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2 years ago

Please help WILL MARK AS BRAINLIEST

Mathematics

1 answer:

adoni [48]2 years ago

7 0

Which question first or second??

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If b+c-a/a, c+a-b/CA are in A.P., prove that a, b, c are in<br>Н.Р.

SOVA2 [1]

Answer:I love thee fact that I am a straight a Student but I don't wanna help you dumb people

Step-by-step explanation:

3 0

2 years ago

How do I solve this problem

erma4kov [3.2K]

10√3 (-7 - √5)

10√3 × -7 + 10√3 × -√5

-70√3 + 10√3 × -√5

-70√3 - 10√3√5

-70√3 -10√3 × 5

-70√3 - 10√15

hope that helps, God bless!

6 0

3 years ago

Order these numbers from least to greatest.

Lorico [155]

Answer:

1. -√19 2. 4.56 3. √21 4. 23/5

Step-by-step explanation:

6 0

2 years ago

The image of the point (3, -4) under a translation is (0, –5). Find the coordinatesof the image of the point (8,7) under the sam

Sav [38]

We will have the following:

*First: We determine the translation rule:

We have that the point went from (3, -4) to (0, -5), thus the translation rule would be:

$(x,y)\to(x-3,x-1)$

Second: We apply the translation rule to the other point to determine it's image, that is:

$(8,7)\to(8-3,7-1)\to(5,6)$

So, the image of the second point is (5, 6).

6 0

1 year ago

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