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2 years ago

Could you please help me?

Mathematics

1 answer:

slamgirl [31]2 years ago

6 0

Answer:

x = 31

Step-by-step explanation:

AC=BD

3x+7 = 131-x

Subtract 7 from both sides

3x = 124-x

Add x to 3x

4x = 124

Divide by 4

x = 31

Check:

3(31)+7 = 131 - 31

Mulitply 3 times 31 / subtract 131 and 31

93+7 = 100

Add 93 to 7

100 = 100

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Thank you guys fir the help

Lubov Fominskaja [6]

The correct answer to that would be b

4 0

2 years ago

A boy needs to grill 13 pounds of meat for a cookout. Determine the Maximum cost per pound that he can afford if he wants to spe

marissa [1.9K]

Answer:

$5.38 per pound

Step-by-step explanation:

70/13=5.384615

5.38x13=69.94

6 0

2 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

2 years ago

If Sam earns $97.50 for 15 hours of work,How many hours will he need to work to earn $130?

svp [43]

Answer:

it would be 20

Step-by-step explanation:

because you divide to find the unit rate like 97.50/15 and then you multiply a number to the answer you got there until those two numbers have a product of 130

8 0

3 years ago

Read 2 more answers

I dont understand the problem

Roman55 [17]

Answer:

equation of a line:

y = mx+c

1) find the gradient, m

$m = \frac{y2 - y1}{x2 - x1}$