All categories

3 years ago

Help!!!

Mathematics

1 answer:

Anika [276]3 years ago

4 0

It is the third option y+2=-3(x-1)

You might be interested in

In 2010 the Internal Revenue Service allowed a $5,150 deduction for each dependent. In 2005 it was a $3,200. If a family has fiv

Aleks04 [339]

5x5,150 =$<span>25,750 = Answer</span>

Deductions per dependent $5,150 times 5 in 2010

6 0

3 years ago

Fractions to decimals like 5/37 and 2/16

Zina [86]

Answered respectively;
1) 0.13514
2) 0.125

5 0

3 years ago

Read 2 more answers

When the sum of \, 528 \, and three times a positive number is subtracted from the square of the number, the result is \, 120. F

aleksandr82 [10.1K]

Let $x$ be the unknown number. So, three times that number means $3x$ , and the square of the number is $x^2$

We have to sum 528 and three times the number, so we have $528+3x$

Then, we have to subtract this number from $x^2$ , so we have

$x^2-(3x+528)$

The result is 120, so the equation is

$x^2 - 3x - 528 = 120 \iff x^2 - 3x - 648 = 0$

This is a quadratic equation, i.e. an equation like $ax^2+bx+c=0$ . These equation can be solved - assuming they have a solution - with the following formula

$x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

If you plug the values from your equation, you have

$x_{1,2} = \dfrac{3\pm\sqrt{9-4\cdot(-648)}}{2} = \dfrac{3\pm\sqrt{9+2592}}{2} = \dfrac{3\pm\sqrt{2601}}{2} = \dfrac{3\pm51}{2}$

So, the two solutions would be

$x = \dfrac{3+51}{2} = \dfrac{54}{2} = 27$

$x = \dfrac{3-51}{2} = \dfrac{-48}{2} = -24$

But we know that x is positive, so we only accept the solution $x = 27$

6 0

3 years ago

Of the students in Ms.Danver's class, 6 walk to school.This represents 30% of her students.How many Students are in Ms.Damver's

Kisachek [45]

There is 20 students. To find the answer you need to use porportionals and cross multiply.

6 0

4 years ago

Read 2 more answers

How do you do this problem?

zhenek [66]

Answer:

A) g is increasing, and the graph of g is concave up.

Step-by-step explanation:

g'(x) = ∫₀ˣ e^(-t³) dt

Since e^(-t³) is always positive, ∫₀ˣ e^(-t³) dt is positive when x > 0. So the function is increasing.

Find g"(x) by taking the derivative using second fundamental theorem of calculus:

g"(x) = e^(-x³)

g"(x) is always positive, so the function is always concave up.

5 0

3 years ago