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alexira [117]
4 years ago
7

What are the coordinates of the image of vertex D after a reflection across the x-axis?

Mathematics
1 answer:
Elena-2011 [213]4 years ago
4 0

Answer:

(-3,5) ia the answer

Step-by-step explanation:

It is because when you reflect it across the x axis the first coordinate will be negative and the second number is positive since its in the second square of the graph, so it is more of a resonable guess.

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Awnser photo quick please #3
eduard

Answer:

#1

Step-by-step explanation:

Hope this helps! <3

4 0
3 years ago
4a. In a book store, for every 9 books sold, 13 magazines were sold. If they had a 572 sales, how many of the sales were for boo
topjm [15]

Answer:

234 books.

Step-by-step explanation:

You need to set up a proportion. Be careful to note what it looks like and the explanation that follows.

9 books            13 magazines

=======     =     ===========

x                        572 - x                Cross multiply

What you want is the number of book sales from the proportion.

However all you have is the ratio of books to magazines. So you have to set up the total sales to magazines, which is a bit awkward.  572 - x is the number of magazine sales.

9*(572 - x) = 13x                        Remove the  brackets on the left.

5148- 9x = 13x                           Add 9x to both sides.

5148 = 13x + 9x                         Transpose and add the right.

22x = 5148                                Divide by 22

x = 5148/22

x = 234                    

234 books were sold.

5 0
3 years ago
A restaurant purchased kitchen equipment on January​ 1, 2017. On January​ 1, 2019, the value of the equipment was ​$14 comma 550
timama [110]

Answer:

\frac{dV(t)}{dt} = - 1675.38

Step-by-step explanation:

In 2017, the vakue of the kitchen equipment was $14550

V(0)=$14550

Its value after then was modelled by V(t)=14550e^{-0.158t

We are required to find the rate of change in value on January 1, 2019

V(t)=14550e^{-0.158t

\frac{dV(t)}{dt} =\frac{d}{dt}14550e^{-0.158t

\frac{dV(t)}{dt} =14550 \frac{d}{dt}e^{-0.158t

\\Let u= -0.158t,\frac{du}{dt}=-0.158

\frac{dV(t)}{dt} =14550 \frac{d}{du}e^u\frac{du}{dt}

\frac{dV(t)}{dt} =14550 X -0.158 e^{-0.158t}=-2298.9e^{-0.158t}

In 2019, i.e. 2 years after, t=2

The rate of change of the value

\frac{dV(t)}{dt} =-2298.9e^{-0.158X2}

=\frac{dV(t)}{dt} =-2298.9e^{-0.316}= - 1675.38

3 0
3 years ago
what is the smallest positive integer a such that the intermediate value theorem guarantees a zero exists between 0 and a?
liberstina [14]

The smallest positive integer that the intermediate value theorem guarantees a zero exists between 0 and a is 3.

What is the intermediate value theorem?

Intermediate value theorem is theorem about all possible y-value in between two known y-value.

x-intercepts

-x^2 + x + 2 = 0

x^2 - x - 2 = 0

(x + 1)(x - 2) = 0

x = -1, x = 2

y intercepts

f(0) = -x^2 + x + 2

f(0) = -0^2 + 0 + 2

f(0) = 2

(Graph attached)

From the graph we know the smallest positive integer value that the intermediate value theorem guarantees a zero exists between 0 and a is 3

For proof, the zero exists when x = 2 and f(3) = -4 < 0 and f(0) = 2 > 0.

<em>Your question is not complete, but most probably your full questions was</em>

<em>Given the polynomial f(x)=− x 2 +x+2 , what is the smallest positive integer a such that the Intermediate Value Theorem guarantees a zero exists between 0 and a ?</em>

Thus, the smallest positive integer that the intermediate value theorem guarantees a zero exists between 0 and a is 3.

Learn more about intermediate value theorem here:

brainly.com/question/28048895

#SPJ4

4 0
2 years ago
I need help with 3,4,5<br> Please!!
gladu [14]
3. g(x) = x^2 - 5x + 3
    g(-3) = -3^2 - 5(-3) + 3
    g(-3) = 9 + 15 + 3
    g(-3) = 27 <==

4. 4(7b + 15f - 8f) =
    4(7b + 7f) =
     28b + 28f <===

5. C = 2.50r + 15
    for 5 rolls...r = 5
    C = 2.50(5) + 15
    C = 12.50 + 15
    C = 27.50...its A
3 0
3 years ago
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