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charle [14.2K]
3 years ago
12

What is 7 yd 2 ft = ___________ft PLEASE EXPLAIN worth 19 points

Mathematics
2 answers:
ElenaW [278]3 years ago
8 0
23 ft.  There are 3 feet in each yard so 7 X 3 = 21 + 2 = 23 ft.
Aleonysh [2.5K]3 years ago
5 0
1 yd=3 ft so 7 yd *3=21 ft+2ft=23 ft

Hope that helps.  Feel free to ask any questions
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Please help me out <br> Thank you!!
Roman55 [17]
Y=2x

2x3 = 6
2x5 = 10
Etc.
4 0
3 years ago
Read 2 more answers
How do I solve this ?
eduard

Look at the picture.

ΔZYX and ΔDYC are similar. Therefore the lengths of sides are in proportion:

\dfrac{DC}{ZX}=\dfrac{DY}{ZY}

DC = x

ZX = 16

DY = a

ZY = a + a = 2a

Substitute:

\dfrac{x}{16}=\dfrac{a}{2a}

\dfrac{x}{16}=\dfrac{1}{2}       <em>multiply both sides by 16</em>

\boxed{x=8}

<h3>Answer: CD = 8</h3>

6 0
4 years ago
Solve the inequality 2x2 + 10x &lt; –8
BigorU [14]

Answer:

-4<x<-1

Step-by-step explanation:

To solve the problem, we divide the whole expression by 2:

2x^2 + 10x < –8 → x^2 + 5x < –4

→  x^2 + 5x + 4 < 0

Factorizing

→ (x+4)(x+1) < 0

The expression is ONLY negative when:

x>-4 and x<-1

Therefore, the solution is:

-4<x<-1

5 0
3 years ago
Integrate cosx/sqrt(1+cosx)dx
Step2247 [10]
<span>Take the integral: integral (cos(x))/sqrt(cos(x)+1) dx
For the integrand (cos(x))/sqrt(1+cos(x)), substitute u = 1+cos(x) and du = -sin(x) dx: = integral (u-1)/(sqrt(2-u) u) du
For the integrand (-1+u)/(sqrt(2-u) u), substitute s = sqrt(2-u) and ds = -1/(2 sqrt(2-u)) du: = integral -(2 (1-s^2))/(2-s^2) ds
Factor out constants: = -2 integral (1-s^2)/(2-s^2) ds
For the integrand (1-s^2)/(2-s^2), cancel common terms in the numerator and denominator: = -2 integral (s^2-1)/(s^2-2) ds
For the integrand (-1+s^2)/(-2+s^2), do long division: = -2 integral (1/(s^2-2)+1) ds
Integrate the sum term by term: = -2 integral 1/(s^2-2) ds-2 integral 1 ds
Factor -2 from the denominator: = -2 integral -1/(2 (1-s^2/2)) ds-2 integral 1 ds
Factor out constants: = integral 1/(1-s^2/2) ds-2 integral 1 ds
For the integrand 1/(1-s^2/2), substitute p = s/sqrt(2) and dp = 1/sqrt(2) ds: = sqrt(2) integral 1/(1-p^2) dp-2 integral 1 ds
The integral of 1/(1-p^2) is tanh^(-1)(p): = sqrt(2) tanh^(-1)(p)-2 integral 1 ds
The integral of 1 is s: = sqrt(2) tanh^(-1)(p)-2 s+constant
Substitute back for p = s/sqrt(2): = sqrt(2) tanh^(-1)(s/sqrt(2))-2 s+constant
Substitute back for s = sqrt(2-u): = sqrt(2) tanh^(-1)(sqrt(1-u/2))-2 sqrt(2-u)+constant
Substitute back for u = 1+cos(x): = sqrt(2) tanh^(-1)(sqrt(sin^2(x/2)))-2 sqrt(1-cos(x))+constant
Factor the answer a different way: = sqrt(1-cos(x)) (csc(x/2) tanh^(-1)(sin(x/2))-2)+constant
Which is equivalent for restricted x values to:
Answer: | | = (2 cos(x/2) (2 sin(x/2)+log(cos(x/4)-sin(x/4))-log(sin(x/4)+cos(x/4))))/sqrt(cos(x)+1)+constant</span>
3 0
3 years ago
Calculate the distance between the points L = (1, -1) and J=(4,-9) in the coordinate plane.
Blizzard [7]

Answer:

8.54

Step-by-step explanation:

distance formula is d = \sqrt{(x_2 - x_1)^2 + (y_2-y_1)^2}

plug in the coordinates the answer is 8.54

7 0
3 years ago
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