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emmainna [20.7K]
3 years ago
10

I need help I don't understand this very well

Mathematics
1 answer:
kupik [55]3 years ago
4 0
The triangles are similar because they are the same shape. :) Good luck

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Rita purchased two loaves of wheat bread. Each loaf was $1.89. How much
Amanda [17]

Answer: 3.78

Step-by-step explanation:

Because she bought 2 $1.89 loaves, she spent 2 * 1.89, or 3.78 dollars.

Hope it helps <3

8 0
3 years ago
Read 2 more answers
Which of the following functions is graphed below?
svetoff [14.1K]
<h3>Answer:</h3>

C (see attached)

<h3>Step-by-step explanation:</h3>

The linear portion of the graph is defined for x ≥ 1, with the point x=1 included. Only selections A and C do that.

The quadratic portion of the graph is defined for x < 1. Only selection C does that. (Selection A is doubly-defined for x > 1, so is not a function. It is undefined for x < 1.)

5 0
3 years ago
QUESTION ATTACHED PLZ HELP NEEDED
saw5 [17]
Y = a (x + 3)² - 1

Let's take the point (-1;3)

3 = a ( -1 + 3)² - 1
3 = 4a - 1
4a = 4

Therefore a = 4/4 = 1

(x+3)² - 1
= x² + 6x + 9 - 1
= x² + 6x + 8

The answer would be y = x² + 6x + 8.

Hope this helps !

Photon
4 0
3 years ago
HELP I NEED HELP ASAP
Tpy6a [65]

Answer:

I think it's -14

Step-by-step explanation:

x = 0 so the x won't be included but

-2 • x = -2x

-12 • x = -12x

-2x + (-)12x = -14

hope this helps :)

7 0
3 years ago
A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat
jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.  

We are given a function (corrected)

N(t) = 20(t^2-lnt^2)+ 30

N(t) = 20(t^2-2lnt)+ 30

(a)

First, we take its derivative

N'(t) = 20(2t-\frac{2}{t})

Solve N'(t)=0

20(2t-\frac{2}{t})=0

Simplifying

2t^2-2=0

Solving for t

t=1\ ,t=-1

Only t=1 belongs to the valid interval 1\leqslant t\leqslant 15

Taking the second derivative

N''(t) = 20(2+\frac{2}{t^2})

Which is always positive, so t=1 is a minimum

(b)

N(1)=20(1^2-2ln1)+ 30

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

N(15)=20(15^2-2ln15)+ 30

N(15)=4391.7 is a maximum

7 0
3 years ago
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