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2 years ago

PLEASE HELP WILL GIVE BRAINLY AND 5.0 RATING

Mathematics

1 answer:

Natalka [10]2 years ago

7 0

Answer:

Step-by-step explanation:

(-2,5)

-3 on y-axis

(-2,2)

+2 on x-axis

(0,2)

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Tom jogged from 10.30 to 12.15. He traveled 7 miles. What was his average rate of jogging?

maks197457 [2]

From 1030 to 1100 is 30 minutes
From 1100 to 1200 is 60 minutes
From 1200 to 1215 is 15 minutes
Total Time is 105 minutes
105/7=15
Tom completes 1 mile every 15 minutes

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3 years ago

You need $3,000 to buy a new stereo for your car. If you have $1,200 to invest at

DiKsa [7]

Answer:

15 years and 8 months

Step-by-step explanation:

I used the formula for compound interest as shown below and solved for the unknown, time.

Since our interest is compounded annually (So once a year) our m value is 1.

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2 years ago

Solve Systems of Equations Algebracallyy=x+2 y=-3x

Sphinxa [80]

y = x + 2

y = -3x

Do y = -3x in y = x + 2

y = x + 2

-3x = x + 2

-3x - x = 2

-4x = 2

x = -2/4

x = -1/2

Now put x = -1/2 in y = -3x

y = -3x

y = -3.(-1/2)

y = 3/2

5 0

3 years ago

The diagram shows a sector of a circle radius 10cm

Yuki888 [10]

Answer:

$\text{Perimeter of sector = 43.6 (Rounded to the nearest tenth.)}$

Step-by-step explanation:

$\text{Perimeter of the sector = Radius + arc length + radius}$

$\text{Radius = 10 cm}$

$\text{Perimeter of the sector = 10 + arc length + 10 }$

$\text{Perimeter of the sector = 20 + arc length }$

$\text{arc length = (sector angle/ 360 ) * perimeter of the circle}$

$\text{Perimeter of a circle }=2\pi r$

$\text{sector angle} = 135$

$\text{arc length} = (135/ 360 ) * 2\pi (10)$

$= (3/8) 20\pi$

$= 15\pi /2$

$\text{Using }\pi =3.14$

$= 15 * 3.14 /2$

$= 23.55$

$\text{Perimeter of the sector = 20 + 23.55}$

$= 43.55$

$= 43.6 cm$

$\text{perimeter of the sector = 43.6 cm }$

8 0

2 years ago

Solving fp<br> Algebra 2<br> I do not understand how to solve this, could someone please explain?

Olenka [21]

$f(x) = \sqrt{x + 2}$
What's The Fixed Point? Well, Let's Assume X=0, this is our point.

$f(0) = \sqrt{0 + 2} \\ f(0) = \sqrt{2} \\ this \: is \: our \: fixed \: point \: \\ (0, \sqrt{2} )$
OR

$f(p) = \sqrt{p + 2} \\ \\ or \: \sqrt{x + 2} = 0 \\ ( \sqrt{x + 2} ) ^{2} = {0}^{2} \\ x + 2 = 0 \\ x = - 2$
I'm unaware of the nature of the question, so here are some different ways a fixed point is found based on the merits of the question.

4 0

3 years ago