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hram777 [196]
3 years ago
6

in order to qualify for a role in a play, an actor must be taller than 64 inches but shorter than 68 inches. the inequality 64 &

lt; x < 68, where x represents height, can be used to represent the height range. which is another way of writing the inequality?x > 64 and x < 68 x > 64 or x < 68 x < 64 and x < 68 x < 64 or x < 68
Mathematics
2 answers:
fenix001 [56]3 years ago
8 0

Answer:

The correct answer is A, x > 64 and x < 68.

Step-by-step explanation:

This happens because the actor <u>has to be</u> taller than 64 inches, and <u>less than</u> 68 inches. We know we are using greater than or less than signs and not greater than or equal to or less than or equal to because it says <u>less than.</u>

I know this was kind of a weird "Step-by-step" explanation but sorry in a rush, just trying to help out of fellow classmates.

Irina18 [472]3 years ago
3 0
X > 64 and x < 68 <== ur answer
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Write the first 6 terms of the arithmetic sequence whose first term is 8 and has a common difference of 2
soldier1979 [14.2K]

Answer: The first 6 terms are = 8, 10, 12,14,16,18

Step-by-step explanation:

The NTH term of an Arithmetic Sequence is given as

an = a1 + (n - 1 ) d

where a1 = First term  given as 8 and

d=  common difference given as 2

Therefore  We have that

the first term

an = a1 + (n - 1 ) d = 8+(1-1) 2

a1= 8

second term=

an = a1 + (n - 1 ) d= a2= 8 + (2-1) 2

= 8+ 2(1) = 10

3rd term

an = a1 + (n - 1 ) d= a3= 8 + (3-1) 2

= 8+ 2(2)= 8 + 4=12

4th term

an = a1 + (n - 1 ) d= a4= 8 + (4-1) 2

= 8+ 2(3)= 8+6=14

5th term

an = a1 + (n - 1 ) d= a5= 8 + (5-1) 2

= 8+ 2(4)=8+ 8=16

6th term

an = a1 + (n - 1 ) d= a6= 8 + (6-1) 2

= 8+ 2(5)=8 +10 =18

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3 years ago
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Annie is framing a photo with a length of 6 inches and a width of 4 inches. The distance from the edge of the photo to the edge
Ede4ka [16]

Answer:

Part a) The quadratic function is 4x^{2} +20x-39=0

Part b) The value of x is 1.5\ in

Part c) The photo and frame together are 7\ in wide

Step-by-step explanation:

Part a) Write a quadratic function to find the distance from the edge of the photo to the edge of the frame

Let

x----> the distance from the edge of the photo to the edge of the frame

we know that

(6+2x)(4+2x)=63\\24+12x+8x+4x^{2}=63\\ 4x^{2} +20x+24-63=0\\4x^{2} +20x-39=0

Part b) What is the value of x?

Solve the quadratic equation 4x^{2} +20x-39=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem

we have

4x^{2} +20x-39=0

so

a=4\\b=20\\c=-39

substitute in the formula

x=\frac{-20(+/-)\sqrt{20^{2}-4(4)(-39)}} {2(4)}

x=\frac{-20(+/-)\sqrt{1,024}} {8}

x=\frac{-20(+/-)32} {8}

x=\frac{-20(+)32} {8}=1.5\ in  -----> the solution

x=\frac{-20(-)32} {8}=-6.5\ in

Part c) How wide are the photo and frame together?

(4+2x)=4+2(1.5)=7\ in

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3 years ago
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Alex Ar [27]

Answer:

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Step-by-step explanation:

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2 years ago
Use the Commutative Property to write an expression equivalent to –5d + 1/2.
Jet001 [13]

Answer: 12-5d

because all you do is flip the expression

7 0
2 years ago
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