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solniwko [45]
2 years ago
7

Please help, if I get this wrong, my grade will plummet :( /srs

Mathematics
2 answers:
nata0808 [166]2 years ago
4 0
The correct answer is 12
KIM [24]2 years ago
4 0
I believe the answer is 12
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Urgent…. I need help
Firdavs [7]

A.

Step-by-step explanation:

To find out the answer, there is a method that can be used to find out if a line is a function. It is called the vertical line method. And it also has to be per x, One y value has to be there. Analyzing the graphs, the answer should be option A.

8 0
2 years ago
To print tickets, a printer chargers a $70 setup fee plus $1.25 per ticket. (a) Write an algebraic expression for the cost of t
ioda
A. 70+1.25t
b. Plug in 650 for t
70+1.25(650)
70+812.5
882.5

Final answer: $882.50
8 0
3 years ago
A fruit basket is filled with 888 bananas, 333 oranges, 555 apples, and 666 kiwis.For every 3 kiwis, there are 4_
Novosadov [1.4K]

Answer:

I think it's bananas

Step-by-step explanation:

4 0
2 years ago
Write the standard form equation of the ellipse shown in the graph, and identify the foci.
vlada-n [284]

Answer option A

From the given graph is a Vertical ellipse

Center of ellipse = (-2,-3)

Vertices are (-2,3)  and (-2,-9)

Co vertices are (-6,-3) and (2,-3)

The distance between center and vertices = 6, so a= 6

The distance between center and covertices = 4 , so b= 4

The general equation of vertical ellipse is

\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2}=1

(h,k) is the center

we know center is (-2,-3)

h= -2, k = -3 , a= 6  and b = 4

The standard equation  becomes

\frac{(x+2)^2}{4^2} + \frac{(y+3)^2}{6^2}=1

\frac{(x+2)^2}{16} + \frac{(y+3)^2}{36}=1

Foci  are (h,k+c)  and (h,k-c)

c=\sqrt{a^2-b^2}

Plug in the a=6  and b=4

c=\sqrt{6^2-4^2}

 c=\sqrt{20}

  c=2\sqrt{5}, we know h=-2  and k=-3

Foci  are   (-2,-3+2\sqrt{5})  and  (-2,-3-2\sqrt{5})

Option A is correct

6 0
3 years ago
X &lt; 36 and x &gt; -36<br>x &lt; 36 or x&gt;-36<br>x&lt; 36 or x &gt; 0<br>x &lt; 36 and x &gt; 0​
hjlf

Answer:

D,x < 36 and x > 0, negative number cannot be in sq.root

3 0
3 years ago
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