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3 years ago

6

.05×+.25=1.0 idk how to do this

1 answer:

Ede4ka [16]3 years ago

3 0

.05 x 1.5 + .25 = 1.0

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4(-3x-5)-(10+4x)<br> Help?

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-12x-20-40plus 16x
-4x plus -60

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3 years ago

1. Which of the following linear equations has

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where m is the slope.
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2 years ago

Henry is giving a 5.5% tip on his bill ,b and his final cost is $60. Which of the following represents that

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2 years ago

Read 2 more answers

Molly was on a long 136 mile road trip. The first part of the trip there was lots of traffic, she only averaged 16 mph. The seco

Mazyrski [523]

Answer:

In traffic, she drove for 3 hours

and After the traffic cleared she drove for 2 hours.

Explanation:

Given that the road trip was 136 miles;

$d=136$

The first part of the trip there was lots of traffic, she only averaged 16 mph;

$v_1=16$

The second part of the trip there was no traffic so she could drive 44 mph;

$v_2=44$

She traveled for a total of 5 hours;

$t=5$

let x represent the time in traffic when she traveled at 16 mph

$t_1=x$

the time the traffic is clear would be;

$t_2=t-t_1=5-x$

Recall that distance equals speed multiply by time;

$d=v_1t_1_{}_{}^{}+v_2t_2$

substituting the values;

$136=16x+44(5-x)$

solving for x;

$\begin{gathered} 136=16x+220-44x \\ 44x-16x=220-136 \\ 28x=84 \\ x=\frac{84}{28} \\ x=3 \end{gathered}$

So;

$\begin{gathered} t_1=3\text{ hours} \\ t_2=5-x=5-3=2 \\ t_2=2\text{ hours} \end{gathered}$

Therefore, In traffic, she drove for 3 hours

and After the traffic cleared she drove for 2 hours.

7 0

1 year ago

The Hudson Bay tides vary between 3 feet and 9 feet. The tide is at its lowest point when time (t) is 0 and completes a full cyc

BaLLatris [955]

Answer:

$y(t)= 6-3cos(\dfrac{2\pi}{14}t )$

Step-by-step explanation:

The function that could model this periodic phenomenon will be of the form

$y(t) = y_0+Acos(wt)$

The tide varies between 3ft and 9ft, which means its amplitude $A$ is

$A =\dfrac{(9-3)ft}{2} \\\\\boxed{A = 3ft}$

and its midline $y_0$ is

$y_o=3+3 \\\\\boxed{y_o= 6ft}$ .

Furthermore, since at $t=0$ the tide is at its lowest ( 3 feet ), we know that the trigonometric function we must use is $-cos(\omega t)$ .

The period of the full cycle is 14 hours, which means

$\omega t =2\pi$

$\omega (t+14)= 4\pi$

giving us

$\boxed{\omega = \dfrac{2\pi}{14}.}$

With all of the values of the variables in place, the function modeling the situation now becomes

$\boxed{y(t)= 6-3cos(\dfrac{2\pi}{14}t ).}$

8 0

3 years ago