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Ghella [55]
3 years ago
5

The given triangles are similar. Solve for x.

Mathematics
1 answer:
Maru [420]3 years ago
8 0
The first one is 21.93 rounded up to 22.
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PLEASE HURRY I NEED HELP :( The area of a regular hexagon inscribed in a circle with radius 2 is
NISA [10]

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Step-by-step explanation:

4 I think not sure

8 0
3 years ago
The figure shows two parallel lines PQ and ST cut by the transversals PT and QS.
wariber [46]


answer is 

<span>Triangle PQR is similar to triangle TSR because measure of angle 3 equals measure of angle 4 and measure of angle 1 equals measure of angle 5</span>
3 0
3 years ago
Read 2 more answers
In 2000, the circulation of a local newspaper was 3,250. In 2001, its circulation was 3,640. In 2002, the circulation was 4,100.
vodka [1.7K]

Percent increase from 2000 to 2001 is 12% and from 2001 to 2002 is 12.6%. Thus period of 2001 to 2002 has higher increase percentage.

<u>Solution:</u>

Given that,

Total circulation of local newspaper in 2000 = 3250

Total circulation of local newspaper in 2001 = 3640

Total circulation of local newspaper in 2002 = 4100

<em><u>Finding percent of increase in the newspaper’s circulation from 2000 to 2001:</u></em>

\text { Percent increase from } 2000 \text { to } 2001=\frac{\text { change in circulation from 2001 - 2000 }}{\text { circulation in } 2000} \times 100

\begin{array}{l}{=\frac{3640-3250}{3250} \times 100} \\\\ {=\frac{390}{3250} \times 100=12 \%}\end{array}

<em><u>Finding percent of increase in the newspaper’s circulation from 2001 to 2002:</u></em>

\begin{array}{l}{\text { Percent increase from } 2001 \text { to } 2002=\frac{\text { change in circulation }}{\text {circulation in } 2001} \times 100} \\\\ {=\frac{4100-3640}{3640} \times 100} \\\\ {=\frac{460}{3640} \times 100=12.6 \%}\end{array}

As we can see 12% < 12.6%

So period of 2001 to 2002 has higher increase percentage.

7 0
3 years ago
Directions: Find the value of x.​
Umnica [9.8K]
I think it would be 12.207… use the Pythagorean theorem (a^2+b^2=c^2) so 10^2+7^2
100+49=149
Then take square root of 149 to undo the the squaring of c and it should be around 12.207 :)
4 0
2 years ago
Read 2 more answers
Performance task A parade route must start And and at the intersections shown on the map. The city requires that the total dista
Flauer [41]

Answer:

  A: the proposed route is 3.09 miles, so exceeds the city's limit

Step-by-step explanation:

The length of the route in grid squares can be found using the Pythagorean theorem on the two parts of the route. Let 'a' represent the length of the route to the park from the start, and 'b' represent the route length from the park to the finish. Then we have (in grid squares) ...

  a^2 = (12-6)^2 +3^2 = 45

  a = √45 = 3√5

and

  b^2 = (6 -2)^2 +4^2 = 32

  b = √32 = 4√2

Then the total length, in grid squares, is ...

  3√5 + 4√2 = 6.7082 +5.6569 = 12.3651

If each grid square is 1/4 mile, then 12.3651 grid squares is about ...

  (12.3651 squares) · (1/4 mile/square) = 3.0913 miles

The proposed route is too long by 0.09 miles.

5 0
3 years ago
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