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3 years ago

11

How much longer was the world record than Sheila first jump

1 answer:

Nadusha1986 [10]3 years ago

8 0

The answer would be 58

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M +4- 2m<br> add like terms

Yuri [45]

Answer:

-m+4

Step-by-step explanation:

$m+4-2m\\\\=(m-2m)+4\\\\=(-1m)+4\\\\=-1m+4\\\\=-m+4$

8 0

3 years ago

Read 2 more answers

Someone help me please<br><br> Find the length of side AC.

kykrilka [37]

Answer:

D

Step-by-step explanation:

a sqrt 2 = 5

a = 5 / (sqrt 2)

a=5sqrt2 / 2

6 0

3 years ago

If RX=4 and XS=9, then XT=<br> And how do you get it?

konstantin123 [22]

Answer:

XT=6 units

Step-by-step explanation:

The picture of the question is the attached figure

step 1

In the right triangle RST

Applying the Pythagorean theorem

$RS^2=RT^2+TS^2$

we have

$RS=RX+XS=4+9=13\ units$ ---> by segment addition postulate

substitute

$RT^2+TS^2=169$ ----> equation A

step 2

In the right triangle RTX

Applying the Pythagorean theorem

$RT^2=RX^2+XT^2$

we have

$RX=4\ units$

substitute

$RT^2=4^2+XT^2$

$RT^2=16+XT^2$

$XT^2=RT^2-16$ ----> equation B

step 3

In the right triangle XTS

Applying the Pythagorean theorem

$TS^2=XS^2+XT^2$

we have

$XS=9\ units$

substitute

$TS^2=9^2+XT^2$

$TS^2=81+XT^2$

$XT^2=TS^2-81$ ----> equation C

step 4

equate equation B and equation C

$TS^2-81=RT^2-16$

$TS^2-RT^2=81-16$

$TS^2-RT^2=65$ ----> equation D

step 5

Solve the system

$RT^2+TS^2=169$ ----> equation A

$TS^2-RT^2=65$ ----> equation D

Solve by elimination

Adds equation A and equation D

$RT^2+TS^2=169\\TS^2-RT^2=65\\---------\\TS^2+TS^2=169+65\\2TS^2=234\\TS^2=117$

Find the value of RT^2

$RT^2+117=169\\RT^2=52$

step 6

Find the value of XT

equation C

$XT^2=117-81\\XT^2=36\\XT=6\ units$

7 0

3 years ago

99 POINTS!!!! PLEASE HELP!! WILL MARK BRAINLIST!!

RUDIKE [14]

The answer is reduction. This can be found by the size and number differences in the two quadrilaterals.

8 0

4 years ago

Find all solutions in the interval [0, 2π). <br> 2 sin2x = sin x

tester [92]

Answer:

The solutions on the given interval are :

$0$

$\pi$

$\cos^{-1}(\frac{1}{4})$

$-\cos^{-1}(\frac{1}{4})+2\pi$

Step-by-step explanation:

We will need the double angle identity $\sin(2x)=2\sin(x)\cos(x)$ .

Let's begin:

$2\sin(2x)=\sin(x)$

Use double angle identity mentioned on left hand side:

$2\cdot 2\sin(x)\cos(x)=\sin(x)$

Simplify a little bit on left side:

$4\sin(x)\cos(x)=\sin(x)$

Subtract $\sin(x)$ on both sides:

$4\sin(x)\cos(x)-\sin(x)=0$

Factor left hand side:

$\sin(x)[4\cos(x)-1]=0$

Set both factors equal to 0 because at least of them has to be 0 in order for the equation to be true:

$\sin(x)=0 \text{ or } 4\cos(x)-1=0$

The first is easy what angles $\theta$ are $y$ -coordinates on the unit circle 0. That happens at $0$ and $\pi$ on the given range of $x$ (this $x$ is not be confused with the $x$ -coordinate).

Now let's look at the second equation:

$4\cos(x)-1=0$

Isolate $\cos(x)$ .

Add 1 on both sides:

$4 \cos(x)=1$

Divide both sides by 4:

$\cos(x)=\frac{1}{4}$

This is not as easy as finding on the unit circle.

We know $\arccos( )$ will render us a value between $0$ and $2\pi$ .

So one solution on the given interval for x is $x=\cos^{-1}(\frac{1}{4})$ .

We know cosine function is even.

So an equivalent equation is:

$\cos(-x)=\frac{1}{4}$

Apply $\cos^{-1}$ to both sides:

$-x=\cos^{-1}(\frac{1}{4})$

Multiply both sides by -1:

$x=-\cos^{-1}(\frac{1}{4})$

This going to be negative in the 4th quadrant but if we wrap around the unit circle, $2\pi$ , we will get an answer between $0$ and $2\pi$ .

So the solutions on the given interval are :

$0$

$\pi$

$\cos^{-1}(\frac{1}{4})$

$-\cos^{-1}(\frac{1}{4})+2\pi$

5 0

3 years ago