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3 years ago

Please help 8th grade math ez

Mathematics

2 answers:

34kurt3 years ago

7 0

I believe the answer is 22?

sp2606 [1]3 years ago

4 0

Answer

I think the answer is 22 minutes.

Step-by-step explanation:

Hope this helps!

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you invest $2800 in an account that pays an interest of 5.5%, compounded continuously. calculate the balance of your account aft

LuckyWell [14K]

Hi, hope this helps you. Have a good day. :)

Answer:

A = $5,417.42 (balance of the account)

I = $2,617.42

5 0

3 years ago

Read 2 more answers

Find the value of x. Please help!!!!

Murrr4er [49]

Answer:

9°

Step-by-step explanation:

In a ∆, equal sides make equal angle to the third side. So the angle that's not mentioned is same '2'.

Sum of angles in a ∆ = 180°

80° + m/_2 + m/_2 = 180°

80° + (5x + 5) + (5x + 5) = 180°

10x + 90° = 180°

10x = 90°

x = 9°

6 0

3 years ago

Can anyone help me with Qs 9-14 pls

lions [1.4K]

Answer:

Step-by-step explanation:

angle 2 and angle 3 are linear pair

angle 1 and angle 7 are alternate exterior angle

angle 4 + angle 7=a80 degree

124 + angle 7=180

angle 7=180-124

angle 7=56 degree

angle 1 is supplementary to angle 2 ,angle 4, angle 6, angle 8 .In supplementary angle sum of two angle is always 180 degree.

angle 4 and angle 5 are supplementary angles because their sum is equal to 180 degree.

angle 4 + angle 5(being co interior angle)

3x+17+x+23=180

4x+40=180

4x=19=180-40

x=140/4

x=35

3 0

3 years ago

If a runner was traveling with a velocity of 4 m/s north and had a displacement of 2500 m, how long was he running?

Mademuasel [1]

Answer:

625

Step-by-step explanation:

t= d/s

2500/4

4 0

2 years ago

Solve the equation <img src="https://tex.z-dn.net/?f=%2835%20x%5E%7B4%7D%20y%2B14%20x%5E%7B5%7D%20y-2%20y%5E%7B3%7D-4x%20y%5E%7B

jeka94

$\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0$

$M_y=35x^4+14x^5-6y^2-12xy^2$
$N_x=35x^4-6y^2$

$\dfrac{N_x-M_y}N=\dfrac{-14x^5+12xy^2}{7x^5-6xy^2}=-2$

This suggests an integrating factor depending on $x$ only is possible, and given by

$\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^{2x}$

Distributing across the ODE, we end up with

$\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^{2x}}_{M^*}\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^{2x}}_{N^*}\,\mathrm dy=0$

The equation is now exact, with

${M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^{2x}$

Now we find the solution:

$F_x=M^*$
$F=\displaystyle\int(35x^4+14x^5-2y^3-4xy^3)e^{2x}\,\mathrm dx$
$F=(7x^5y-2xy^3)e^{2x}+f(y)$

$F_y=N^*$
$(7x^5-6xy^2)e^{2x}+f'(y)=(7x^5-6xy^2)e^{2x}$
$f'(y)=0$
$\implies f(y)=C$

The general solution is then

$F(x,y)=(7x^5y-2xy^3)e^{2x}=C$

7 0

3 years ago