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deff fn [24]
3 years ago
9

In 1999 Daniel was 146 cm tall. He grew to be 176 cm by the year 2006. What was Daniel's rate of growth over this period of his

life? (Find growth each year)
Mathematics
1 answer:
andreyandreev [35.5K]3 years ago
3 0

Answer:

he grows by 5 cm every year between 1999 and 2006

Step-by-step explanation:

This is a arithmetic progression problem with the formula;

T_n = a + (n - 1)d

We are told that In 1999 Daniel was 146 cm tall. He grew to be 176 cm by the year 2006.

Thus;

a = 146

d = 2006 - 1999 = 7

Thus;

176 = 146 + (7 - 1)d

176 - 146 = 6d

30 = 6d

d = 30/6

d = 5 cm

Thus, he grows by 5 cm every year between 1999 and 2006

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QF Q6.) Find the function for a.
irakobra [83]
The function for a is (f o g) (x) = 8x^2-10.



Explanation

=========================
f(x)=2x-4, g(x)=4x^2-3, (f o g)=f(g(x))  

I already knew that the g of x is equal to 4x^2-3 so I substitute inside of the g of x

g(x)=4x^2-3 =======> f(x)=4x^2-3 =============> 2(4x^2-3)-4 

simplify it 
 
2(4x^2-3)-4  =======> 8x^2-6-4 =========> 8x^2-10

(f o g) = 8x^2-10


5 0
3 years ago
Read 2 more answers
Find the locus of a point such that the sum of its distance from the point ( 0 , 2 ) and ( 0 , -2 ) is 6.
jok3333 [9.3K]

Answer:

\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1

Step-by-step explanation:

We want to find the locus of a point such that the sum of the distance from any point P on the locus to (0, 2) and (0, -2) is 6.

First, we will need the distance formula, given by:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Let the point on the locus be P(x, y).

So, the distance from P to (0, 2) will be:

\begin{aligned} d_1&=\sqrt{(x-0)^2+(y-2)^2}\\\\ &=\sqrt{x^2+(y-2)^2}\end{aligned}

And, the distance from P to (0, -2) will be:

\displaystyle \begin{aligned} d_2&=\sqrt{(x-0)^2+(y-(-2))^2}\\\\ &=\sqrt{x^2+(y+2)^2}\end{aligned}

So sum of the two distances must be 6. Therefore:

d_1+d_2=6

Now, by substitution:

(\sqrt{x^2+(y-2)^2})+(\sqrt{x^2+(y+2)^2})=6

Simplify. We can subtract the second term from the left:

\sqrt{x^2+(y-2)^2}=6-\sqrt{x^2+(y+2)^2}

Square both sides:

(x^2+(y-2)^2)=36-12\sqrt{x^2+(y+2)^2}+(x^2+(y+2)^2)

We can cancel the x² terms and continue squaring:

y^2-4y+4=36-12\sqrt{x^2+(y+2)^2}+y^2+4y+4

We can cancel the y² and 4 from both sides. We can also subtract 4y from both sides. This leaves us with:

-8y=36-12\sqrt{x^2+(y+2)^2}

We can divide both sides by -4:

2y=-9+3\sqrt{x^2+(y+2)^2}

Adding 9 to both sides yields:

2y+9=3\sqrt{x^2+(y+2)^2}

And, we will square both sides one final time.

4y^2+36y+81=9(x^2+(y^2+4y+4))

Distribute:

4y^2+36y+81=9x^2+9y^2+36y+36

The 36y will cancel. So:

4y^2+81=9x^2+9y^2+36

Subtracting 4y² and 36 from both sides yields:

9x^2+5y^2=45

And dividing both sides by 45 produces:

\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1

Therefore, the equation for the locus of a point such that the sum of its distance to (0, 2) and (0, -2) is 6 is given by a vertical ellipse with a major axis length of 3 and a minor axis length of √5, centered on the origin.

5 0
3 years ago
Read 2 more answers
What ides it mean by simplify?
MrRissso [65]
Putting it in simpler terms
7 0
2 years ago
Which point is not on the graph of y =2x + 3?
Nostrana [21]

Answer:

i'm pretty sure its D

Step-by-step explanation:

if you make a graph yourself you'll see all the rel;relationship match except for d

3 0
3 years ago
Factor the quadratic by checking factor pairs.<br> -15x2 – 21x + 18<br> -15x2 – 218 + 19 = -3
d1i1m1o1n [39]

Answer:

-3(x + 2)(5x - 3)

Step-by-step explanation:

-15x² – 21x + 18

-3(5x² + 7x - 6)

-3[5x² + 10x - 3x - 6]

-3[5x(x + 2) - 3(x + 2)]

-3(x + 2)(5x - 3)

3 0
3 years ago
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