Ask question

Ask question

All categories

2 years ago

5

I NEED HELP AGAIN PLEASE

2 answers:

Troyanec [42]2 years ago

8 0

Answer:106

Step-by-step explanation:

d1i1m1o1n [39]2 years ago

3 0

The answer is 106 your welcome

You might be interested in

I need help with 50. I don't know how to figure it out

Nikolay [14]

The answer is in the picture

8 0

3 years ago

What is the equation of a line parallel to y=(3)/(4)x-5 and passes through (12,-2)?

Vika [28.1K]

A parallel line has the same slope as the original line. So in this case the slope of the line is also 3/4. Now how do we know if it intersects the point? We need to adjust the y intercept.

Currently, we know the equation of the line is y= 3/4 x + b, where b is the thing we are looking for. We also have a point, which supplies the x and y. Plug that in and solve for b

-2 = (3/4)*(12) + b

You'll get b= -11

So the equation of the parallel line intersecting the point given is y= 3/4x -11.

I am assuming that the slope is 3/4 based on the way you formatted the original equation, but it's the same steps if the slope is different.

3 0

1 year ago

Find the sum of x-1 and 6x-8

amm1812

Answer:

7x-9

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Pleasantburg has a population growth model of P(t)=at2+bt+P0 where P0 is the initial population. Suppose that the future populat

yulyashka [42]

Answer:

The population will reach 34,200 in February of 2146.

Step-by-step explanation:

Population in t years after 2012 is given by:

$P(t) = 0.8t^{2} + 6t + 19000$

In what month and year will the population reach 34,200?

We have to find t for which P(t) = 34200. So

$P(t) = 0.8t^{2} + 6t + 19000$

$0.8t^{2} + 6t + 19000 = 34200$

$0.8t^{2} + 6t - 15200 = 0$

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$0.8t^{2} + 6t - 15200 = 0$

So $a = 0.8, b = 6, c = -15200$

Then

$\bigtriangleup = 6^{2} - 4*0.8*(-15100) = 48356$

$t_{1} = \frac{-6 + \sqrt{48356}}{2*0.8} = 134.14$

$t_{2} = \frac{-6 - \sqrt{48356}}{2*0.8} = -141.64$

We only take the positive value.

134 years after 2012.

.14 of an year is 0.14*365 = 51.1. The 51st day of a year happens in February.

So the population will reach 34,200 in February of 2146.

6 0

2 years ago

Help please help help please please please

AnnZ [28]

Answer:

C. $\frac{5}{4}$

D. $\frac{6}{7}$

Step-by-step explanation:

C. $2 - \frac{12}{16}$

$= 2 - \frac{3}{4}$

$= \frac{8 - 3}{4}$

$= \frac{5}{4}$

D. $\frac{20}{22} \times \frac{33}{35}$

$= \frac{10}{11} \times \frac{33}{35}$

$= 10 \times \frac{3}{35}$

$= 2 \times \frac{3}{7}$

$= \frac{6}{7}$

3 0

2 years ago