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1 year ago

Rewrite in polar form: x=12. I have r=12 sec theta

Mathematics

2 answers:

Wittaler [7]1 year ago

7 0

Answer:

C. r=12Secθ

Step-by-step explanation:

romanna [79]1 year ago

3 0

To solve this problem you must apply the proccedure shown below:
1. You have to rewrite x=12 in polar form. Then, you have:
12=rCosθ
2. Then, you must solve for r, as following:
r=12/Cosθ
 3. You have that 1/Cosθ=Secθ, therefore:
 r=12(1/Cosθ)
 r=12Secθ
 Therefore, as you can see, the answer is: r=12Secθ

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Which congruence theorem can be used to prove △BDA ≅ △DBC? Triangles B D A and D B C share side D B. Angles C B D and A D B are

Tomtit [17]

Answer:

A. Hypotenuse-leg (HL) congruence.

HL, when you have 2 right triangles and their hypotenuses are congruent you are able to say HL

Step-by-step explanation:

We know that the hypotenuse-leg theorem states that if the hypotenuse and one leg of a right triangle are congruent to hypotenuse and corresponding leg of another right triangle, then the triangles are congruent.

hypotenuse(AB) of △BDA equals to hypotenuse (CD) of △DBC.

BDA and DBC share a common side DB.

Using Pythagorean theorem we will get,

$CD^{2}=DB^{2}+BC^{2}...(1) \\\\AB^{2}=DB^{2}+AD^{2}...(2)$

We have been given that CD=AB, Upon using this information we will get,

$DB^{2}+BC^{2}=DB^{2}+AD^{2}$

Upon subtracting $DB^{2}$ from both sides of our equation we will get,

$BC^{2}=AD^{2}\\\\BC=AD$

<h3>Therefore, by HL congruence △BDA ≅ △DBC.</h3>

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1 year ago

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Arnold borrowed $7890 at 11.5 percent for five years. How much did Arnold pay in interest?

Rudiy27

937.5 is what I got.

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2 years ago

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Forty students are in the science club. Of those, 45% are girls. This percent increases to 56% after new girls join the club. Ho

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1 year ago

Find the GCF of 44j5k4 and 121j2k6.

BartSMP [9]

Answer:

D. $11j^2k^4$

Step-by-step explanation:

We are asked to find the GCF of $44j^5k^4\text{ and }121j^2k^6$ .

Since we know that GCF of two numbers is the greatest number that is a factor of both of them.

First of all we will GCF of 44 and 121.

Factors of 44 are: 1, 2, 4, 11, 22, 44.

Factors of 121 are: 1, 11, 11, 121.

We can see that greatest common factor of 44 and 121 is 11.

Now let us find GCF of $j^5\text{ and }j^2$ .

Factors of $j^5$ are: $j*j*j*j*j$

Factors of $j^2$ are: $j*j$

We can see that greatest common factor of $j^5\text{ and }j^2$ is $j*j=j^2$ .

Now let us find GCF of $k^4\text{ and }k^6$ .

Factors of $k^4$ are: $k*k*k*k$

Factors of $k^6$ are: $k*k*k*k*k*k$

We can see that greatest common factor of $k^4\text{ and }k^6$ is $k*k*k*k=k^4$ .

Upon combining our all GCFs we will get,

$11j^2k^4$

Therefore, GCF of $44j^5k^4\text{ and }121j^2k^6$ is $11j^2k^4$ and option D is the correct choice.

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1 year ago

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tankabanditka [31]

Answer:

If we assume that the bottle is cylindrical and we take the same radius (3.26) of both the bottles (bottles only differ in heights) then the larger bottle will hold approximately 701.14 ml of fluid (the answer says 700ml which is very close)

Step-by-step explanation:

Step 1: Formula of volume of a cylinder is pi*r^2*h

where value of pi is 3.14

r is the radius

h is the height of the bottle (height is different for both bottles)

After putting the values and estimated radius for both as 3.26, we get the volume of the taller bottle.

You can extract the radius by following this method:

Volume of a cylinder = pi*r^2*h (now put the value of the known volume and height of the smaller cylinder)

500 = 3.14 * (r)^2 *15

500/(3.14*15) = r^2

10.616 = r^2

Taking sqrt. on both sides

We get r = 3.26

Now put the same value in the formula of volume with the radius and height. You will get the answer for second bottle.

V= 3.14 * (3.26)^2 *21

V= 701.14 (approx)

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1 year ago

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