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3 years ago

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There are 20 girls on the basketball team. of these, 17 are over 16 years old, 12 are taller than 170 cm, and 9 are both older t

han 16 and taller than 170 cm. how many of the girls are older than 16 or taller than 170 cm?

1 answer:

Oduvanchick [21]3 years ago

5 0

17 are older than 16 years old and 12 are taller than 170 cm

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Find the standard form of the equation for the conic section represented by x^2 + 10x + 6y = 47.

Levart [38]

Answer:

The standard form of the equation for the conic section represented by $x^2\:+\:10x\:+\:6y\:=\:47$ is:

$4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2$

Step-by-step explanation:

We know that:

$4p\left(y-k\right)=\left(x-h\right)^2$ is the standard equation for an up-down facing Parabola with vertex at (h, k), and focal length |p|.

Given the equation

$x^2\:+\:10x\:+\:6y\:=\:47$

Rewriting the equation in the standard form

$4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2$

Thus,

The vertex (h, k) = (-5, 12)

Please also check the attached graph.

Therefore, the standard form of the equation for the conic section represented by $x^2\:+\:10x\:+\:6y\:=\:47$ is:

$4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2$

where

vertex (h, k) = (-5, 12)

7 0

2 years ago

13. For the arithmetic sequence: 6, 12, 18, 24,...

serg [7]

a_15 = 90

a_n = 6n

Step-by-step explanation:

Given sequence is:

6, 12, 18, 24,...

First of all we have to find the common difference. The common difference is the difference between consecutive terms of an arithmetic sequence.

Here,

$d=a_2-a_1 = 12-6 = 6\\d=a_3-a_2= 18-12 = 6$

The general form for arithmetic sequence is:

$a_n=a_1+(n-1)d$

Putting the values for a_1 and d

$a_n=6+(n-1)(6)\\a_n=6+6n-6\\a_n=6n$

<u>a) Find a15</u>

$a_{15} = 6(15)\\=90$

<u>b) Write an equation for the nth term.</u>

The equation is: a_n=6n

Keywords: Arithmetic sequence, Common Difference

Learn more about arithmetic sequence at:

#LearnwithBrainly

4 0

3 years ago

Determine if the given mapping phi is a homomorphism on the given groups. If so, identify its kernel and whether or not the mapp

shtirl [24]

Answer:

(a) No. (b)Yes. (c)Yes. (d)Yes.

Step-by-step explanation:

(a) If $\phi: G \longrightarrow G$ is an homomorphism, then it must hold

that $b^{-1}a^{-1}=(ab)^{-1}=\phi(ab)=\phi(a)\phi(b)=a^{-1}b^{-1}$ ,

but the last statement is true if and only if G is abelian.

(b) Since G is abelian, it holds that

$\phi(a)\phi(b)=a^nb^n=(ab)^{n}=\phi(ab)$

which tells us that $\phi$ is a homorphism. The kernel of $\phi$

is the set of elements g in G such that $g^{n}=1$ . However,

$\phi$ is not necessarily 1-1 or onto, if $G=\mathbb{Z}_6$ and

n=3, we have

$kern(\phi)=\{0,2,4\} \quad \text{and} \quad\\\\Im(\phi)=\{0,3\}$

(c) If $z_1,z_2 \in \mathbb{C}^{\times}$ remeber that

$|z_1 \cdot z_2|=|z_1|\cdot|z_2|$ , which tells us that $\phi$ is a

homomorphism. In this case

$kern(\phi)=\{\quad z\in\mathbb{C} \quad | \quad |z|=1 \}$ , if we write a

complex number as $z=x+iy$ , then $|z|=x^2+y^2$ , which tells

us that $kern(\phi)$ is the unit circle. Moreover, since

$kern(\phi) \neq \{1\}$ the mapping is not 1-1, also if we take a negative

real number, it is not in the image of $\phi$ , which tells us that

$\phi$ is not surjective.

(d) Remember that $e^{ix}=\cos(x)+i\sin(x)$ , using this, it holds that

$\phi(x+y)=e^{i(x+y)}=e^{ix}e^{iy}=\phi(x)\phi(x)$

which tells us that $\phi$ is a homomorphism. By computing we see

that $kern(\phi)=\{2 \pi n| \quad n \in \mathbb{Z} \}$ and

$Im(\phi)$ is the unit circle, hence $\phi$ is neither injective nor

surjective.

7 0

3 years ago

The circumference of a circle is 50.24 centimeters. Find the radius of the circle. (Use 3.14 for π.)

Deffense [45]

2piR=circumference

2(3.14)R=50.24

R=8 cm

4 0

3 years ago

Read 2 more answers

How many real solutions does a quadratic equation have if its discriminant is negative?

kvasek [131]

Answer:

A

Step-by-step explanation:

Discriminant is given by

$D = b^{2} -4ac$

If it is negative then in quadratic formula the square root part becomes negative which makes the solution complex.

so there cannot be any real solution if the Discriminant is negative or less than 0 ,

The correct option for the given question is

A

3 0

3 years ago