I need help with number 11. Can someone please help me.

the temperature in winter night was -23 degrees Fahrenheit the temperature Rose by 5° when the sun rose when the sunset again th

expeople1 [14]

So we have -23 degrees to start

Then add 5 to it

It becomes -23+5=-18

Now subtract 7 from that so -18-7=-25 degrees

Hope this helps

8 0

3 years ago

BRAINLIESTT ASAP! PLEASE HELP ME :)

antiseptic1488 [7]

Step-by-step explanation:

$\displaystyle y = Acos(Bx - C) + D$

According to this trigonometric function, −C gives you the OPPOSITE terms of what they really are, so be EXTREMELY CAREFUL:

$\displaystyle Phase\:[Horisontal]\:Shift → \frac{-\frac{2}{3}π}{2} = -\frac{π}{3} \\ Period → \frac{2}{2}π = π$

Therefore we have our answer.

Extended Information on the trigonometric function

$\displaystyle Vertical\:Shift → D \\ Phase\:[Horisontal]\:Shift → \frac{C}{B} \\ Period → \frac{2}{B}π \\ Amplitude → |A|$

NOTE: Sometimes, your vertical shift might tell you to shift your graph below or above the <em>midline</em><em> </em>where the amplitude is.

I am joyous to assist you anytime.

3 0

3 years ago

Read 2 more answers

Help me pleaseeeeeeeeeeeeeeeeeeeeeeeee

ArbitrLikvidat [17]

Eduardo is packing 12 pairs of mismatched socks for his trip.

3 0

3 years ago

Read 2 more answers

The length of a 200 square foot rectangular vegetable garden is 4feet less than twice the width. Find the length and width of th

inna [77]

Answer:

Length = 18.099 ft

Width = 11.049 ft

Step-by-step explanation:

let the length of the field be x ft

and the width be y ft

as per the condition given in problem

x=2y-4 -----------(A)

Also the area is given as 200 sqft

Hence

xy=200

Hence from A we get

y(2y-4)=200

taking 2 as GCF out

2y(y-2)=200

Dividing both sides by 2 we get

$y(y-2)=100$

$y^2-2y=100$

subtracting 100 from both sides

$y^2-2y-100=0$

Now we solve the above equation with the help of Quadratic formula which is given in the image attached with this for any equation in form

$ax^2+bx+c=0$

Here in our case

a=1

b=-1

c=-100

Putting those values in the formula and solving them for y

$y=\frac{-(-2)+\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

$y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

Solving first

$y=\frac{2+\sqrt{4+400}{2}$

$y=\frac{2+\sqrt{404}{2}$

$y=\frac{2+20.099}{2}$

$y=\frac{22.099}{2}$

$y=11.049$

Solving second one

$y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

$y=\frac{2-\sqrt{4+400}{2}$

$y=\frac{2-\sqrt{404}{2}$

$y=\frac{2-20.099}{2}$

$y=\frac{-18.99}{2}$

$y=-9.045$

Which is wrong as the width can not be in negative

Our width of the field is

y=11.099

Hence the length will be

x=2y-4

x=2(11.049)-4

x=22.099-4

x=18.099

Hence our length x and width y :

Length = 18.099 ft

Width = 11.049 ft

4 0

3 years ago

Based on the table, what is Nick's rate?

Cerrena [4.2K]

Nicks rate is 7 yards per second.

6 0

3 years ago