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never [62]
3 years ago
13

Hellpppppppo asappppppppppopp plzzzzz

Mathematics
2 answers:
astra-53 [7]3 years ago
6 0

Answer:

Step-by-step explanation:

radius = d/2=22/2=11

area=pi*r*r=22/7*11*11

=380.28

Anna35 [415]3 years ago
3 0

Answer:

380.13271

Step-by-step explanation:

Area of a circle is πr2

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How do I do this Q11
SashulF [63]
Look at one of the vertices of the heptagon where two squares meet. The angles within the squares are both of measure 90 degrees, so together they make up 180 degrees.

All the angles at one vertex must clearly add up to 360 degrees. If the angles from the squares contribute a total of 180 degrees, then the two remaining angles (the interior angle of the heptagon and the marked angle) must also be supplementary and add to 180 degrees. This means we can treat the marked angles as exterior angles to the corresponding interior angle.

Finally, we know that for any convex polygon, the exterior angles (the angles that supplement the interior angles of the polygon) all add to 360 degrees (recall the exterior angle sum theorem). This means all the marked angles sum to 360 degrees as well, so the answer is B.
5 0
3 years ago
You are buying a party pack of chips for $14.50 with a 3% tax. How much do you pay in total?
myrzilka [38]

Answer:

Step-by-step explanation:

The answer is .50

6 0
2 years ago
Read 2 more answers
Image explains everything worth alot of points! Thanks​
asambeis [7]

Answer:

11

Step-by-step explanation:

55/5 = 11

7 0
2 years ago
Determine the standard form of the equation for the circle with center (h,k) = (-5 ½) and radius r =1
wariber [46]
Asuuing yo meant (h,k)=(-5,1/2)

equation is (x-h)^2+(y-k)^2=r^2
(h,k) is center and r=radious
given that (h,k)=(-5,1/2) and r=1
h=-5
k=1/2

(x-(-5))^2+(y- \frac{1}{2} )^2=1^2
(x+5)^2+(y- \frac{1}{2} )^2=1
if you wanted us to expand
x²+y^2+10x-y+25.25=1
8 0
3 years ago
Please help with any of this Im stuck and having trouble with pre calc is it basic triogmetric identities using quotient and rec
german

How I was taught all of these problems is in terms of r, x, and y. Where sin = y/r, cos = x/r, tan = y/x, csc = r/y, sec = r/x, cot = x/y. That is how I will designate all of the specific pieces in each problem.

#3

Let's start with sin here. \frac{2\sqrt{5}}{5} = \frac{2}{\sqrt{5}} Therefore, because sin is y/r, r = \sqrt{5} and y = +2. Moving over to cot, which is x/y, x = -1, and y = 2. We know y has to be positive because it is positive in our given value of sin. Now, to find cos, we have to do x/r.

cos = \frac{-1}{\sqrt{5}} = \frac{-\sqrt{5}}{5}

#4

Let's start with secant here. Secant is r/x, where r (the length value/hypotenuse) cannot be negative. So, r = 9 and x = -7. Moving over to tan, x must still equal -7, and y = 4\sqrt{2}. Now, to find csc, we have to do r/y.

csc = \frac{9}{4\sqrt{2}} = \frac{9\sqrt{2}}{8}

The pythagorean identities are

sin^2 + cos^2 = 1,

1 + cot^2 = csc^2,

tan^2 + 1 = sec^2.

#5

Let's take a look at the information given here. We know that cos = -3/4, and sin (the y value), must be greater than 0. To find sin, we can use the first pythagorean identity.

sin^2 + (-3/4)^2 = 1

sin^2 + 9/16 = 1

sin^2 = 7/16

sin = \sqrt{7/16} = \frac{\sqrt{7}}{4}

Now to find tan using a pythagorean identity, we'll first need to find sec. sec is the inverse/reciprocal of cos, so therefore sec = -4/3. Now, we can use the third trigonometric identity to find tan, just as we did for sin. And, since we know that our y value is positive, and our x value is negative, tan will be negative.

tan^2 + 1 = (-4/3)^2

tan^2 + 1 = 16/9

tan^2 = 7/9

tan = -\sqrt{7/9} = \frac{-\sqrt{7}}{3}

#6

Let's take a look at the information given here. If we know that csc is negative, then our y value must also be negative (r will never be negative). So, if cot must be positive, then our x value must also be negative (a negative divided by a negative makes a positive). Let's use the second pythagorean identity to solve for cot.

1 + cot^2 = (\frac{-\sqrt{6}}{2})^{2}

1 + cot^2 = 6/4

cot^2 = 2/4

cot = \frac{\sqrt{2}}{2}

tan = \sqrt{2}

Next, we can use the third trigonometric identity to solve for sec. Remember that we can get tan from cot, and cos from sec. And, from what we determined in the beginning, sec/cos will be negative.

(\frac{2}{\sqrt{2}})^2 + 1 = sec^2

4/2 + 1 = sec^2

2 + 1 = sec^2

sec^2 = 3

sec = -\sqrt{3}

cos = \frac{-\sqrt{3}}{3}

Hope this helps!! :)

3 0
3 years ago
Read 2 more answers
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