Which explains how to find the question of the division below<br><br><br>-3 1/3 divided by 4/9

Could someone help me please if you don’t mind thank you so much!

Marina86 [1]

Answer:

m = 11

Step-by-step explanation:

Given

$\frac{2}{5}$ (m + 4) = 6 ← multiply both sides by 5 to clear the fraction

2(m + 4) = 30 ( divide both sides by 2 )

m + 4 = 15 ( subtract 4 from both sides )

m = 11

This method avoids having to deal with awkward fractions

6 0

3 years ago

If a cannonball is shot directly upward with a velocity of 160 ft per second, its height above the ground after t seconds is gi

Contact [7]

Step-by-step explanation:

If a cannonball is shot directly upward with a velocity of 160 ft per second. Its height as a function of time is given by :

$h(t)=160t-16t^2$ .......(1)

t is in seconds

(a) Velocity is given by :

$v=\dfrac{dh}{dt}\\\\v=\dfrac{d(160t-16t^2)}{dt}\\\\v=(160-32t)\ ft/s$

Acceleration is given by :

$a=\dfrac{dv}{dt}\\\\a=\dfrac{d(160-32t)}{dt}\\\\a=-32\ ft/s^2$

(b) For maximum height put $\dfrac{dh}{dt}=0$

i.e.

$160-32t=0\\\\t=5\ s$

Put t = 5 s in equation (1). So,

$h(5)=160t-16t^2\\\\h(5)=160(5)-16(5)^2\\\\h(5)=400\ ft$

(c) When the ball reaches ground, its height is equal to 0. So,

h(t) = 0

$160t-16t^2=0\\\\t(160-16t)=0\\\\t=0,t=10\ s$

Hence, this is the required solution.

4 0

3 years ago

Help with this assignment

Irina-Kira [14]

Answer:

i think the answer is 21

hope it will help you

6 0

3 years ago

Examine whether each of the following pair of lines AB and CD are parallel or not

MatroZZZ [7]

Answer:

Yes

Step-by-step explanation:

For lines to be parallel alternate interior angles should be equal

Angle GHD=180-angle FHD

=180-115 = 65°

Therefore angle AGH = angle GHD

So lines are parallel

3 0

2 years ago

Read 2 more answers

What is the other square root of 119 + 120i?

MakcuM [25]

Answer:

$\sqrt{119+120 i}=\pm (12.24+i 4.90)$

Step-by-step explanation:

Given

z = 119 + 120 i

Let $\sqrt{119+120 i}=p+iq$

Squaring both sides

$119+120 i=p^2-q^2+2ipq$

Comparing real and imaginary part

Re(LHS)=Re(RHS)

$119=p^2-q^2$ ...........................(1)

comparing Im(LHS)=Im(RHS)

120=2pq

$q=\frac{60}{p}$

Substitute q in 1

$119=p^2-(\frac{60}{p})^2$

$p^4-119p^2-(68)^2=0$

Let $x=p^2$

$x^2-119x-4624=0$

$x=\dfrac{119\pm \sqrt{119^2+4\times 4624}}{2}$

$x=\frac{119\pm 180.71}{2}$

we take only Positive value because $p^2=x$

x=149.85

$p^2=149.85$

thus $p=\pm 12.24$

$q=\pm 4.90$

thus,

$\sqrt{119+120 i}=\pm (12.24+i 4.90)$

8 0

3 years ago

Which explains how to find the question of the division below-3 1/3 divided by 4/9