5/2
70 is the constant number that doesn’t change, but anytime you increase t you are raising it above 5/2. So every time you grow 5/2.
It can be handy to start with a version of the point-slope form of the equation for a line. For a line of slope m through point (h, k), an equation can be written as
... y = m(x -h) +k
The given line can be solved for y to get
... y = (3x -6)/2
Then the slope is the x-coefficient, 3/2.
The parallel line through (2/5, -1) will have the same slope, so its equation can be written
... y = (3/2)(x -2/5) -1
... y = (3/2)x -8/5 . . . . parallel line
The perpendicular line will have a slope that is the negative reciprocal of 3/2, that is, -2/3. Its equation can be written as
... y = (-2/3)(x -2/5) -1
... y = (-2/3)x -11/15 . . . . perpendicular line
Evaluate exponents first
64-10/5
Then evaluate division
64-2
Finally, subtract
62
Final answer: 62
Answer:
See how we can multiply or divide two functions to create a new function. Just like we can multiply and divide numbers, we can multiply and divide functions. For example, if we had functions f and g, we could create two new functions: f ⋅ g f\cdot g f⋅g and gfstart fraction, f, divided by, g, end fraction .
Step-by-step explanation:
Two solutions were found :<span> t = 8 t = 0</span>
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "t2" was replaced by "t^2".
Step by step solution :Skip Ad
<span>Step 1 :</span><span>Step 2 :</span>Pulling out like terms :
<span> 2.1 </span> Pull out like factors :
<span> t2 - 8t</span> = t • (t - 8)
<span>Equation at the end of step 2 :</span> t • (t - 8) = 0
<span>Step 3 :</span>Theory - Roots of a product :
<span> 3.1 </span> A product of several terms equals zero.<span>
</span>When a product of two or more terms equals zero, then at least one of the terms must be zero.<span>
</span>We shall now solve each term = 0 separately<span>
</span>In other words, we are going to solve as many equations as there are terms in the product<span>
</span>Any solution of term = 0 solves product = 0 as well.
Solving a Single Variable Equation :
<span> 3.2 </span> Solve : t = 0<span>
</span> Solution is t = 0
Solving a Single Variable Equation :
<span> 3.3 </span> Solve : t-8 = 0<span>
</span>Add 8 to both sides of the equation :<span>
</span> t = 8
Two solutions were found :<span> t = 8<span> t = 0</span></span>
