All categories

4 years ago

Help me asap only 24 mins left.answer question 1 ok pls help me

Mathematics

1 answer:

vovikov84 [41]4 years ago

5 0

Answer:

Step-by-step explanation:

If it is linear, or has a slope, or is in a straight line, then the line will be proportional.

You might be interested in

What is the ratio of people who prefer pistachio to the people who prefer vanilla?

Brrunno [24]

Answer:

3/2

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

A parking garage charges 21.00 for 6 hours how much does it charge per hour

alexdok [17]

Answer:

If a parking garage charges $21 for 6 hours in order to find out how much it charges per hour you will need to divide 21 by 6:

21÷6=3.50

So in conclusion a parking garage charges

$3.50/hour

5 0

3 years ago

Solve each equation. I don't know this

MArishka [77]

1. $(2x - 3)(x + 7) = 0$
   $2x^{2} + 14x - 3x - 21 = 0$
   $2x^{2} + 11x - 21 = 0$

   $x = \frac{-11 +/- \sqrt{11^{2} - 4(2)(-21)}}{2(2)}$

   $x = \frac{-11 +/- \sqrt{121 + 168}}{4}$

   $x = \frac{-11 +/- \sqrt{289}}{4}$

   $x = \frac{-11 +/- 17}{4}$

   $x = -2.75 +/- 4.25$
   $x = -2.75 + 4.25$    $x = -2.75 - 4.25$
   $x = 1.5$    <u></u> $x = -7$
---------------------------------------------------------------------------------------------------------- 2. $8x(2x - 5) = 0$
   $8x(2x) - 8x(5) = 0$
   $16x^{2} - 40x = 0$
   $16x^{2} - 4x + 0 = 0$

   $x = \frac{-(-40) +/- \sqrt{(-40)^{2} - 4(16)(0)}}{2(16)}$

   $x = \frac{40 +/- \sqrt{1600 - 0}}{32}$

   $x = \frac{40 +/- \sqrt{1600}}{32}$

   $x = \frac{40 +/- 40}{32}$

   $x = 1.25 +/- 1.25$
   $x = 1.25 + 1.25$       $x = 1.25 - 1.25$
   $x = 2.5$          $x = 0$
----------------------------------------------------------------------------------------------------------
3. $x^{2} + 3x - 10 = 0$

   $x = \frac{-3 +/- \sqrt{3^{2} - 4(1)(-10)}}{2(1)}$

   $x = \frac{-3 +/- \sqrt{9 + 40}}{2}$

   $x = \frac{-3 +/- \sqrt{49}}{2}$

   $x = \frac{-3 +/- {7}}{2}$

   $x = -1.5 +/- 3.5$
   $x = -1.5 + 3.5$    $x = -1.5 - 3.5$
   $x = 2$        $x = -5$
----------------------------------------------------------------------------------------------------------
4. $x^{2} = 13x - 36$
    $x^{2} - 13x + 36 = 13x - 13x - 36 + 36$
    $x^{2} - 13x + 36 = 0$

    $x = \frac{-(-13) +/- \sqrt{(-13)^{2} - 4(1)(36)}}{2(1)}$


    $x = \frac{13 +/- \sqrt{169 - 144}}{2}$

    $x = \frac{13 +/- \sqrt{25}}{2}$

    $x = \frac{13 +/- 5}{2}$

    $x = 6.5 +/- 2.5$
    $x = 6.5 + 2.5$ $x = 6.5 - 2.5$
    $x = 9$    $x = 4$
----------------------------------------------------------------------------------------------------------
5. $3x^{2} - 7x + 2 = 0$

   $x = \frac{-(-7) +/- \sqrt{(-7)^{2} - 4(1)(2)}}{2(3)}$

   $x = \frac{7 +/- \sqrt{49 - 8}}{6}$

   $x = \frac{7 +/- \sqrt{41}}{6}$

   $x = \frac{7 +/- 6.403124237432849}{6}$

   $x = 1.167+/- 1.06718737290547$
   $x = 1.67 + 1.06718737290547$ $x = 1.167 - 1.06718737290547$
   $x = 2.73718737290547$    $x = 0.60281262709453$
----------------------------------------------------------------------------------------------------------
6. $10x^{2} - 10x + 9 = 5x^{2} + 4x + 1$
   $10x^{2} - 5x^{2} - 10x + 10x + 9 - 1 = 5x^{2} - 5x^{2} + 4x + 10x + 1 - 1$
   $5x^{2} + 8 = 14x$
   $5x^{2} - 14x + 8 = 14x - 14x$
   <u /> $5x^{2} - 14x + 8 = 0$

   $x = \frac{-(-14) + \sqrt{(-14)^{2} - 4(5)(8)}}{2(5)}$

   $x = \frac{14 +/- \sqrt{196 - 160}}{10}$

   $x = \frac{14 +/- \sqrt{36}}{10}$ .

   $x = \frac{14 +/- 6}{10}$

   $x = 1.4 +/- 0.6$
   $x = 1.4 + 0.6$       $x = 1.4 - 0.6$
   <u /> $x = 2$ $x = 0.8$

5 0

4 years ago

Determine the second derivative evaluated at t=0

enyata [817]

$M(t) = e^{3(e^t-1)}$