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2 years ago

Here is a base of the prism from the previous screen.

Mathematics

1 answer:

adelina 88 [10]2 years ago

7 0

Answer:

you have to break it into 3 shapes tge first is 5×3=15. tge second is 4×2=8, tge 3rd is 1/2 (4)×3=6, 15+8+6=20 units

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A rectangular house measuring 50 feet by 24 feet is to have a gravel pathway of constant width built all around it. There is eno

Maksim231197 [3]

First, we determine the area of the rectangular house by multiplying the length by the width. That is,
area of the house = (50 ft)(24 ft) = 1200 ft²
Given that all sides of the house should have the pathway, the length of the house plus the pathway should be 50+2x and the new width would be 24+2x. The new area minus the original area of the house should not be more than 345 ft². Thus, the equation would be,
(50+2x)(24+2x) - 1200 ≤ 345

3 0

2 years ago

Meredith feeds her 3 dogs twice a day. How many times does she feed the dogs in 3 days?

Maru [420]

<span>She would feed them 6 times in 3 days.</span>

8 0

3 years ago

Read 2 more answers

Wendy kept track of the number of text messages she sent each day for three weeks. Drag and drop each number into the boxes to c

NeX [460]

Answer:

0-9 = 4

10-19 = 3

20-29 = 9

30-39 = 5

Step-by-step explanation:

8 0

2 years ago

Which expression is equivalent to the expression -3 (4r - 2) -2r?

liubo4ka [24]

Answer:

-14r + 6

Step-by-step explanation:

-12r + 6 - 2r

3 0

2 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.

That is:

$19900 - 4900 - 2842 + 784 = 12942$ .

8 0

3 years ago