Answer:
If each input value leads to only one output value, classify the relationship as a function. If any input value leads to two or more outputs, do not classify the relationship as a function.
Step-by-step explanation:
Answer:
B
Step-by-step explanation:
The photo is unclear but based on what I saw, -61 is smaller than +61.
Answer:
a. The value of the constant k is 21
b. The equation is y = k * x, where k is the proportionality constant, "x" is the number of terraced houses and "y" is the width of a row of identical houses.
Step-by-step explanation:
a.
<em>A proportional relationship satisfies the equation y = k * x, where k is a positive constant and is called a proportionality constant. In this case "x" is the number of terraced houses and "y" is the width of a row of identical houses.
</em>
The data you have is that the width of 5 townhouses are 105 feet. This means that the value of "x" is 5 houses and the value of "y" is 105 feet. By replacing in the equation y = k * x and isolating the constant k, you get:
<em>105=k*5
</em>

<em>k=21
</em>
<u><em>So the value of the constant k is 21.</em></u>
b.
<em>As mentioned, the equation is y = k * x, where k is the proportionality constant, "x" is the number of terraced houses and "y" is the width of a row of identical houses.</em>
This means that just as "x" increases, "y" increases. And that if "x" decreases, "y" will decrease. And this relationship between "x" e "and" will always be the same, determined by the value of the constant "k".
Answer:

Step-by-step explanation:
The given trigonometric equation is
.
We can either use the Pythagorean identity or the right angle triangle to solve for
.
According to the Pythagorean identity,

Recall that, the cosine function is an even function, therefore

.
We substitute this value in to the above Pythagorean identity to get;






But we were given that,
, so we choose the negative value.

The correct answer is B
Area of ∆=1/2bh
80yd^2=1/2(b)(10yd)
80yd^2=5yd(b)
80yd^2÷5yd=b
16yd=b