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kodGreya [7K]
3 years ago
12

These Questions is worth 20 Points. No Explanation Required.

Mathematics
1 answer:
avanturin [10]3 years ago
7 0

13. five plus <em>y</em> equal to negative two. y = -7

14. eight plus <em>h</em> equal to twelve. h = 4

15. negative thirteen plus four equal <em>n</em>. n = -9

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PLS HELP ILL GIVE BRAINLIEST!
lesantik [10]

Answer:

B

Step-by-step explanation:

-the base angles are 34, so the total value of both base angles is 68

-an isosceles triangle is equal to 180

-subtract 68 from 180 for the answer, so the equation is 68+x=180

hope this helps, comment in you have a question :)

8 0
3 years ago
Read 2 more answers
In a group of 60 numbers, average of
Arada [10]

Answer:

c. 18

Step-by-step explanation:

First, you would find the total sum of the 60 numbers. The average is always the sum of the numbers divided by the number of numbers.

35*18=630

15*10=150

Now we do not know how many numbers the last group has. However, we can find it out doing 60-35-15 to see how many numbers are left. There would be 10 numbers left

10*30=300

630+150+300=1080

1080/60=18

Hence, the answer is C(18)

4 0
3 years ago
Lim x-&gt; vô cùng ((căn bậc ba 3 (3x^3+3x^2+x-1)) -(căn bậc 3 (3x^3-x^2+1)))
NNADVOKAT [17]

I believe the given limit is

\displaystyle \lim_{x\to\infty} \bigg(\sqrt[3]{3x^3+3x^2+x-1} - \sqrt[3]{3x^3-x^2+1}\bigg)

Let

a = 3x^3+3x^2+x-1 \text{ and }b = 3x^3-x^2+1

Now rewrite the expression as a difference of cubes:

a^{1/3}-b^{1/3} = \dfrac{\left(a^{1/3}-b^{1/3}\right)\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)}{\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)} \\\\ = \dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}

Then

a-b = (3x^3+3x^2+x-1) - (3x^3-x^2+1) \\\\ = 4x^2+x-2

The limit is then equivalent to

\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}}

From each remaining cube root expression, remove the cubic terms:

a^{2/3} = \left(3x^3+3x^2+x-1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3}

(ab)^{1/3} = \left((3x^3+3x^2+x-1)(3x^3-x^2+1)\right)^{1/3} \\\\ = \left(\left(x^3\right)^{1/3}\right)^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1x\right)\left(3-\dfrac1x+\dfrac1{x^3}\right)\right)^{1/3} \\\\ = x^2 \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3}

b^{2/3} = \left(3x^3-x^2+1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}

Now that we see each term in the denominator has a factor of <em>x</em> ², we can eliminate it :

\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}} \\\\ = \lim_{x\to\infty} \frac{4x^2+x-2}{x^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}\right)}

=\displaystyle \lim_{x\to\infty} \frac{4+\dfrac1x-\dfrac2{x^2}}{\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}}

As <em>x</em> goes to infinity, each of the 1/<em>x</em> ⁿ terms converge to 0, leaving us with the overall limit,

\displaystyle \frac{4+0-0}{(3+0+0-0)^{2/3} + (9+0-0+0+0-0)^{1/3} + (3-0+0)^{2/3}} \\\\ = \frac{4}{3^{2/3}+(3^2)^{1/3}+3^{2/3}} \\\\ = \frac{4}{3\cdot 3^{2/3}} = \boxed{\frac{4}{3^{5/3}}}

8 0
3 years ago
8. A paper bag contains one pink crayon, four yellow crayons, three black crayons and two red crayons. What is the probability o
vivado [14]

Answer:

C 1/2

Add all the crayons together and then see how many red and black crayons there are. which in this case it's half. That is the probability

8 0
3 years ago
Find the value of x and y
Lesechka [4]

Answer:

x = 33, y = 17

Step-by-step explanation:

1) the sum of all the angles in any triangle is 180 degrees

2) when 2 lines intersect, the angles opposite each other are equal

3) the sum of any angles that break up one side of a straight line is 180

7 0
3 years ago
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