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2 years ago

1/2x+40=50 What is the answer for this problem??

Mathematics

2 answers:

Mama L [17]2 years ago

8 0

Answer:

x = 20.

Step-by-step explanation:

1/2x + 40 = 50

1/2x + 40 - 40 = 50 - 40

1/2x = 10

1/2x * 2 = 10 * 2

x = 20

SIZIF [17.4K]2 years ago

7 0

Answer:

x=20

Step-by-step explanation:

50-40= 10 and 10 x 2 = 20

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Slope-intercept of 5x-7y=-45

bagirrra123 [75]

Slope-intercept form is y=ax+b where a is the slope and b is the y-intercept.
Start by getting y alone on one side.
5x-7y=-45
5x+45=7y
Then divide both sides by 7 to get y alone.
y= 5/7x + 45/7

3 0

2 years ago

A computer cost £900. It is on sale for 25% off. You also need to pay 6% sales tax. What is the final price of the computer?

Vesna [10]

First we have to find 25% of 900 pounds (i hope that's the right currency). so 900 x .25 = 225

so we subtract 225 from 900 and get 675 pounds.

to find sales tax we have to find 6% of 675, so 675 x .06 = 40.50

675 + 40.50 = 715.5 pounds

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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.