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Talja [164]
3 years ago
6

What do u do this please explain

Mathematics
2 answers:
Aneli [31]3 years ago
5 0
Its blurry cant help sorry
LenKa [72]3 years ago
5 0
The value of the 6 is related to the value of the 6 because the 6 = 60 which is 10 times greater than the other 6.

I hope this helps! :D
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Please help on this one plzzzz
deff fn [24]

We know that Angle Bisector Divides an Angle into Two Equal Angles.

As UP is the Angle Bisector of Angle U, It Divides Angle U into two Equal Parts they are Angle(1) and Angle(2)

⇒ Angle(1) = Angle(2)

Given Angle(1) = 5x + 10 and Angle(2) = 3x + 14

⇒ 5x + 10 = 3x + 14

⇒ 2x = 4

⇒ x = 2

⇒ Angle(1) = 5x + 10 = 5(2) + 10 = 10 + 10 = 20

So : Measure of Angle(1) is 20

8 0
3 years ago
If yvaries inversely as x and y=18 when x=2, find y when x=4
Sergio039 [100]

I believe the answer would be 36.

Hope this helped you. ;}

7 0
3 years ago
Choose any positive integer. Powers of two here are not very interesting, so choose something else. If the number you have chose
Yakvenalex [24]

Answer:

Step-by-step explanation:

Let the integer be 6 for even and 7 for odd (say)

For 6, we divide by 2, now get 3.  Now we multiply by 3 and add 1 to get 10. Now since 10 is even divide by 5, now multiply by 3 and add 1 to get 16.  Now divide by 2 again by 2 again by 2 again by 2 till we get rid of even numbers.

The result is 1, so multiply by 3 and add 1 we get 4 now divide 2 times by 2 to get 1, thus this result now again repeats after 2 times.

Say if we select off number 3, multiply by 3 and add 1 to get 10 now divide by 5, now repeat the same process as above for 5 until we get 1 and it gets repeated every third time.

Thus whether odd or even after some processes, we get 1 and the process again and again returns to 1.

5 0
3 years ago
CALC- limits<br> please show your method
gladu [14]
A. Factor the numerator as a difference of squares:

\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6

c. As x\to\infty, the contribution of the terms of degree less than 2 becomes negligible, which means we can write

\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4

e. Let's first rewrite the root terms with rational exponents:

\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}

Next we rationalize the numerator and denominator. We do so by recalling

(a-b)(a+b)=a^2-b^2
(a-b)(a^2+ab+b^2)=a^3-b^3

In particular,

(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3
(x^{1/2}-x)(x^{1/2}+x)=x-x^2

so we have

\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}

For x\neq0 and x\neq1, we can simplify the first term:

\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x

So our limit becomes

\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43
3 0
3 years ago
Dominique spent $31.41 on 9 bags of Takis. How much did each bag cost?
suter [353]
Each one cost $3.49
8 0
3 years ago
Read 2 more answers
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