Ask question

Ask question

All categories

3 years ago

15

-2|3-2r|<-34 how do you solve this?

1 answer:

Alexxandr [17]3 years ago

4 0

$|a| -b\\\\|a|>b\iff a>b\ or\ a$

$-2|3-2r|34\qquad\text{divide both sides by 2}\\\\|3-2r|>17\iff3-2r>17\ or\ 3-2r14\ or\ -2r$

You might be interested in

The length of a sticky note measures 77 mm express this length in meters

katen-ka-za [31]

If a sticky note measures 77mm then when converted into meters it is 0.077

8 0

3 years ago

Read 2 more answers

How to find the bin size

labwork [276]

To find the area of a rectangular prism multiply length by width by height.

If the problem has a variable take the two other dimensions and multiply them together. Then take that number and subtract it from the total to find the number that the variable equals.

3 0

3 years ago

These two polygons are similar.

Jet001 [13]

Answer:

$\huge\boxed{z=3}$

Step-by-step explanation:

If two polygons are similar, then corresponding sides are in proportion.

The corresponding sides:

4 → x

y → 15

3 → w

2 → 6

z → 9

therefore:

$\dfrac{z}{9}=\dfrac{2}{6}$ <em>cross multiply</em>

$(z)(6)=(9)(2)$

$6z=18$ <em>divide both sides by 6</em>

$z=3$

4 0

3 years ago

The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 -57, where t is the time

cupoosta [38]

A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?

Answer:

$1\leq t\geq \sqrt{2}$

Step-by-step explanation:

Given:

A stuntman jumping off a 20-m-high building is modeled by the equation

$h =20-5t^{2}$ -----------(1)

A high-speed camera is ready to making film between 15 m and 10 m above the ground

when the stuntman is 15m above the ground.

height $h = 15m$

Put height value in equation 1

$15 =20-5t^{2}$

$5t^{2} =20-15$

$5t^{2} =5$

$t^{2} =1$

$t =\pm1$

We know that the time is always positive, therefore $t=1$

when the stuntman is 10m above the ground.

height $h = 10m$

Put height value in equation 1

$10 =20-5t^{2}$

$5t^{2} =20-10$

$5t^{2} =10$

$t^{2} =\frac{10}{5}$

$t^{2} =2$

$t=\pm\sqrt{2}$

$t=\sqrt{2}$

Therefore ,time interval of camera film him is $1\leq t\geq \sqrt{2}$

7 0

3 years ago

Solve for x. Round to the nearest tenth, if necessary.

Ray Of Light [21]

Answer:

22.6

Step-by-step explanation:

because in this one you are using cos , adjacent/hypotenuse

24cos20

x = 22.552622899

3 0

2 years ago