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timama [110]
3 years ago
11

Starting at 1:00 p.m., the temperature changes −4 degrees Fahrenheit per hour. It is 35°F at 1:00 p.m.. Write and solve an equat

ion to find how many hours x it will take to reach −1°F.
An equation is

It will take _ hours to reach −1°F.
Mathematics
1 answer:
kondor19780726 [428]3 years ago
7 0
Start at 35 and start subtracting 4 (-4) then count each number so (35,31,27,23,19,15,11,7,3,-1) that’s 9 numbers so 1:00Pm +9 hours to 10:00PM and there you go
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The base of an isosceles triangle is 12 cm and its perimeter is 48 cm find its area​
NikAS [45]

I hope it helps you ❤️❤️❤️❤️❤️❤️

8 0
3 years ago
Read 2 more answers
Ralph went on a shopping trip. He bought a pair of shoes and a sweater. Ralph spent 12​% more on the sweater than on the shoes.
IgorLugansk [536]

Answer: 112%

 

Step-by-step explanation:

Let x = the cost of the shoes

If the cost of the sweater is 12% more than the cost of the shoes, then 1.12x is the cost of the sweater.

To convert the 1.12x into a percentage, multiply 1.12 by 100

This gives you:  112%

4 0
2 years ago
HELP!!!!<br><br> x ÷ 5.35 = 1.2
zloy xaker [14]

Answer:

x=6.42

Step-by-step explanation:

Multiply both sides by 5.35.

x=1.2×5.35

Simplify 1.2×5.35  to  6.42.

x=6.42

3 0
3 years ago
GEOMETRY find the angle of PRQ NO LINKS!!
anzhelika [568]

Answer:

55

Step-by-step explanation:

Remark

This is a really good question to know the answer to. <PRQ = 1/2 POQ (the central angle. )

The central angle for this question is 110o. So any angle that has its vertex on the circumference is 1/2 110 = 55. The central angle and the angle on the circumference must be related as they are as this question is. (Both are on the same side of PQ which is not drawn but you can draw it).

4 0
3 years ago
A horizontal trough is 16 m long, and its end are isosceles trapezoids with an altitude of 4 m, a lower base of 4 m, and an uppe
Ganezh [65]

Answer:

0.28cm/min

Step-by-step explanation:

Given the horizontal trough whose ends are isosceles trapezoid  

Volume of the Trough =Base Area X Height

=Area of the Trapezoid X Height of the Trough (H)

The length of the base of the trough is constant but as water leaves the trough, the length of the top of the trough at any height h is 4+2x (See the Diagram)

The Volume of water in the trough at any time

Volume=\frac{1}{2} (b_{1}+4+2x)h X H

Volume=\frac{1}{2} (4+4+2x)h X 16

=8h(8+2x)

V=64h+16hx

We are not given a value for x, however we can express x in terms of h from Figure 3 using Similar Triangles

x/h=1/4

4x=h

x=h/4

Substituting x=h/4 into the Volume, V

V=64h+16h(\frac{h}{4})

V=64h+4h^2\\\frac{dV}{dt}= 64\frac{dh}{dt}+8h \frac{dh}{dt}

h=3m,

dV/dt=25cm/min=0.25 m/min

0.25= (64+8*3) \frac{dh}{dt}\\0.25=88\frac{dh}{dt}\\\frac{dh}{dt}=\frac{0.25}{88}

=0.002841m/min =0.28cm/min

The rate is the water being drawn from the trough is 0.28cm/min.

3 0
4 years ago
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