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Doss [256]
3 years ago
8

PLZZ help me with this I will give brilliantis to whoever gets it right

Mathematics
1 answer:
blagie [28]3 years ago
8 0

Answer:

C

Step-by-step explanation:

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Cos 45 is option B! :)
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In 1911, the temperature in Rapid City,
Sliva [168]

The 3.1 °F/min rate of change of the temperature and 15 minutes change duration gives the change in temperature as 46.5 °F

<h3>How can the change in temperature be found from the rate of change?</h3>

The rate at which the temperature changed = 3.1 °F/min

The duration of the change in temperature = 15 minutes

The relationship between the change in temperature, the rate of change in temperature and the time can be presented as follows;

3.1  ^{\circ} F/min =  \frac{ \delta T }{\delta t}  =  \frac{ \Delta T }{ \Delta t}

Where;

∆T = The required change in temperature

∆t = The duration of the change = 15 minutes

Which gives;

∆T = 3.1°F/min × 15 minutes = 46.5 °F

  • The change in temperature, ∆T = 46.5 °F

Learn more about the rate of change of a variable here:

brainly.com/question/10208814

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8 0
1 year ago
Two lines that intersect at 90 degree angles
Molodets [167]
That would be a right angle 
3 0
3 years ago
Find the total surface area of this cuboid
Harman [31]

Step-by-step explanation:

☄ \underline{ \underline{ \sf{Given}}}:

  • Length ( l ) = 4 cm
  • Width ( w ) = 2 cm
  • Height ( h ) = 5 cm

☄ \underline{ \underline{ \sf{To \: find}}} :

  • Total surface area of a cuboid

❅\underline{ \underline{ \text{Solution}}}:

✑ \boxed{ \text{TSA = 2lw + 2lh + 2hw}}

Plug the known values :

⟼ \sf{2 \times 4 \times 2 + 2 \times 4 \times 5 + 2 \times 5 \times 2}

⟼ \sf{16 + 40 + 20}

⟼ \boxed{\sf{76 \:  {cm}}^{2} }

\red{ \boxed{ \boxed{ \tt{⟿ \: Our \: final \: answer : 76 \:  {cm}}^{2} }}}

Hope I helped !♡

Have a wonderful day / night ! ツ

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4 0
2 years ago
A football player is running the length of a 100 yard long football field. For the first part of the field, he runs at a rate of
notka56 [123]

Answer:

\frac{2}{5}th part of the field is covered in the second sprint.

Step-by-step explanation:

A football player is running the length of a 100 yard long football field.

Let player sprints x yards with the speed = 2 yards per second.

So time taken to cover x yards player will take time = \frac{x}{2} seconds

Now rest distance (100 - x) yards when covered with the speed of 4 yards per second, so time taken to cover this distance = \frac{Distance}{Speed}

= \frac{100-x}{4} seconds

Now total time taken by the player can be represented by the equation

\frac{x}{2}+\frac{100-x}{4}=40

Now we can solve this equation for the value of x.

\frac{2x+100-x}{4}=40

x + 100 = 40×4

x + 100 = 160

x = 160 - 100 = 60 yards

And length of the second part will be = 100 - 60 = 40 yards

Now the fraction of the field covered by the player in second sprint will be

= \frac{40}{100}

= \frac{2}{5} or 40%

Therefore, \frac{2}{5}th part of the field was covered in second sprint.

5 0
3 years ago
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