All categories

2 years ago

Select the three inequalities that have n = 3 in the solution set

Mathematics

2 answers:

kherson [118]2 years ago

6 0

Answer: A, C, E

Solve for all the inequalitites.

n + 5 < 9

n < 9 - 5

n < 4

∴ n = 3 is a part of the solution set.

13n < 1

n < 1/13

∴ This definitely doesn't include n = 3.

3n + 1 > 7

3n > 6

n > 2

∴ n = 3 is part of the solution because n is larger than 2.

3n > 9.5

n > 9.5 ÷ 3

n > 3.16666666

∴ n = 3 is not part of the solution set.

n - 2 < 5

n < 5 + 2

n < 7

∴ n = 3 is a part of the solution set.

SCORPION-xisa [38]2 years ago

4 0

Answer:

Its just E

I'm sorry if I'm wrong

You might be interested in

Which expression demonstrates the distributive property applied to the expression 4(3 + 7)

GREYUIT [131]

4(3 + 7)

= 4(3) + 4(7) //Demonstrating distributive property

= 12 + 28

= 40

7 0

2 years ago

Please help me with this fast cus i need to take a nap asap

GREYUIT [131]

Answer:

66 inches

Step-by-step explanation:

:))

6 0

2 years ago

Read 2 more answers

On a coordinate plane, triangle A B C is shifted 4 units to the right and 2 units up to form triangle A prime B prime C prime. I

zheka24 [161]

Answer:

(x, y) ⟶ (x + 4, y + 2)

Step-by-step explanation:

If you shift a point four units to the right, x ⟶ x + 4.

If you shift it up two units, y ⟶ y + 2.

If you combine the two shifts into one transformation, the rule becomes

(x, y) ⟶ (x + 4, y + 2)

3 0

2 years ago

For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by (−1)(k+1)∗(12k). If T is the sum of

Katena32 [7]

Answer:

Option D. is the correct option.

Step-by-step explanation:

In this question expression that represents the kth term of a certain sequence is not written properly.

The expression is $(-1)^{k+1}(\frac{1}{2^{k}})$ .

We have to find the sum of first 10 terms of the infinite sequence represented by the expression given as $(-1)^{k+1}(\frac{1}{2^{k}})$ .

where k is from 1 to 10.

By the given expression sequence will be $\frac{1}{2},\frac{(-1)}{4},\frac{1}{8}.......$

In this sequence first term "a" = $\frac{1}{2}$

and common ratio in each successive term to the previous term is 'r' = $\frac{\frac{(-1)}{4}}{\frac{1}{2} }$

r = $-\frac{1}{2}$

Since the sequence is infinite and the formula to calculate the sum is represented by

$S=\frac{a}{1-r}$ [Here r is less than 1]

$S=\frac{\frac{1}{2} }{1+\frac{1}{2}}$

$S=\frac{\frac{1}{2}}{\frac{3}{2} }$

S = $\frac{1}{3}$

Now we are sure that the sum of infinite terms is $\frac{1}{3}$ .

Therefore, sum of 10 terms will not exceed $\frac{1}{3}$

Now sum of first two terms = $\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$

Now we are sure that sum of first 10 terms lie between $\frac{1}{4}$ and $\frac{1}{3}$

Since $\frac{1}{2}>\frac{1}{3}$

Therefore, Sum of first 10 terms will lie between $\frac{1}{4}$ and $\frac{1}{2}$ .

Option D will be the answer.

3 0

3 years ago

Solve for x and then find the measure of

bonufazy [111]

The answer should be 150

7 0

2 years ago

Read 2 more answers