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3 years ago

Help with this problem^

Mathematics

1 answer:

Fittoniya [83]3 years ago

6 0

This formula is the exposition of 8 and equals to 6 how u ask ? well if u 8 minus 2 u get 6 than add 6

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Please help me with this! I will mark you as brainliest!

erastovalidia [21]

Step-by-step explanation:

$\frac{ {5}^{3} }{ {5}^{7} }$

$\frac{1}{ {5}^{7 - 3} }$

$\frac{1}{ {5}^{4} }$

${5}^{ - 4}$

Hope it will help :)

4 0

4 years ago

What is the equation of the quadratic graph with a focus of (3, 6) and a directrix of y = 4? f(x) = one fourth (x − 3)2 + 1 f(x)

mina [271]

Let point (x, y) be any point on the graph, than the distance between (x, y) and the focus (3, 6) is sqrt((x - 3)^2 + (y - 6)^2) and the distance between (x, y) and the directrix, y = 4 is |y - 4|

Thus sqrt((x - 3)^2 + (y - 6)^2) = |y - 4|
(x - 3)^2 + (y - 6)^2 = (y - 4)^2
x^2 - 6x + 9 + y^2 - 12y + 36 = y^2 - 8y + 16
x^2 - 6x + 29 = -8y + 12y = 4y
(x - 3)^2 + 20 = 4y
y = 1/4(x - 3)^2 + 5

Required answer is f(x) = one fourth (x - 3)^2 + 5

8 0

3 years ago

LCM of two numbers is 1134 and HCF is 18. if one of the numbers is 162, find the other. number.<br>

Brilliant_brown [7]

<h3>Answer: 126</h3>

=====================================================

Work Shown:

Let x and y be the two numbers.

We're given x = 162 and the variable y is unknown.

We're also given LCM = 1134 and HCF = 18

So,

LCM = (x*y)/HCF

1134 = 162*y/18

1134 = (162/18)y

1134 = 9y

9y = 1134

y = 1134/9

y = 126

The other number is 126

---------------------

Notice that

162 = 18*9
126 = 18*7

showing that 18 is the highest common factor (HCF) of the numbers 162 and 126. This partially confirms the answer.

Now let,

A = multiples of 162
B = multiples of 126

So,

A = 162, 324, 486, 648, 810, 972, 1134, 1296, ...
B = 126, 252, 378, 504, 630, 756, 882, 1008, 1134, 1260, ...

We see that 1134 is in each list of multiples and the smallest such common item. So the lowest common multiple (LCM) of 162 and 126 is 1134. This helps fully confirm the answer.

7 0

2 years ago

Match each interval with its corresponding average rate of change for q(x) = (x + 3)2. 1. -6 ≤ x ≤ -4 1 2. -3 ≤ x ≤ 0 -4 3. -6 ≤

MrMuchimi

The average rate of change of a function f(x) in an interval, a < x < b is given by
$\frac{f(b) - f(a)}{b - a}$

Given q(x) = (x + 3)^2

1.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -4 is given by $\frac{q(-4)-q(-6)}{-4-(-6)} = \frac{(-4+3)^2-(-6+3)^2}{-4+6} = \frac{1-9}{2} = \frac{-8}{2} =-4$

2.) The average rate of change of q(x) in the interval -3 ≤ x ≤ 0 is given by $\frac{q(0)-q(-3)}{0-(-3)} = \frac{(0+3)^2-(-3+3)^2}{0+3} = \frac{9-0}{3} = \frac{9}{3} =3$

3.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-6)}{-3-(-6)} = \frac{(-3+3)^2-(-6+3)^2}{-3+6} = \frac{0-9}{3} = \frac{-9}{3} =-3$

4.) The average rate of change of q(x) in the interval -3 ≤ x ≤ -2 is given by $\frac{q(-2)-q(-3)}{-2-(-3)} = \frac{(-2+3)^2-(-3+3)^2}{-2+3} = \frac{1-0}{1} = \frac{1}{1} =1$

5.) The average rate of change of q(x) in the interval -4 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-4)}{-3-(-4)} = \frac{(-3+3)^2-(-4+3)^2}{-3+4} = \frac{0-1}{1} = \frac{-1}{1} =-1$

6.) The average rate of change of q(x) in the interval -6 ≤ x ≤ 0 is given by $\frac{q(0)-q(-6)}{0-(-6)} = \frac{(0+3)^2-(-6+3)^2}{0+6} = \frac{9-9}{6} = \frac{0}{6} =0$

3 0

3 years ago

Read 2 more answers

Your baseball team has a goal to collect at least 160 blankets for a shelter. Team members brought 42 blankets on Monday and 65

Ad libitum [116K]

107 that’s the number added up if your asking that

3 0

3 years ago