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valkas [14]
3 years ago
7

Determine the number of gallons of gas that your car holds. Find the cost of the type of gasoline that your car uses at a local

gas station. Determine the amount of money you would spend to fill up your tank if it was empty.
Mathematics
1 answer:
Alex17521 [72]3 years ago
3 0

16 gallon tank, 2.75 $/gal so roughly 53.25


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Y= x^2 -5 solve for x
dexar [7]

Y= x^2 -5

We need to solve for x, we need to get x alone

Y= x^2 -5

Lets start by removing -5

Add 5 on both sides

y + 5= x^2 -5 + 5

y + 5= x^2

Now to isolate x , we need to remove the square from x

To remove square , take square root on both sides

+-\sqrt{y+5} = \sqrt{x^2}

square and square root will get cancelled

+-\sqrt{y+5} = x

So x = +\sqrt{y+5} and  x = -\sqrt{y+5}


7 0
3 years ago
How are the points (8,7) and (-8,7) related?
melisa1 [442]

Answer:

The point (-8,7) is the reflection of the point (8,7)

8 0
3 years ago
Please help with this!!
maw [93]

Answer:

gcf=2

12-2=4(3-1)

hope this is right!

8 0
2 years ago
Pls help! It's kinda urgent :/
erastova [34]

Answer:

What are you trying to do to my brain?

5 0
3 years ago
Given:
diamong [38]

Answer:

10-5\sqrt{2}

Step-by-step explanation:

As per the attached figure, right angled \triangle MDL has an inscribed circle whose center is I.

We have joined the incenter I to the vertices of the \triangle MDL.

Sides MD and DL are equal because we are given that \angle M = \angle L = 45 ^\circ.

Formula for <em>area</em> of a \triangle = \dfrac{1}{2} \times base \times height

As per the figure attached, we are given that side <em>a = 10.</em>

Using pythagoras theorem, we can easily calculate that side ML = 10\sqrt{2}

Points P,Q and R are at 90 ^\circ on the sides ML, MD and DL respectively so IQ, IR and IP are heights of  \triangleMIL, \triangleMID and \triangleDIL.

Also,

\text {Area of } \triangle MDL = \text {Area of } \triangle MIL +\text {Area of } \triangle MID+ \text {Area of } \triangle DIL

\dfrac{1}{2} \times 10 \times 10 = \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10\sqrt2\\\Rightarrow r = \dfrac {10}{2+\sqrt2} \\\Rightarrow r = \dfrac{5\sqrt2}{\sqrt2+1}\\\text{Multiplying and divinding by }(\sqrt2 +1)\\\Rightarrow r = 10-5\sqrt2

So, radius of circle = 10-5\sqrt2

8 0
3 years ago
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