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3 years ago

9

A circle with radius 3 with a central angle of 17/9pi radians. What is the area of the sector?

2 answers:

inn [45]3 years ago

8 0

It can maybe be um 3 times 17 55 um maybe 88

erik [133]3 years ago

4 0

Answer:

17pi/2

Step-by-step explanation:

You might be interested in

Use linear approximation to approximate √25.3 as follows.

Sophie [7]

The idea is to use the tangent line to $f(x)=\sqrt x$ at $x=25$ in order to approximate $f(25.3)=\sqrt{25.3}$ .

We have

$f(x)=\sqrt x\implies f(25)=\sqrt{25}=5$

$f'(x)=\dfrac1{2\sqrt x}\implies f'(25)=\dfrac1{10}$

so the linear approximation to $f(x)$ is

$L(x)=f(5)+f'(5)(x-5)=5+\dfrac{x-5}{10}=\dfrac x{10}+\dfrac92$

Hence $m=\frac1{10}$ and $b=\frac92$ .

Then

$f(25.3)\approx L(25.3)=\dfrac{25.3}{10}+\dfrac92=\boxed{7.03}$

4 0

3 years ago

Bas_tet [7]

Answer:

A

Step-by-step explanation:

using formula

area of triangle=1/2 × AB×AC×sinA

WHERE AB=10 andAC=15 and sinA=37○

7 0

4 years ago

Read 2 more answers

A cylinder has a base diameter of 16 feet and a height of 19 feet. What is its volume in cubic feet, to the nearest tenths place

qaws [65]

Answer:

3820.18

Step-by-step explanation:

5 0

3 years ago

What's additive inverse? Like for example this expression 2-16 what's the additive inverse??

Stels [109]

Answer: The multiplicative inverse is 1/2 the addition inverse is -2

Step-by-step explanation:

Since the inverse of multiplication is division, use a fraction. The inverse of addition is subtraction.

4 0

3 years ago

Read 2 more answers

Please help me on this problem

Anna71 [15]

Answer:

The pairs of integer having two real solution for $ax^{2} -6x+c = 0$ are

$a = -4, c = 5$
$a = 1, c = 6$
$a = 2, c = 3$
$a = 3, c = 3$

Step-by-step explanation:

Given

$ax^{2} -6x+c = 0$

Now we will solve the equation by putting all the 6 pairs so we get the following

$-3x^{2} -6x-5 = 0$ for $a = -3 , c=-5$

$-4x^{2} -6x+5 = 0$ for $a = -4 , c=5$

$1x^{2} -6x+6 = 0$ for $a = 1 , c=6$

$2x^{2} -6x+3 = 0$ for $a = 2 , c=3$

$3x^{2} -6x+3 = 0$ for $a = 3 , c=3$

$5x^{2} -6x+4 = 0$ for $a = 5 , c=4$

The above all are Quadratic equations inn general form $ax^{2} +bx+c=0$

where we have a,b and c constant values

So for a real Solution we must have

$Disciminant , b^{2} -4\timesa\timesc \geq 0$

for $a = -3 , c=-5$ we have

$Discriminant =-24$ which is less than 0 ∴ not a real solution.

for $a = -4 , c=5$ we have

$Discriminant = 116$ which is greater than 0 ∴ a real solution.

for $a = 1 , c=6$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 2 , c=3$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 3 , c=3$ we have

$Discriminant =0$ which is equal to 0 ∴ a real solution.

for $a = 5 , c=4$ we have

$Discriminant =-44$ which is less than 0 ∴ not a real solution.

7 0

3 years ago