All categories

3 years ago

When you owe money to a lender, you are a___?

Mathematics

2 answers:

Arturiano [62]3 years ago

8 0

You are in debt

Thank you

snow_tiger [21]3 years ago

5 0

I'm pretty sure the answer is loanee but debt is a good answer too

You might be interested in

How to solve 2a^b-4ab^

Maslowich

Simplifying

2ab + 4ab = 0

Combine like terms: 2ab + 4ab = 6ab

6ab = 0

Solving

6ab = 0

Solving for variable 'a'.

Move all terms containing a to the left, all other terms to the right.

Divide each side by '6'.

ab = 0

Simplifying

ab = 0

6 0

3 years ago

A student says that the x-intercept of a line is (0,-10,). Is the student correct? Explain.

nataly862011 [7]

Answer:

Yes, the student is correct

Step-by-step explanation:

Given

$(0,-10)$

Required

Does it represent the x intercept

At the x intercept, the value of x is 0.

So, the function $(x,y)$ becomes $(0,y)$

$(0,-10)$ can be compared to $(0,y)$

Hence, the student is correct

5 0

2 years ago

The equation of a line is 4x+y=13 solve the equation for y

Salsk061 [2.6K]

Answer:

y = 13 - 4x

Why?

4y +y = 13

Subtract 4x from both sides

4x+y-4x = 13 - 4x

Simplify

Y= 13 - 4x

Have a good day

3 0

3 years ago

What is the solution to he equation (y/y-4)-(4/y+4)=3^2/y^2-16

kicyunya [14]

Answer:

$y=\pm i \sqrt{7}$

Step-by-step explanation:

Given equation is $\frac{y}{\left(y-4\right)}-\frac{4}{\left(y+4\right)}=\frac{3^2}{(y^2-16)}$

Factor denominators then solve by making denominators equal

$\frac{y}{\left(y-4\right)}-\frac{4}{\left(y+4\right)}=\frac{3^2}{(y^2-16)}$

$\frac{y}{\left(y-4\right)}-\frac{4}{\left(y+4\right)}=\frac{9}{(y+4)\left(y-4\right)}$

$\frac{y\left(y+4\right)-4\left(y-4\right)}{\left(y-4\right)\left(y+4\right)}=\frac{9}{(y+4)\left(y-4\right)}$

$y\left(y+4\right)-4\left(y-4\right)=9$

$y^2+4y-4y+16=9$

$y^2=9-16$

$y^2=-7$

take squar root of both sides

$y=\pm \sqrt{-7}$

$y=\pm i \sqrt{7}$

Hence final answer is $y=\pm i \sqrt{7}$ .

7 0

3 years ago

Look at the triangle show on the right. The Pythagorean Theorem states that the sum of the squares of the legs of a right triang

Vladimir [108]

Answer:

$cos^2\theta + sin^2\theta = 1$

Step-by-step explanation:

Given

$(\frac{b}{r})^2 + (\frac{a}{r})^2$

Required

Use the expression to prove a trigonometry identity

The given expression is not complete until it is written as:

$(\frac{b}{r})^2 + (\frac{a}{r})^2 = (\frac{r}{r})^2$

Going by the Pythagoras theorem, we can assume the following.

a = Opposite
b = Adjacent
r = Hypothenuse

So, we have:

$Sin\theta = \frac{a}{r}$

$Cos\theta = \frac{b}{r}$

Having said that:

The expression can be further simplified as:

$(\frac{b}{r})^2 + (\frac{a}{r})^2 = 1$

Substitute values for sin and cos

$(\frac{b}{r})^2 + (\frac{a}{r})^2 = 1$ becomes

$cos^2\theta + sin^2\theta = 1$

3 0

3 years ago