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2 years ago

WHICH NUMBERS ARE SOLUTIONS OF EACH INEQUALITY?:

Mathematics

1 answer:

love history [14]2 years ago

4 0

Answer:

I don't know if i am right but i think it is number 7 because it like goes in order

Step-by-step explanation:

i am so unsure so i think it is 7 i don't know for sure

You might be interested in

Find the measures of an interior angle and an exterior angle of a regular 16-gon

Anastasy [175]

Answer:

2520°

Step-by-step explanation:

The sum of the angles in any triangle is 180°

shape. sides. triangles. sum int angles.

triangle. 180×1=180°

quad. 180×2=360°

pentagon. 180×3=540°

n-sides. n( n−2) 180(n−2)

16-sides. 180×14=2520°

7 0

3 years ago

Find the 10th term of the sequence if the explicit function/formula is f(n)=100+10n

Nutka1998 [239]

Answer:

The 10th term of the sequence is 200.

Step-by-step explanation:

Given that,

The sequence is as follows :

f(n)=100+10n

We need to find 10th term of the sequence.

Put n = 10 in the given expression.

f(10)=100+10(10)

= 100 + 100

= 200

Hence, the 10th term of the sequence is 200.

8 0

2 years ago

Use the graph to state the solution for the system.

Alenkasestr [34]

Your answer would be (3,0)

4 0

2 years ago

Solve: 2 + = 2. Give solutions in interval [0, 2)

Dovator [93]

$\because\cos 2x+\sin x=\sin 2x$ $\because\sin 2x=2\sin xcosx$ $\because\cos 2x=\cos ^2x-\sin ^2x$

→ Substitute cos 2x and sin 2x by their expressions

$\therefore\cos ^2x-\sin ^2+\sin x=2\sin x\cos x$

→ Subtract both sides by sin x

$\cos ^2x-\sin ^2x=2\sin xcosx-\sin x$

→ Take sin x as a common factor in the right side

$\therefore\cos ^2x-\sin ^2x=\sin x(2\cos x-1)$

From the graph, the equation has 4 solutions, the intersection points between the 2 graphs

7 0

10 months ago

Simplify the expression.

Anna71 [15]

Since we have a cubic root, we're interested in factoring cubes inside the root, so that we can take them out. If we factor 648, we have

$648=2^3\times 3^4=2^3\times 3^3\times 3 = (2\times 3)^3\times 3 = 3\times 6^3$

So, we have

$3x\sqrt[3]{648 x^4 y^8} = \sqrt[3]{3\times 6^3\cdot x^3\cdot x \cdot y^6\cdot y^2}=3x\cdot 6\cdot x\cdot y^2\sqrt[3]{3\cdot x\cdot y^2}$

And the result simplifies to

$18x^2y^2\sqrt[3]{3xy^2}$

5 0

2 years ago