#1 {(3, 7), (2, 4), (-1, 7), (0, 2)} The domain is... The range is....

Using the standard 28/36 guidelines, what is the maximum mortgage payment allowed for someone with an annual salary of $82,175?

stich3 [128]

Answer: 28/36 Rule calculator tells you whether your debt is too high for your income.

Step-by-step explanation:

4 0

2 years ago

Need help plz very difficult

Oksanka [162]

Answer:

D - 9x^(2)+2x-5

Step-by-step explanation:

6 0

3 years ago

WILL GIVE BRAINLIEST!!!!!!!!!!!!!

Phantasy [73]

Writing the domain in words: negative infinity to three halves and three halves to infinity

So, Option A is correct.

Step-by-step explanation:

We need to find domain of the function: $f(x)=\frac{x-5}{2x-3}$

Domain:

The domain of function is defined as all the input values for which the function is defined.

In other words we can say that any value of x that makes the denominator zero is not considered as domain of function.

In the function given: $f(x)=\frac{x-5}{2x-3}$

if x= 3/2 then the denominator will be zero and the function will not be defined.

So, value of x should be x>3/2 or x<3/2

We can write domain as: (-∞,3/2)(3/2,∞)

3/2 can be written as three halves.

Writing the domain in words: negative infinity to three halves and three halves to infinity

So, Option A is correct.

Keywords: Domain of Function

Learn more about Domain of function at:

#learnwithBrainly

4 0

4 years ago

Find cos y and tan y if csc y = -√6/2 and cot y >0.

fomenos

Answer:

$\cos y = -\dfrac{\sqrt{3} }{3}$

$\tan y = \sqrt{2}$

Step-by-step explanation:

Recall that

$\boxed{\csc y := \dfrac{1}{\sin y}}$

$\boxed{\cot y := \dfrac{\cos y}{\sin y}}$

We know that

$\csc y = \dfrac{-\sqrt{6} }{2}$

Note that according to the definition of $\csc y$ it is true that both sine and cosine are negative, once $\csc y = \dfrac{-\sqrt{6} }{2}$ . Because $\cot y > 0$ , this conclusion is true. We basically have

$\boxed{(-a)(1/-b)=a/b \text{ such that } a,b\in\mathbb{R}_{\geq 0}}$

Sure it is true $\forall y\in\mathbb{R}$ but perhaps this way is better to understand.

In order to find sine, we can use the definition and manipulate the rational expression.

$\csc y = \dfrac{-\sqrt{6} }{2} = \dfrac{-\sqrt{6} / -\sqrt{6} }{2/-\sqrt{6} } = \dfrac{1 }{-\dfrac{2}{\sqrt{6} } }$

Therefore,

$\sin y =-\dfrac{2}{\sqrt{6} }$

Here I just divided numerator and denominator by $-\sqrt{6}$ .

Now, to find cosine we can use the identity

$\boxed{\sin^2y +\cos ^2y =1}$

Thus,

$\left(-\dfrac{2}{\sqrt{6} }\right)^2 + \cos ^2y =1 \implies \dfrac{4}{6 } +\cos ^2y =1$

$\implies \cos ^2y =1 - \dfrac{4}{6 } \implies \cos ^2y =\dfrac{1}{3 } \implies \cos y = \pm \dfrac{\sqrt{1} }{\sqrt{3} } = \pm \dfrac{\sqrt{1} \sqrt{3} }{3} = \pm \dfrac{\sqrt{3} }{3}$

$\cos y = \pm\dfrac{\sqrt{3} }{3}$

Once we have $\cot y > 0$ , we just consider

$\cos y = -\dfrac{\sqrt{3} }{3}$

FInally, for tangent, just consider

$\boxed{\tan y := \dfrac{\sin y}{\cos y}}$

thus,

$\tan y = \dfrac{\sin y}{\cos y} = \dfrac{-\frac{2}{\sqrt{6} }}{-\frac{\sqrt{3} }{3}} = \dfrac{6}{\sqrt{18} } =\dfrac{6}{3\sqrt{2} } =\dfrac{2}{\sqrt{2} } = \sqrt{2}$

5 0

3 years ago

How many Triangles are in this Triangle

Alex17521 [72]

Answer:

7 triangles

Step-by-step explanation:

6 0

3 years ago

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