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nikklg [1K]
3 years ago
15

Please help me :) thank u.

Mathematics
2 answers:
kap26 [50]3 years ago
7 0
Since the figures are congruent, the sides and angles are the same. So, the answer is:
IJ ~= TU
GJ ~=VU
miss Akunina [59]3 years ago
5 0

Answer:

IJ corresponds to TU

GJ corresponds to VU

Step-by-step explanation:

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jenny went to the mall with $120 in her wallet.she spent $25.65 on a shirt, $7.98 on a pair of socks, $16.23 on a cd , and $4.58
Effectus [21]
Total money=$120
on shirt=$25.65
on socks=$7.98
on a cd=$16.23
on snack=$4.58
so total money left=total money-total money spent 
=$120-($25.65+$7.98+$16.23+$4.58)
=$120-$54.44
so total money left=$65.56
4 0
3 years ago
Read 2 more answers
Lucy is a dress maker. She sews 4/7 of a dress in 3/4 hour. Lucy sews at a constant rate. At this rate, how many dresses does Lu
MAVERICK [17]

Answer:

16/21 dresses

Step-by-step explanation:

3/4 · k hours: 4/7 · k dresses for any constant k

If we want 1 hour, then k will need to be 4/3

4/3 × 4/3 hours: 4/7 × 4/3

1 hour = 16/21 dresses

7 0
3 years ago
If Esteban counted the number of steps it took him to school .And he counted 1,138 steps how many steps does he take walking to
Semmy [17]

Answer:

1,138 steps

Step-by-step explanation:

According to the scenario, computation of the given data are as follows,

Steps counted to school = 1,138

So, we can calculate the steps he take walking to school by using following formula,

Steps take walking to school = Steps he counted to school

So, Steps take walking to school = 1,138 steps

Hence, Esteban take 1,138 steps while walking to school.

7 0
3 years ago
A baseball diamond is a square that is 90feet on each side. What is the distance a catcher has to throw the ball from home to se
Crank

Answer:

well, since a real baseball diamond is 90 feet on each side then the distance from homebase to second is 127 feet  3 3/8 inches

hope this helps :D


4 0
3 years ago
Read 2 more answers
Particle 1 of charge q1 �� ��5.00q and particle 2 of charge q2 �� ��2.00q are fixed to an x axis. (a) as a multiple of distance
Naya [18.7K]

<span>Assuming that the particle is the 3rd particle, we know that it’s location must be beyond q2; it cannot be between q1 and q2 since both fields point the similar way in the between region (due to attraction). Choosing an arbitrary value of 1 for L, we get </span>

<span>
k q1 / d^2 = - k q2 / (d-1)^2 </span>

Rearranging to calculate for d:

<span> (d-1)^2/d^2 = -q2/q1 = 0.4 </span><span>
<span> d^2-2d+1 = 0.4d^2 </span>
0.6d^2-2d+1 = 0  
d = 2.72075922005613 
d = 0.612574113277207 </span>

<span>
We pick the value that is > q2 hence,</span>

d = 2.72075922005613*L

<span>d = 2.72*L</span>

3 0
3 years ago
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