The variables are: x, y
The constant is: 4
The coefficients are: 5, 3
The exponent is: y^2
Answer:
7,366,327.50 pascals
Step-by-step explanation:
we know that
1 atm= 101,325 pascals
we have
72.7 am
Convert to pascals
72.7 atm=72.7*101,325=7,366,327.50 pascals
I would say B because of KAO keep add opposite.
Answer:
8
Step-by-step explanation:
we know that
the triangle ABC and triangle EDF are congruent triangles-----> given problem
therefore

we know that
the distance between two points is equal to

Step 
<u>Find the distance AB</u>

substitute in the formula of the distance



Step 
<u>Find the distance BC</u>

substitute in the formula of the distance



Step 
<u>Find the distance AC</u>
we know that
the triangle ABC is a right triangle
so
Applying the Pythagorean Theorem

substitute the values in the formula


round to the nearest tenth

therefore



<u>the answer is</u>
The length of the hypotenuse is equal to 