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Helga [31]
2 years ago
7

Can you please help me do this?

Mathematics
1 answer:
nlexa [21]2 years ago
4 0

Answer: U=39

Step-by-step explanation:

6=u -15/4

cross multiply ✖

U-15=6*4

U-15=24

U=24+15

U=39

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Write the standard form of the equation of the parabola with the given focus or directrix and
horsena [70]

Answer:

x = 1/20 y^2

Step-by-step explanation:

This is a rightward opening  (sideways ) parabola

Directrix = x = -5

Distance formula form point x,y  to directrix and focus are equal:  

(x - -5)^2 = (x-5)^2 + y^2

x^2 + 10x + 25 = x^2 - 10x  + 25 + y^2

20x = y^2

x = 1/20 y^2

3 0
2 years ago
Identify the two prime numbers that are greater than 25 and less than 35
sdas [7]
<span>Identify the two prime numbers that are greater than 25 and less than 35.
29 and 31</span>

8 0
3 years ago
Will give brainliest. Find the length and measure of each arc. Show your work.
o-na [289]

Problem 1

The circumference is the full perimeter around the circle. You can think of it as the combination of "circle" and "fence" to get "circumference", but there might be other tricks to remember the term.

Anyways, the formula to get the circumference of a circle is

C = 2*pi*r

In this case, r = 14 is our radius so,

C = 2*pi*r

C = 2*pi*14

C = 28pi .... exact circumference in terms of pi

We only want a portion of this circumference as shown by the piece of the circle darkened. The fractional portion we want is 135/360 of a circle. Divide the angle by 360 to get the fractional portion you want. If the angle was say 180 degrees, then 180/360 = 1/2 is the fractional portion.

So we take 135/360 and multiply it by the value of C found earlier

arc length = (fractional portion)*(circumference)

arc length = (135/360)*28pi

arc length = 10.5pi

That's the exact arc length in terms of pi. Use a calculator to find that

10.5pi = 32.9867228626929

Or you could use pi = 3.14 to say

10.5*pi = 10.5*3.14 =  32.97

Which is fairly close to what the calculator is saying

-----------------

<h3>Summary:</h3>

Exact arc length = 28pi

Approximate arc length (using calculator) = 32.9867228626929

Approximate arc length (using 3.14 for pi) = 32.97

Units are in feet

When I write "using calculator", I mean using the calculator's stored version of pi, instead of pi = 3.14

======================================================

Problem 2

We could use the same idea as problem 1, or we could use the formula below. The formula is just a quick way of encapsulating what I discussed earlier.

L = arc length

x = central angle

L = (x/360)*2*pi*r

L = (150/360)*2pi*13

L = (65/6)pi .... exact arc length

L = 34.0339204138894 .... approx arc length (using calculator)

L = 34.0166666666667 .... approx arc length (using 3.14 for pi)

-----------------

<h3>Summary:</h3>

Exact arc length = (65/6)pi

Approximate arc length (using calculator) = 34.0339204138894

Approximate arc length (using 3.14 for pi) = 34.0166666666667

Units are in meters

8 0
4 years ago
Geometry What’s the answer to 5x-6y= 33 and 2x+y = -14
V125BC [204]

Answer:

X=-3 y=-8

Step-by-step explanation:

5x - 6y = 33 \\ y  =  - 14 - 2x \\

Put the second one in first

5x - 6( - 14 - 2x) = 33 \\ 5x + 84 + 12x = 33 \\ 17x = -  51 \\ x =  - 3

y =  - 14 - 2 \times ( - 3) \\ y =  - 8

7 0
3 years ago
50 POINTSSS
leonid [27]

Answer:

After 2 minutes the temperature of the hot chocolate will be 149.46 degrees Fahrenheit.

Step-by-step explanation:

We are going to use the Newton's law of cooling to solve this exercise. The Newton's law of cooling states that the amount of heat lost by a body is proportional to the difference of temperature between the body and the enviroment. We are going to use the following function :

T(t)=T_{0}+(T_{i}-T_{0}).e^{-kt}T(t)=T

0

+(T

i

−T

0

).e

−kt

Where ''T(t)'' is the temperature of the body that depends of the variable ''t'' which is time.

Where T_{0}T

0

is the temperature of the surroundings

In this case T_{0}T

0

is the temperature of the freezer

Where T_{i}T

i

is the initial temperature of the body which is cooling. In this case, T_{i}T

i

is the temperature of the hot chocolate

And where ''k'' is a constant. In this case, k=0.12k=0.12 is a data of the exercise

If we replace all the values in the equation and replacing t=2minutest=2minutes

⇒

T(2minutes)=0+(190-0).e^{-(0.12).(2)}T(2minutes)=0+(190−0).e

−(0.12).(2)

T(2minutes)=(190).(e^{-0.24})=149.46T(2minutes)=(190).(e

−0.24

)=149.46

We find that the temperature of the hot chocolate after 2 minutes is 149.46 degrees Fahrenheit

8 0
3 years ago
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