Need help ASAP. Last two problems on my study guide, can't seem to solve them?

Mathematics

1 answer:

NNADVOKAT [17]2 years ago

6 0

Answer:

9.)b=O or B=-1 or b=1

Step-by-step explanation:

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Explain how to write a function rule from the table below.<br><br> x :2 4 6<br> y :1 0 –1

nydimaria [60]

Our "y-intercept" (where x = 0) acts as the foundation of our function here.
We can extrapolate from what we have here...if x is 0, y is going to be 2.
So, we start at 2 for y, and every time we decrease x by 2, y increases by 1.
If we decrease x by just 1, y increases by ½.
This also goes in the opposite direction. Increase x by 1, decrease y by ½.
We can write the change in y as -½x.
That change is being added to our original value of 4, of course.
The equation of this line is $y=-\frac{1}2x+2$ .

5 0

3 years ago

Read 2 more answers

How do you solve this problem

Kryger [21]

Answer:

$sin(A) = \frac{21}{{29} }$

$cos(A)=\frac{20}{29 }$

$tan(A)=\frac{21}{20}$

$sin(C) = \frac{20}{{29} }$

$cos(C)=\frac{21}{29 }$

$tan(C)=\frac{20}{21}$

Step-by-step explanation:

So first we need to make sure we know the trig identities to solve this problem.

$sin(\alpha )=\frac{opposite}{hypotenuse}$
$cos(\alpha )=\frac{adjacent}{hypotenuse}$
$tan(\alpha )=\frac{opposite}{adjacent}$

Here we only have two legs of the triangle, so we will need to use the Pythagorean Theorem a² + b² = c² to solve for the missing leg, the hypotenuse in this case.

Solving for the hypotenuse, c, we get $c = \sqrt{a^{2} +b^{2} }$
Here a = 20 and b = 21, so plugging in these values to the equation we get: $c= \sqrt{(20)^{2}+(21)^{2} } =\sqrt{400+441} =\sqrt{841}=29$