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m_a_m_a [10]
3 years ago
14

The polygons to the right are similar. Find the values of each variable! ( simplify ur answer)

Mathematics
1 answer:
Sever21 [200]3 years ago
4 0

Answer:

hi! i'm so sorry about your headache, hope it gets better !

still need help?

Step-by-step explanation:

You might be interested in
Solve 9,159 divided by 7
aniked [119]

Answer:

(9159 / 7 = 1308.429)

Step-by-step explanation:

Simply multiply the last digit by 2 and then subtract the product from the remaining digits.

If that difference is divisible by 7, then 9159 is divisible by 7.

The last digit in 9159 is 9 and the remaining digits are 915. Thus, the math to determine if 9159 is divisible by 7 using our alternate method is:

915 - (9 x 2) = 897

Since 897 is not divisible by 7, 9159 is also not divisible by 7.

Therefore, the answer to "Is 9159 Divisible By 7?" is no.

(9159 / 7 = 1308.429)

5 0
2 years ago
Read 2 more answers
When we graph a line using slope and y-intercept, the first point that you graph is
k0ka [10]

Answer:

y=mx+b is the formula starting with b say it is 4 you put a dot on 4  on the y axis m is 1/2 you go up 1 right 2 on the 4 points then do the same on the next point you get then you get your line. hope this helps  

4 0
2 years ago
PLEASE ANSWER!!
Elodia [21]
The answer is A I believe
6 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
A log raft is moving downstream at a speed of 3 km/h. To insure that there are no river jams a supervisor travels along the raft
skad [1K]

Answer:

2000 meters

Step-by-step explanation:

When the motorboat is moving downstream, its speed relative to the ground is 18 km/h, so its speed relative to the raft is 15 km/h.

When the motorboat is moving upstream, its speed relative to the ground is 12 km/h, so its speed relative to the raft is again 15 km/h.

Converted to m/min, the relative speed is:

15 km/h × (1000 m/km) × (1 h / 60 min) = 250 m/min

It takes the motorboat 16 minutes to travel to the front of the raft and back.  Since the speed is the same in both directions, the motorboat takes 8 minutes to travel the length of the raft.

So the length of the raft is 250×8 = 2000 meters.

8 0
3 years ago
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