / 28 + 1 = 7$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}$ .

The requested sum will be equal to:

the sum of all integers from $1$ to $199$ ,
minus the sum of all integer multiples of $4$ between $1\!$ and $199\!$ , and the sum integer multiples of $7$ between $1$ and $199$ ,
plus the sum of all integer multiples of $28$ between $1$ and $199$ - these numbers were subtracted twice in the previous step and should be added back to the sum once.