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Gelneren [198K]

3 years ago

7

In Beijing in 2008, the Women's 3,000-meter Steeplechase became an Olympic event. What is this distance in kilometers? Quick ple

ase

1 answer:

laiz [17]3 years ago

7 0

Answer:

3 km

Step-by-step explanation:

3000/1000=3

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Does (-1, 2), (1, 1),( 1 -1), (2,1), (4,2) represent y as a function

Paha777 [63]

No. A function maps one input to exactly one output. The given relation maps 1 to both 1 and -1, as indicated by the pairs (1, 1) and (1, -1).

8 0

3 years ago

PLEASE HELP I GIVE 29 POINTS TO WHOEVER GETS THIS RIGHT

-Dominant- [34]

Direction: Isolate the variable by dividing each side by factors that don't contain the variable.

Here are the answers for all 15 questions:

1) x < - 2

2) x < 4

3) x > - 15/7

4) y < 28/5

5) x > 6

6) y > - 10/9

7) 0 < 333

8) x > 2

9) x > - 10

10) y > - 3

11) y > - 4

12) x > 3

13) x < 9/8

14) x > - 2

15) h < 7/2

4 0

3 years ago

Read 2 more answers

3. if kx² + 2x + k = -kx have equal roots, find the values of k.<br>

crimeas [40]

Answer:

k = - $\frac{2}{3}$ , k = 2

Step-by-step explanation:

Using the discriminant Δ = b² - 4ac

The condition for equal roots is b² - 4ac = 0

Given

kx² + 2x + k = - kx ( add kx to both sides )

kx² + 2x + kx + k = 0 , that is

kx² + (2 + k)x + k = 0 ← in standard form

with a = k, b = 2 + k and c = k , thus

(2 + k)² - 4k² = 0 ← expand and simplify left side

4 + 4k + k² - 4k² = 0

- 3k² + 4k + 4 = 0 ( multiply through by - 1 )

3k² - 4k - 4 = 0 ← in standard form

(3k + 2)(k - 2) = 0 ← in factored form

Equate each factor to zero and solve for k

3k + 2 = 0 ⇒ 3k = - 2 ⇒ k = - $\frac{2}{3}$

k - 2 = 0 ⇒ k = 2

5 0

4 years ago

A. Evaluate ∫20 tan 2x sec^2 2x dx using the substitution u = tan 2x.

irakobra [83]

Answer:

The integral is equal to $5\sec^2(2x)+C$ for an arbitrary constant C.

Step-by-step explanation:

a) If $u=\tan(2x)$ then $du=2\sec^2(2x)dx$ so the integral becomes $\int 20\tan(2x)\sec^2(2x)dx=\int 10\tan(2x) (2\sec^2(2x))dx=\int 10udu=\frac{u^2}{2}+C=10(\int udu)=10(\frac{u^2}{2}+C)=5\tan^2(2x)+C$ . (the constant of integration is actually 5C, but this doesn't affect the result when taking derivatives, so we still denote it by C)

b) In this case $u=\sec(2x)$ hence $du=2\tan(2x)\sec(2x)dx$ . We rewrite the integral as $\int 20\tan(2x)\sec^2(2x)dx=\int 10\sec(2x) (2\tan(2x)\sec(2x))dx=\int 10udu=5\frac{u^2}{2}+C=5\sec^2(2x)+C$ .

c) We use the trigonometric identity $\tan(2x)^2+1=\sec(2x)^2$ is part b). The value of the integral is $5\sec^2(2x)+C=5(\tan^2(2x)+1)+C=5\tan^2(2x)+5+C=5\tan^2(2x)+C$ . which coincides with part a)

Note that we just replaced 5+C by C. This is because we are asked for an indefinite integral. Each value of C defines a unique antiderivative, but we are not interested in specific values of C as this integral is the family of all antiderivatives. Part a) and b) don't coincide for specific values of C (they would if we were working with a definite integral), but they do represent the same family of functions.

3 0

3 years ago

Literal Equations: 8c + 6j = 5p, for c

bixtya [17]

8c + 6j = 5p Subtract 6j from both sides
8c = 5p - 6j Divide both sides by 8
c = $\frac{5p - 6j}{8}$

3 0

3 years ago