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belka [17]
3 years ago
10

Find radius from this sentence plz

Mathematics
1 answer:
Citrus2011 [14]3 years ago
3 0

Answer:

1.2 inches

Step-by-step explanation:

Area of sphere

A = 4πr²

17.6 = 4(3.14)r²

17.6 = 12.56r²

Divide both sides by 12.56

1.4012738854 = r²

Take the sqaure root of both sides

1.1837541491 = r

Rounded

r = 1.2

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AABC is reflected across the x-axis and then translated 4 units up to create AABC! What are the coordinates of the vertices of A
Greeley [361]

Answer:

d

Step-by-step explanation:

3 0
3 years ago
What is the radius of a circle that has a diameter of 34<br> inches?
Zepler [3.9K]

Answer:

17 !!!!!!!!!!!!!!!!!!!!

Step-by-step explanation:

34 ÷ 2

7 0
3 years ago
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What is the standard form of a quadratic function passing through the points (2,4) , (3,5) and (4,3)
Sladkaya [172]

The standard form of a quadratic function is y = 3/2x² + 17/2x - 7

<h3>How to determine the quadratic equation?</h3>

From the question, the points on the quadratic equation are given as

(2,4), (3,5) and (4,3)

The standard form of a quadratic equation is represented as

y = ax² + bx + c

So, we have

Point (2,4):

4a + 2b + c = 4

Point (3,5):

9a + 3b + c = 5

Point (4,3):

16a + 4b + c = 3

Solving using an online calculator. we have

a = 3/2, b = 17/2 and c = -7

Substitute a = 3/2, b = 17/2 and c = -7 in y = ax² + bx + c

y = 3/2x² + 17/2x - 7

Hence, the equation is y = 3/2x² + 17/2x - 7

Read more about quadratic equation at

brainly.com/question/1214333

#SPJ1

5 0
9 months ago
Let v1=(3,5) and v2=(-4,7).
Nat2105 [25]
PART A

The given vectors are,

v_1 = \: < \: 3 , \: 5 \: >

v_2 = \: < \: - 4 , \: 7 \: >

The magnitude of the vector

v= \: < \: x , \: y \: > \:

is given by:

|v| = \sqrt{ {x}^{2} + {y}^{2} }

This implies that,

|v_1| = \sqrt{ {3}^{2} + {5}^{2} }

|v_1| = \sqrt{9 + 25}

|v_1| = \sqrt{34}

|v_2| = \sqrt{ {( - 4)}^{2} + {7}^{2} }

|v_2| = \sqrt{ 16+ 49}

|v_2| = \sqrt{65}

PART B

To find the unit vector in the direction of a given vector, we divide by the magnitude of that vector.

^{ - } _{v_1} = \: < \: \frac{3}{ \sqrt{34} } , \: \frac{5}{ \sqrt{34} } \: >

Rationalize the denominator.

^{ - } _{v_1} = \: < \: \frac{3\sqrt{34}}{ 34 } , \: \frac{5\sqrt{34}}{ 34 } \: >

Also,

^{ - } _{v_2} = \: < \: \frac{ - 4}{ \sqrt{65} } , \: \frac{7}{ \sqrt{65} } \: >

^{ - } _{v_2} = \: < \: \frac{ - 4\sqrt{65}}{ 65 } , \: \frac{7\sqrt{65}}{ 65 } \: >

PART C

The sketch of the given vectors as well as their unit vectors are shown in the attachment.

3 0
2 years ago
Huh heyrhd Rhfhgrysamwoaj
kow [346]

Answer:

what ? You Must Be Real Bored Kid

8 0
3 years ago
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