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horrorfan [7]
3 years ago
13

I need helo with this test

Mathematics
2 answers:
tia_tia [17]3 years ago
8 0

Answer:its 5i might be wrong

Step-by-step explanation:

Alecsey [184]3 years ago
4 0
Hello Did you mean help or just a hello
You might be interested in
In an arithmetic sequence, find a6 when a1 = 13 and d = 4
jolli1 [7]

a_{6} = 33

the n th term of an arithmetic sequence is

a_{n} = a_{1} + (n - 1 )d, hence

a_{6} = 13 + (5 × 4 ) = 13 + 20 = 33


6 0
3 years ago
4 out of 5 dentists prefer Crest. If they asked 30 dentists what tooth paste they prefer, how many did NOT PREFER Crest.
ExtremeBDS [4]

Answer:

24 would prefer CREST

6 woul NOT prefer Crest

Step-by-step explanation:

4:5=x:30

8 0
3 years ago
Need help on problem 40 part b for integrating in respect to y! Thanks!
igomit [66]

Answer:   \bold{(a)\quad \dfrac{32}{3}\qquad (b)\quad \dfrac{32}{3}}

<u>Step-by-step explanation:</u>

(a) First, find the x-coordinates where the two equations cross

                            y = -1    and   y = 3 - x²

  -1 = 3 - x²

 -4 =     -x²

  4 =       x²

± 2 =       x       → These are the upper and lower limits of your integral

Then subtract the two equations and integrate with upper bound of x = 2 and lower bound of x = -2

<u />

\int_{-2}^{+2}[(3-x^2)-(-1)]dx\\\\\\=\int_{-2}^2(4-x^2)dx\\\\\\=4x-\dfrac{x^3}{3}\bigg|_{-2}^{+2}\\\\\\=\bigg(8-\dfrac{8}{3}\bigg)-\bigg(-8+\dfrac{8}{3}\bigg)\\\\\\=\large\boxed{\dfrac{32}{3}}

(b) We know the upper and lower bounds of the y-axis as y = 3 and y = -1

Next, find the equation that we need to integrate by solving for x.

      y = 3 - x²

x² + y = 3

x²       = 3 - y

x         =\pm\sqrt{3-y}\\

\rightarrow \qquad x=\sqrt{3-y}\quad and \quad x=-\sqrt{3-y}

Now, subtract the two equations and integrate with upper bound of y = 3 and lower bound of y = -1

\int_{-1}^{+3}[(\sqrt{3-y})-(-\sqrt{3-y})]dy\\\\\\=\int_{-1}^{+3}(2\sqrt{3-y})dy\\\\\\=\dfrac{-4\sqrt{(3-y)^3}}{3}\bigg|_{-1}^{+3}\\\\\\=\bigg(0\bigg)-\bigg(-\dfrac{32}{3}\bigg)\\\\\\=\large\boxed{\dfrac{32}{3}}

8 0
4 years ago
Are the following ratios equivalent 4:8 and 36:72
astraxan [27]

4 : 8

x 9 => 36 : 72  <=== equivalent

8 0
4 years ago
Read 2 more answers
A ball made out of a special material is inflated such that its diameter changes from 14 inches to 18 inches. What is the approx
Sedbober [7]
The volume of the ball is:
 V = (4/3) * (pi) * (r ^ 3)
 Where r is the radius.
 We have then:
 diameter 14 inches:
 V1 = (4/3) * (pi) * ((14/2) ^ 3)
 V1 = 1436.75504 in ^ 3
 diameter 18 inches:
 V2 = (4/3) * (pi) * ((18/2) ^ 3)
 V2 = 3053.628059 in ^ 3
 The volume change is:
 V = V2-V1
 V = 3053.628059 - 1436.75504
 V = 1616.873019 in ^ 3
 Answer:
 The approximate change in the volume of the ball is:
 V = 1616.9 in ^ 3
7 0
4 years ago
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