Ask question

Ask question

All categories

3 years ago

6

the temperature dropped by 7.f each hour from 5:00 am to 9:00 am. What was the beginning temperature at 5:00 am if the temperatu

re at 9:00 am was -10.f

1 answer:

Sati [7]3 years ago

6 0

4 * 7 = 28.
28 + (-10) = 18
Therefore, the answer is 18.

If you want to check this, you can subtract 7 from 18 four times.

You might be interested in

Kevin has 14 buttermilks how many buttermilk is he need to make a pancakes7

Luda [366]

Answer:

dont know redo the question

Step-by-step explanation:

7 0

3 years ago

Sam wants to find the best buy on cans of tuna. He reads these four ads in the newspaper.$12.53 for seven cans$5.97 for three ca

grandymaker [24]

In each case divide the price by the number of cans.
The best buy is the lowest price.
$12.53/7 = $1.79
$5.97/3 = $1.99
$7.95/5 = $1.59
$8.34/6 = $1.39

The fourth ad, $8.34 for 6 cans results in the lowest price per can, $1.39.
The fourth ad is the best buy.

8 0

3 years ago

NO LINKS OR FILES!

Archy [21]

(a) If the particle's position (measured with some unit) at time <em>t</em> is given by <em>s(t)</em>, where

$s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}$

then the velocity at time <em>t</em>, <em>v(t)</em>, is given by the derivative of <em>s(t)</em>,

$v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}$

(b) The velocity after 3 seconds is

$v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}$

(c) The particle is at rest when its velocity is zero:

$\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}$

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

$\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|$

In interval notation, this happens for <em>t</em> in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt$

By definition of absolute value, we have

$|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)$

In part (d), we've shown that <em>v(t)</em> > 0 when -√11 < <em>t</em> < √11, so we split up the integral at <em>t</em> = √11 as

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt$

and by the fundamental theorem of calculus, since we know <em>v(t)</em> is the derivative of <em>s(t)</em>, this reduces to

$s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}$

7 0

3 years ago

PLEASE HELP ASAP. RIGHT ANSWER WILL GET BRAINLIEST

dimulka [17.4K]

Given
P(1,-3); P'(-3,1)
Q(3,-2);Q'(-2,3)
R(3,-3);R'(-3,3)
S(2,-4);S'(-4,2)

By observing the relationship between P and P', Q and Q',.... we note that
(x,y)->(y,x) which corresponds to a single reflection about the line y=x.
Alternatively, the same result may be obtained by first reflecting about the x-axis, then a positive (clockwise) rotation of 90 degrees, as follows:
Sx(x,y)->(x,-y) [ reflection about x-axis ]
R90(x,y)->(-y,x) [ positive rotation of 90 degrees ]
combined or composite transformation
R90. Sx (x,y)-> R90(x,-y) -> (y,x)

Similarly similar composite transformation may be obtained by a reflection about the y-axis, followed by a rotation of -90 (or 270) degrees, as follows:
Sy(x,y)->(-x,y)
R270(x,y)->(y,-x)
=>
R270.Sy(x,y)->R270(-x,y)->(y,x)

So in summary, three ways have been presented to make the required transformation, two of which are composite transformations (sequence).

4 0

4 years ago

Read 2 more answers

One angle in a triangle has a measure that is three times as large as the smallest angle. The measure of the

Nikolay [14]

Step-by-step explanation:

no need.,..................

....

3 0

3 years ago