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Strike441 [17]
3 years ago
12

someone has asked you to find the distance between mile marker fourteen and mile marker twenty-nine while driving on an intersta

te highway. What operation will you need to find the answer (A) Addition (B) Multiplication (C) Division (D) Subtraction
Mathematics
1 answer:
galina1969 [7]3 years ago
4 0

D) Subtraction because in order to find the distance in between the marker you will need to subtract 14 from 29. So the operation will you need to find the answer is subtraction.

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Factor this expression completely.<br> 36y2 - 1
shtirl [24]

Answer:

(6y - 1)(6y + 1)

Step-by-step explanation:

To factorize completely, we may adopt the difference of 2 squares approach. The theory states that difference of two square  is the product of the difference of the number and the sum of the numbers.

Such that

a² - b² = (a - b) (a + b)

Hence 36y² - 1

= 6²y² - 1²

= (6y)² - 1²

= (6y - 1)(6y + 1)

5 0
3 years ago
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If f(x) = -(-2x - 5), what is f(-5)?
sammy [17]

Answer:

f(-5)=-5

Step-by-step explanation:

f(-5)= -((-2)(-5)-5)

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4 0
2 years ago
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What is the distance from C to D?<br><br> A. 5 units<br> B. 1 unit<br> C. 25 units<br> D. 7 units
Vera_Pavlovna [14]

Answer:

Option A

Step-by-step explanation:

According to the distance formula , for any 2 points P & Q whose coordinates are (x¹ , y¹) and (x² , y²) respectively. So ,

PQ = \sqrt{(x^{2} - x^{1} )^{2} + (y^{2} - y^{1})^{2} }

NOTE = HERE x² , y² DOESN'T MEAN THE SQUARES OF x & y . THEY ARE JUST COORDINATES.

According to the question ,

Coordinate of C = (2 , -1)

Coordinate of D = (5 , 3)

Using distance formula ,

CD = \sqrt{(5-2)^{2} + (3 -(-1))^{2} }

=> CD = \sqrt{3^{2} + 4^{2} } = \sqrt{9 + 16} = \sqrt{25} = 5

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3 years ago
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SCORPION-xisa [38]

100 it's like the absolute value

6 0
3 years ago
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Naddik [55]
ANSWER
3\pi sq.\: in.

EXPLANATION

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R:r=3:1

We can rewrite it as fractions to get,

\frac{R}{r} = \frac{3}{1}

We make R the subject to get,

R = 3r

The area of the bigger circle can be found using the formula,

Area=\pi {r}^{2}

This implies that,

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The area of the smaller circle is therefore

= \pi {( \sqrt{3}) }^{2}

= 3\pi
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3 years ago
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