Shawheen puts the cost of a weekend trip⁠

Write a linear equation that passes through the point (2,-9) and has a slope of -5

Doss [256]

Answer:

y=-5x+1

Step-by-step explanation:

y-y1=m(x-x1)

y-(-9)=-5(x-2)

y+9=-5x+10

y=-5x+10-9

y=-5x+1

3 0

3 years ago

An airplane flew 4 hours with a 25 mph tail wind. The return trip against the same wind took 5 hours. Find the speed of the airp

balu736 [363]

Answer:

r = 225 Mil/h speed of the airplane in still air

Step-by-step explanation:

Then:

d is traveled distance and r the speed of the airplane in still air

so the first equation is for a 4 hours trip

as d = v*t

d = 4 * ( r + 25) (1) the speed of tail wind (25 mil/h)

Second equation the trip back in 5 hours

d = 5 * ( r - 25 ) (2)

So we got a system of two equation and two unknown variables d and

r

We solve it by subtitution

from equation (1) d = 4r + 100

plugging in equation 2

4r + 100 = 5r - 125 ⇒ -r = -225 ⇒ r = 225 Mil/h

And distance is :

d = 4*r + 100 ⇒ d = 4 * ( 225) + 100

d = 900 + 100

d = 1000 miles

6 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.