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Sholpan [36]
3 years ago
10

Help ASAP pllllllllllsssssssssssssssssss

Mathematics
1 answer:
valkas [14]3 years ago
5 0

Answer:

15

Step-by-step explanation:

multiply the 6 by 3 and get 18, so do the same for the top, so 5 x3 = 15

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What does the zero on a graph represent
nasty-shy [4]

Answer:

the origin

Step-by-step explanation:

4 0
3 years ago
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An engineer is designing an arch-shaped gate for the entrance to an amusement park. The gate must be 80 feet wide and 25 feet ta
Tom [10]
1) the form of the equation may be written as y = A(X - Xo)(X - X1)

Where Xo and X1 are the two roots of the equation.


2) We can fix the system of coordinates so that the vertex is in the middle of the gate => Xo = - 40 and X1 = +40


=> y = A (X + 40) (X - 40) = A (X^2 - 1600) 


3) The height, at X = 0 is 25

=> A(0 - 1600) = 25


=> -1600A = 25 => A = -25 / 1600 = - 1/64


4) The equation is y =  - [1/64] (X^2 - 1600)


5) You can present it in different equivalent forms.

Some of those other forms are:

1) - 64y = (x^2 - 1600)

2) x^2 = - 64y + 1600

3) X^2 = - 64 (y - 25)
4 0
3 years ago
The coordinate notation for a translation 5 units right and 6 units down is?
wariber [46]
(x+5,y-6) is the coordinate notation.
4 0
3 years ago
I need help please. Thanks!
Karolina [17]

Answer:

A

Step-by-step explanation:

We are given the function f and its derivative, given by:

f^\prime(x)=x^2-a^2=(x-a)(x+a)

Remember that f(x) is decreasing when f'(x) < 0.

And f(x) is increasing when f'(x) > 0.

Firstly, determining our zeros for f'(x), we see that:

0=(x-a)(x+a)\Rightarrow x=a, -a

Since a is a (non-zero) positive constant, -a is negative.

We can create the following number line:

<-----(-a)-----0-----(a)----->

Next, we will test values to the left of -a by using (-a - 1). So:

f^\prime(-a-1)=(-a-1-a)(-a-1+a)=(-2a-1)(-1)=2a+1

Since a is a positive constant, (2a + 1) will be positive as well.

So, since f'(x) > 0 for x < -a, f(x) increases for all x < -a.

To test values between -a and a, we can use 0. Hence:

f^\prime(0)=(0-a)(0+a)=-a^2

This will always be negative.

So, since f'(x) < 0 for -a < x < a, f(x) decreases for all -a < x < a.

Lasting, we can test all values greater than a by using (a + 1). So:

f^\prime(a+1)=(a+1-a)(a+1+a)=(1)(2a+1)=2a+1

Again, since a > 0, (2a + 1) will always be positive.

So, since f'(x) > 0 for x > a, f(x) increases for all x > a.

The answer choices ask for the domain for which f(x) is decreasing.

f(x) is decreasing for -a < x < a since f'(x) < 0 for -a < x < a.

So, the correct answer is A.

3 0
3 years ago
If anyone can help me with answering the pictures above i’d be really grateful
Usimov [2.4K]
No it is not a perfect square
4 0
3 years ago
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