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rusak2 [61]
3 years ago
11

HELP

Mathematics
2 answers:
strojnjashka [21]3 years ago
8 0

Answer: Answer is f(x) = 25x4 - 7x2 + x + 4

Step-by-step explanation:

Leokris [45]3 years ago
6 0

Answer:

d on edg

Step-by-step explanation:

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jenyasd209 [6]

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20

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I NEED HELP ASAP!!People that now the Answer help me please
lilavasa [31]

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xq xw

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Randy wants to run half a marathon. Half a marathon is 13.1 miles. if she wants to run the race in 2 hours what rate(in minutes
IRINA_888 [86]

Answer:

9.16 minutes per mile

At 10 miles, she is short of her intended rate

Step-by-step explanation:

2 hours means 60 * 2 = 120 minutes

Randy aims to run 13.1 miles in 120 minutes. So to get the rate [min/mi], we must divide total minutes by total miles:

120/13.1 = 9.16 minutes per mile

Moreover, she realized at 10 miles, she has taken 90 minutes. So her rate at that point [min/mi] is:

90/10 = 9 min/mi

Well, Randy is short of her intended rate

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What is AB?<br> Please hurry answer
Vera_Pavlovna [14]

Answer:

The Answer is 47

Step-by-step explanation:

Because you do A squared plus B squared equals C squared and Squared is just multiplying both by two.

7 0
3 years ago
CAN SONEONE PLEASE HELP ME WITH MY MATH PLEASEEEE!!!!!​
Marina86 [1]

Answer:

√190

Step-by-step explanation:

In the figure , there are 2 right angled triangles with a common perpendicular & both the triangles combine to form a new right angled triangle.

Let the triangle with 9 as base be T¹ & Let the triangle with base 10 be T². Let the triangle formed by T¹ & T² be T³.

In T² ,

Hypotenuse = y

Base = 10

According to Pythagorean Theorem ,

(Hypotenuse)² = (Base)² + (Perpendicular)²

Hence, (Perpendicular)² = y^2 - 10^2 = y^2 - 100

In T¹ ,

Perpendicular = \sqrt{y^2 - 100}   (∵ Both T¹ & T² have common perpendicular)

⇒(Perpendicular)² = y^2 - 100

Base = 9

⇒ (Base)² = 9²

Hypotenuse =

Using Pythagorean Theorem ,

(Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ (Hypotenuse)² = y^2 - 100 + 9^2 .............................................eqn.2

Now in T³ ,

Base = y

⇒ (Base)² = y²

Perpendicular = \sqrt{(y^2 - 100) + 9^2} (∵Perpendicular of T³ = Hypotenuse of T²)

⇒ (Perpendicular)² = (\sqrt{(y^2 - 100) + 9^2})^2= (y^2 - 100) + 81 = y^2 - 19

Hypotenuse = 9 + 10 = 19

Using Pythagorean Theorem ,

(Hypotenuse)² = (Perpendicular)² + (Base)²

=> 19^2 = y^2 - 19 + y^2\\\\=> 2y^2 = 19^2 + 19 = 19(19 + 1) = 19*20\\\\=> y^2 = \frac{19*20}{2} = 19*10 = 190\\ \\=> y =\sqrt{190}

3 0
3 years ago
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