Ask question

Ask question

All categories

3 years ago

14

8(2w-6)+4(-1-5w)=0 Solve and show work.

1 answer:

Gelneren [198K]3 years ago

7 0

8(2w-6)+4(-1-5w)=0

16w-48-4-20w=0

-4w-52=0

4w=-52

w=(-52)/4

w=-13

You might be interested in

Someone help, what’s 1+1?

Sergeu [11.5K]

It's 2 sir/ Maam Your Welcome

7 0

3 years ago

Yuet solved the equation 6 a minus 2 b = 12 for a. Her steps are shown below. 1. Subtract 2b: 6 a = 12 minus 12 b 2. Divide by 6

Ber [7]

Question:

Yuet solved the equation $6a - 2b = 12$ for a.

Her steps are shown below.

1. Subtract 2b: $6a = 12 - 2b$

2. Divide by 6: $a = 2 - \frac{b}{3}$

Answer:

In step 1 she needed to add 2b to both sides of the equation.

Step-by-step explanation:

Given:

The above steps shows how Yuet solved an equation

Required:

True statement about Yuet's work

The implication of the statement is to state what Yuet should have done instead of what she did.

In step 1, she subtracted 2b from both sides of the equation; this is wrong.

Instead of subtracting, she ought to add 2b to both sides of the equation.

The correct steps and result is as follows:

1. Add 2b: $6a = 12 + 2b$

2. Divide by 6: $a = 2 + \frac{b}{3}$

Hence, we can conclude that in step 1 she needed to add 2b to both sides of the equation.

4 0

3 years ago

Read 2 more answers

So I have 3D figures of nets, and the problem is <br> Triangular prism: 5x5x4x7x6

Elza [17]

5x5x4x7x6 is equal to 4,200

7 0

3 years ago

If x = 4cosa and y = 5tana, find (x^2/16) - (y^2/25)

Cloud [144]

$x = 4 \cos(a)$

$\frac{x}{4} = \cos(a) \\$

$( { \frac{x}{4} })^{2} = ({ \cos(a) })^{2} \\$

$\frac{ {x}^{2} }{16} = {cos}^{2} (a) \\$

_____________________________________________

$y = 5 \tan(a)$

$\frac{y}{5} = \tan(a) \\$

$( { \frac{y}{5} })^{2} = ( { \tan(a) })^{2} \\$

$\frac{ {y}^{2} }{25} = {tan}^{2} (a) \\$

_____________________________________________

Thus ;

$\frac{ {x}^{2} }{16} - \frac{ {y}^{2} }{25} = \\$

${cos}^{2} (a) - {tan}^{2} (a)$

3 0

2 years ago

Can someone help me with all this? Lol idk any of this and E-learning sucks I need to show my work as well

Margarita [4]

Answer:

1a) 230 feet

1b) 3.75s

1c) t=7.54s

2) t=2s

3) t=2.32s

Step-by-step explanation:

The equation that models Brett's last home run is $h(t)=-16t^2+120t+5$

We need to complete the square to obtain the function in the vertex form:

$h(t)=-16(t^2-7.5t)+5$

$h(t)=-16(t^2-7.5t+3.75^2)+5+-16(-3.75)^2$

$h(t)=-16(t-3.75)^2+230$

1a) The vertex is (3.75,230).The maximum value is the y-coordinate of the vertex, which is 230 feet.

1b) The time it takes to reach the maximum value is the x-coordinate of the vertex. It reach the maximum heigth after t=3.75 seconds

1c) To find the time the ball hit the ground, we equate the h(t)=0 and solve for t.

$-16(t-3.75)^2+230=0$

$-16(t-3.75)^2=-230$

$(t-3.75)^2=14.375$

$t-3.75=\pm \sqrt{14.375}$

$t=3.75\pm \sqrt{14.375}$

$t=3.75\pm 3.79$

$t=3.75-3.79\:\:or\:t=3.75+3.79$

$t=3.75-3.79\:\:or\:t=3.75+3.79\\t=-0.04\:\:or\:t=7.54$

The time is positive, so the ball hit the ground after 7.54 seconds.

Question 2)

The function that models the amusement ride is $h(t)=-16t^2+64t+60$

We want to find the time it takes for the riders to reach the maximum height.

This time is given by: $t=-\frac{b}{2a}$

Comparing $h(t)=-16t^2+64t+60$ to $h(t)=at^2+bt+c$ we have a=-16, b=64, c=60.

We substitute to obtain:

$t=-\frac{64}{2*-16} \\t=-\frac{64}{-32} \\t=2$

Hence it took 2 seconds to rech the maximum height.

Question 3)

The equation that models the height f the catridge is $h(t)=-16t^2+35t+5$

To find the time that the catridge will land, we equate the function to zero and solve for t.

$-16t^2+35t+5=0$

This is a quadratic equation with =-16, b=35, an c=5

The solution is given by:

$t=\frac{-b\pm \sqrt{b^2-4ac} }{2a}$

We substitute the values to get:

$t=\frac{-35\pm \sqrt{35^2-4*-16*5} }{2*-16}$

This gives:

$t=\frac{35-\sqrt{1545} }{32} \:or\: t=\frac{35+\sqrt{1545} }{32}$

This simplifies to:

$t=-0.13\:or\: t=2.32$

The time it takes to land must be positive thefore t=2.32 seconds

7 0

3 years ago