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VladimirAG [237]
2 years ago
12

R/2=(r-4)/7 how do i solve this?

Mathematics
2 answers:
rodikova [14]2 years ago
5 0
You put a value in for r, and keep trying until they equal out, let me know if this helps, or if you need more
Alik [6]2 years ago
5 0
R= -8/5
Thx hope it helps
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Which of the following statements is true of chords?
nydimaria [60]
All of those are true, except the one about the radius.  Because alternate definition of diameter uses the idea of chord. So, both ends of a chord have to be on the circle, but one end of a radius is at the center, so a radius can't be a chord.
6 0
3 years ago
S= zh - 2zt^3 solve for z
zysi [14]

Hi ;-)

s=zh-2zt^3\\\\z(h-2t^3)=s \ \ /:(h-2t^3)\\\\z=\dfrac{s}{h-2t^3}

6 0
3 years ago
Read 2 more answers
Find the area of the region enclosed by f(x) and the x axis for the given function over the specified u reevaluate f(x)=x^2+3x+4
lord [1]

Answer:

A = 68 unit^2

Step-by-step explanation:

Given:-

The piece-wise function f(x) is defined over an interval as follows:

                       f(x) =  { x^2+3x+4    , x < 3

                       f(x) =  { x^2+3x+4     , x≥3

                        Domain : [ -3 , 4 ]

Find:-

Find the area of the region enclosed by f(x) and the x axis

Solution:-

- The best way to tackle problems relating to piece-wise functions is to solve for each part individually and then combine the results.

- The first portion of function is valid over the interval [ -3 , 3 ]:      

                       f(x) = x^2+3x+4

- The area "A1" bounded by f(x) is given as:

                      A1 = \int\limits^a_b {f(x)} \, dx

Where,  The interval of the function { -3 , 3 ] = [ a , b ]:

                     A1 = \int\limits^a_b {x^2+3x+4} \, dx\\\\A1 = \frac{x^3}{3} + \frac{3x^2}{2} + 4x |\limits_-_3^3 \\\\A1 = \frac{3^3}{3} + \frac{3*3^2}{2} + 4*3 - \frac{-3^3}{3} - \frac{3(-3)^2}{2} - 4(-3)\\\\A1 = 9 + 13.5 +12 + 9-13.5+12\\\\A1 =42 unit^2

- Similarly for the other portion of piece-wise function covering the interval [3 , 4] :

                     f(x) = 4x+10

- The area "A2" bounded by f(x) is given as:

                      A2 = \int\limits^a_b {f(x)} \, dx

Where,  The interval of the function { 3 , 4 ] = [ a , b ]:

                     A2 = \int\limits^a_b {4x+10} \, dx\\\\A2 = 2x^2 + 10x |\limits_3^4 \\\\A2 = 2*(4)^2 + 10*4 - 2*(3)^2 - 10*3\\\\A2 = 32 + 40 - 18-30\\\\A2 =26 unit^2

- The total area "A" bounded by the piece-wise function over the entire domain [ -3 , 4 ] is given:

                     A = A1 + A2

                     A = 42 + 26

                     A = 68 unit^2

8 0
3 years ago
Please helpppppp!!!!
spayn [35]

Answer 22

step by step explanation

4421.12 = \pi(8 {}^{2})h

\frac{4421.12}{64}  = \pi \frac{64}{64} h

I divided 64 since that's the opposite of multiplying to get h by itself

69.08 = \pi \: h

this could be your answer if you want it as terms of pi but if you say pi is 3.14 then you keep going and divide 69.08 by 3.14 to get h by itself

\frac{69.08}{3.14}  =  \frac{3.14}{3.14} h

then you get the answer

22 = h

hopefully this helps

8 0
3 years ago
Consider the statement: 4 − 3 + 5 = −6 + 8 + 4. This is a mathematically correct
Alla [95]

4 - 3 + 5 =  -6 + 8 + 4 (cancel equal terms)

-3 + 5 = -6 + 8 (calculate)

2 = 2 (check the equality if its equal its true)

Solution is true

3 0
3 years ago
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