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EleoNora [17]
3 years ago
13

If Tim spends $22.00 on 10 gallons of gas, how much did gas cost per gallon?

Mathematics
2 answers:
irinina [24]3 years ago
5 0
22.00 / 10 = 2.20 per gallon
kaheart [24]3 years ago
5 0

The answer is obviously $2.20

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Which function is even? Check all that apply.
iris [78.8K]

im pretty sure its

(A) y=cos x

(B) y=sec x

5 0
2 years ago
What are the steps to 50p+5=225 in an equation
sleet_krkn [62]

Answer:

p = 22/5

Step-by-step explanation:

50p+5=225

Subtract 5 from each side

50p+5-5=225-5

50p = 220

Divide by 50

50p/50 = 220/50

p = 22/5

7 0
3 years ago
Read 2 more answers
Marco has drawn a line to represent the perpendicular cross-section of the triangular prism. Is he correct? Explain. triangular
evablogger [386]

Answer:

The correct option is;

Yes, the line should be perpendicular to one of the rectangular faces

Step-by-step explanation:

The given information are;

A triangular prism lying on a rectangular base and a line drawn along the slant height

A perpendicular bisector should therefore be perpendicular with reference to the base of the triangular prism such that the cross section will be congruent to the triangular faces

Therefore Marco is correct and the correct option is yes, the line should be perpendicular to one of the rectangular faces (the face the prism is lying on).

8 0
2 years ago
Read 2 more answers
-1/225x^2+2/3x what is the vertex of this quadratic function?
Julli [10]
\bf \textit{vertex of a parabola}\\ \quad \\\\

\begin{array}{lccclll}
y=&-\frac{1}{225}x^2&+\frac{2}{3}x\\\\
y=&-\frac{1}{225}x^2&+\frac{2}{3}x&+0\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad  {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)
6 0
3 years ago
PLEASE HELP!! WILL GIVE BRAINLIEST!!!!
Arturiano [62]

Answer:

y=1

Step-by-step explanation:

Given function y=m(x)

The question is to find the y value where x equals  -2 so the easiest way is to plus x=-2 in the graph, we can get y=1.

Hopefully this helps, if you have any questions please feel free to ask them in the comment section below, I'll be more than happy to answer!

5 0
2 years ago
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