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3 years ago

Can someone help me I suck at math I would really appreciate it

Mathematics

1 answer:

andreyandreev [35.5K]3 years ago

8 0

A = 1/2×b×h
= 1/2 × 6 × 8
= 24 sq.cm

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Find f(6) if f(x)=x^2 divided by 3x+x

Alinara [238K]

So, if I'm understanding you correctly, f(x)=x^2 divided by 3x+x, or
f(x) = (x^2)/(3x+x)
f(6) --> plug the 6 in to all values of x --> f(6) = (6^2)/(3×6 + 6) = 36/(18+6)
= 36/24 --> both are divisible by 12, so 36/12=3 and 24/12=2, now we have the reduced (simplified) answer:
f(6) = 3/2, or 1 1/2, or 1.5

3 0

4 years ago

Find the slope of the line below.

morpeh [17]

Answer:

slope = -3

hope this helps :)

6 0

3 years ago

Read 2 more answers

Match each term with the part of the expression 9m4−12m−6−12 it best describes. All terms will be used ONLY once.

RideAnS [48]

Answer:

see below...

Step-by-step explanation:

-12=constant [# doesn't change]

4=degree [exponent]

-6m= term [variable or 3]

9=coefficient [first #]

4 0

3 years ago

Simplify the equation 5(10 – y) – 4

Mrrafil [7]

Answer:

46−5y

Step-by-step explanation:

5(10−y)−4

Use the distributive property to multiply 5 by 10−y.

50−5y−4

Subtract 4 from 50 to get 46.

46−5y

Graph if needed:

5 0

3 years ago

Read 2 more answers

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

3 years ago