Solve -11 > -1 - 2x solve your answer step by step

7 1⁄5 – 6 2⁄5 = ? A. 4⁄5 B. 1 4⁄5 C. 1 1⁄5 D. 13 3⁄5

STatiana [176]

Answer:

A

Step-by-step explanation:

Change the mixed numbers to improper fractions

7 $\frac{1}{5}$ = $\frac{36}{5}$

6 $\frac{2}{5}$ = $\frac{32}{5}$

The subtraction is then

$\frac{36}{5}$ - $\frac{32}{5}$

Both fractions have a common denominator so subtract the numerators leaving the denominator

= $\frac{36-32}{5}$ = $\frac{4}{5}$ → A

4 0

3 years ago

What is the lowest and highest grade average in the class respectively?

sineoko [7]

$\Huge\bf \rightarrow \mid\mathcal{\underline{ \purple{Answer}}} \mid$

 -76 and 95 is the lowest and highest grade average in the class, respectively.

7 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.

That is:

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

8 0

3 years ago

Heather can complete two problems in ten minutes, and when she started she had three problems done. Joel can complete three prob

Hunter-Best [27]

If Heather can do 2 in 10 minutes, then that means that she can do 12 in 1 hour. If Joel can do 3 in 15 minutes, then it means that he can do 12 in 1 hour. Add the number of problems they can do total and that will go between 30 and 50. It is just "less than", not "less than or equal to" because you want to know when the total number goes between the numbers.
Therefore, the correct answer must be 30 < 24x + 5 < 50.

3 0

3 years ago

For this question I need to find 80% of 20

Ipatiy [6.2K]

Answer:

16

Step-by-step explanation:

.8 x 20 =16

7 0

3 years ago