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3 years ago

I dont now this question

Mathematics

1 answer:

Scrat [10]3 years ago

6 0

Answer:

Step-by-step explanation:

3x + 2 = (2x + 13)/3

9x + 6 = 2x + 13

9x - 2x = 13 - 6

7x = 7

x = 7 : 7

x = 1

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Please help?

Mazyrski [523]

Answer:

$23\sqrt{3}\ un^2$

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

$A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\$

2. Note that

$A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2$

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

$A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2$

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

7 0

3 years ago

a subsription is $4.95 per month and 4-by-6 inch prints are $0.18 each.How many prints does .insert person here lol. purchase if

valentinak56 [21]

The number of prints he buys is 36

5 0

3 years ago

Read 2 more answers

How do you solve these?

Murrr4er [49]

Answer:

Step-by-step explanation:

The first and only rule really is to factor these down to their primes and then apply a very simple rule

For every prime, take out 1 prime for every prime under the root sign that equals the index. The rest are thrown away.

That's very wordy. Let's try and see what it means with an example

Take sqrt(27) The index is 1/2 (square root) That means we need two threes in order to apply the rule.

sqrt(27) = sqrt(3 * 3 * 3 ) For every two primes take out 1 and throw one away.

sqrt(27) = 3 sqrt(3) You can't take out that 3rd 3.

64 = 2 * 2 *2 *2 *2 * 2

4th root 64 = 2*2*2 *2 * 2 *2

for every 4th root, you get to take 1 out and throw three away.

4th root 64 = 2 fourth root (2*2)

4th root 64 = 2 fourth root (4)

- 189 = - 3 * 3 * 3 * 7

cuberoot (- 189) = For every 3 roots, you get to pull 1 out and throw the other two away.

3 cube (- 7) is your answer.

72 = 2 * 2 * 2 * 3 * 3

cube root (72) = 2 cube root(9) You don't have enough threes to do any more than what is done.

6 0

3 years ago

Solve the following equation for a number x. -15=8x+1. Verify that the solution is correct.

sasho [114]

Ok so the first step is to -1 from both sides. On the right side it cancels out and on the left side it = -16. So now your left with -16= 8x. Your still not done yet though. So now what you want to do is divide both sides by 8. On the right side it cancels out and you bring down x and on the left side it equals -2. So x=-2. So to verify it you have to plu -2 back into the equation. So -15= 8 Times -2 +1 so 8 Times -2 = -16 plus one is -15. Hoped that helped!!

4 0

4 years ago

Which of the following represents the zeros of f(x)=3x^3-10x^2-81x 28? Answer choices: A.) 7, -4,1/3 B.) 7,-4, -1/3 C.) 7,4,1/3

11Alexandr11 [23.1K]

The right function is f(x)=3x^3-10x^2-81x + 28

You can realize that 7 is a root because it is in all the answers.

So you can divide the polynomial by x - 7. If you do it you can find that the quotient is 3x^2 + 11x - 4

Now you can use the quadratic formula to find the other two roots.

If you do it, you will find they are x = 1/3 and x = -4.

So the answer is option A) 7, -4, 1/3

And the polynomial can be written as (x - 7)(x + 4) (x -1/3)

8 0

3 years ago