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3 years ago

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Mathematics

1 answer:

icang [17]3 years ago

7 0

Answer:

1. D. $n + 4 = 30$

2. A. $5h \ge 15$

3. D. $p + 5 \le 40$

Step-by-step explanation:

1. Initial number of students = n

Number of new students that joined the class = 4

New total number of students after 4 students joined the class (n + 4) = 30

The equation that can be used to find the number of students in the class before new students joined, n, is represented by the equation below:

$n + 4 = 30$

2. The goal of Lucy is to walk at least 15 miles. This means, she must walk a distance that is equal to 15 or greater than 15 (goal: ≥ 15).

Rate of walking = 5 miles per hour

Number of hours = h

To determine how many she would walk in order to attain her goal, the inequality below represents the situation:

$5h \ge 15$

3. Mr Sanchez has $40 as the maximum he can spend. This means the total of all his spending must be equal to or less than $40 (total spending: ≤ 40).

Cost of ticket to the museum = $5

Cost of spending on other things at the museum = p dollars

Total spending = $p + 5$

Since total spending limit is $40, the following inequality can be used to determine the possible values for p:

$p + 5 \le 40$

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What is the product of -15 and -3

gayaneshka [121]

The product would be -18

6 0

4 years ago

Read 2 more answers

What will be the value of

madreJ [45]

The expression as given doesn't make much sense. I think you're trying to describe an infinitely nested radical. We can express this recursively by

$\begin{cases}a_1=\sqrt{42}\\a_n=\sqrt{42+a_{n-1}}\end{cases}$

Then you want to know the value of

$\displaystyle\lim_{n\to\infty}a_n$

if it exists.

To show the limit exists and that $a_n$ converges to some limit, we can try showing that the sequence is bounded and monotonic.

Boundedness: It's true that $a_1=\sqrt{42}\le\sqrt{49}=7$ . Suppose $a_k\le 7$ . Then $a_{k+1}=\sqrt{42+a_k}\le\sqrt{42+7}=7$ . So by induction, $a_n$ is bounded above by 7 for all $n$ .

Monontonicity: We have $a_1=\sqrt{42}$ and $a_2=\sqrt{42+\sqrt{42}}$ . It should be quite clear that $a_2>a_1$ . Suppose $a_k>a_{k-1}$ . Then $a_{k+1}=\sqrt{42+a_k}>\sqrt{42+a_{k-1}}=a_k$ . So by induction, $a_n$ is monotonically increasing.

Then because $a_n$ is bounded above and strictly increasing, the limit exists. Call it $L$ . Now,

$\displaystyle\lim_{n\to\infty}a_n=\lim_{n\to\infty}a_{n-1}=L$

$\displaystyle\lim_{n\to\infty}a_n=\lim_{n\to\infty}\sqrt{42+a_{n-1}}=\sqrt{42+\lim_{n\to\infty}a_{n-1}}$

$\implies L=\sqrt{42+L}$

Solve for $L$ :

$L^2=42+L\implies L^2-L-42=(L-7)(L+6)=0\implies L=7$

We omit $L=-6$ because our analysis above showed that $L$ must be positive.

So the value of the infinitely nested radical is 7.

4 0

3 years ago

HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

VashaNatasha [74]

Short answer
y >-x + 4
y <= 2(x - 1)^2 - 18

Comment
Let's establish what we are talking about. The question says purple and to me purple means the shaded area above the blue and to the upper right of the parabola.

Not so. It is the wine colored area right above the parabola according to the direction of he inequalities.

Second thing: the question is trying to see if you can read all he information you need from a graph.

Step One
Find the equation of the line.
We can read 2 points right away the x and y intercepts.
(4,0) is the x intercept
(0,4) is the y intercept

The general equation for the slope intercept equation is
y = mx + b So when x = 0 then b = y
4 = m*0 + b
b = 4

This far we have
y = mx + 4
when x = 4, y = 0
0 = m*4 + 4 Subtract 4 from both sides.
-4 = m*4 Divide by 4
-4/4 = m
m = - 1

Complete equation is y = -1x + 4
The inequality is y > -x + 4 Top answer block.

Step Two
Find the equation for the parabola

You could just read the minimum point which is (1,-18) There is a catch. (Isn't there always?) This may have a constant in front of the (x - a)^2 term. We'll check that out in the next step.

y = (x - 1)^2 - 18 is the basic vertex solution

Step Three
Find the value of a in
y = a(x - 1)^2 - 18
You have to do this because the answer has a space before the brackets. We need another point. The y intercept might work.

y = a(0 -1)^2 - 18
Use the point (0,-16)
-16 = a(1) - 18
-16 = a - 18 Add 18 to both sides.
2 = a

So the equation of the quadratic is
y <= 2(x - 1)^2 - 18 <<<<<< answer second section down on your question sheet.

5 0

3 years ago

The value of V-16 is not-4 because

disa [49]

Answer:

because V isn't 12, or in symbols, V-16 ≠ -4 because V ≠ 12

Step-by-step explanation:

12- 16 = -4 and that's what we want to avoid.

4 0

3 years ago

Can you help me with 4 through 6 plz

Crazy boy [7]

Example 4:

Let y = ax + b

Where, a = 4

Then,

y = 4x + b

As we have one point = (-2,3)

Replace in the equation:

3 = 4(-2) + b

3 = -8 + b

b = 3 +8

b = 11

So us stay:

y = 4x + 11
________________

Now let's to the 5 example:

Let y = ax + b

Where a = 3/2

Then,

y = 3x/2 + b

As the point is = (4, 7)

Then we will stay:

7 = 3(4)/2 + b

7 = 6 + b

b = 7 - 6

b = 1

Then we will stay:

y = 3x/2 + 1
______________

Now let's to the last example:

Let y = ax + b

Where , a = -4/3

Then we going to stay with:

y = -4x/3 + b

As the point is = (6 , -2)

Then,

-2 = -4(6)/3 + b

-2 = -8 + b

b = -2 + 8

b = 6

So follow:

y = -4x/3 + 6

I hope this helped!

6 0

3 years ago