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horsena [70]
2 years ago
6

Help me thanks please please thanks

Mathematics
1 answer:
enot [183]2 years ago
3 0

Answer:

Area of big circle: 2463.01

Area of little circle: 615.75

Area of shaded space: 1,231.51

Step-by-step explanation:

thats what i came up with

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Factor. 2x2−3x−5 (2x+5)(x−1) (x−5)(2x+1) (2x−5)(x−1) (2x−5)(x+1)
Likurg_2 [28]

Answer:

0

Step-by-step explanation

  1. You do the brackets first
  2. The answer to the brackets is : 0
  3. Because you also times the others by 0 the answer must be 0.
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A formula is given with the value of all but one of the variables in the formula. Find the value of the variable not given.
Svet_ta [14]

Answer:

P = 24

Step-by-step explanation:

P = 2L + 2W ( substitute L = 4, W = 8 into the formula )

P = 2(4) + 2(8) = 8 + 16 = 24

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A spherical balloon is inflated with a gas at the rate of 500 cm3/min. How fast is the radius of the balloon increasing at the i
Burka [1]

The volume of the sphere is expressed in the formula V = 4/3 pi r^3. The rate of change of volume is determined by differentiating the formula: dV/dt = 4pi r^2 dr/dt. When we substitute 500 cm3/min as dV/dt and 30 cm as r. Then dr/dt is equal to 0.0442 cm/min
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Solve |2x|=8<br> Tell me how you got the answer pls
lilavasa [31]

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4

Step-by-step explanation:

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6 0
3 years ago
Sin0=12/37 find tan0
alexira [117]

Answer:

\large{ \tt{❃ \: EXPLANATION}} :

  • We're provide - Sin θ = \frac{12}{37} which means 12 is the perpendicular & 37 is the hypotenuse [ Since Sin θ = \tt{  \frac{p}{h}} ] . We're asked to find out tan θ ].

\large{ \tt{❁ \: USING \: PYTHAGORAS \: THEOREM}} :

\large{ \tt{❊ \:  {h}^{2}  =  {p}^{2}  +  {b}^{2} }}

\large{ \tt{⇢ {p}^{2} +  {b}^{2}   =  {h}^{2} }}

\large{ \tt{⇢ \:  {b}^{2}  =  {h}^{2}  -  {p}^{2} }}

\large{ \tt{⇢ \:  {b}^{2} = {37}^{2} -  {12}^{2}   }}

\large{ \tt{⇢ \:  {b}^{2}  = 1369 - 144}}

\large{ \tt{ ⇢{b}^{2}  = 1225}}

\large{ \tt{⇢ \: b =  \sqrt{1225}}}

\large{ \tt{⇢ \: b = 35  \: \text {units}}}

  • Now , We know - Tan θ= \tt{ \frac{perpendicular}{base} }. Just plug the values :

\large{ \tt{➝ \: Tan  \: \theta =  \frac{p}{b}  = \boxed{ \tt{  \frac{12}{35} }}}}

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

5 0
3 years ago
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