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tatyana61 [14]
4 years ago
12

Please answer the attached question! First CORRECT answer gets Brainliest!

Mathematics
1 answer:
Serhud [2]4 years ago
4 0

the answer is D

y=-4(2)^2=-16

hope this helps you good luck

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In XYZ, XY=XZ<br> Find the length of XY of XYZ if XY=2a, YZ=3a+1, andXZ=5a-12
lorasvet [3.4K]

Answer: XY = 8

Step-by-step explanation:

The diagram of the triangle XYZ is shown in the attached photo.

XY=XZ. This means that two sides of the triangle are equal. Two angles are also equal. It means that the triangle XYZ is an isosceles triangle.

Since XY=XZ, then

2a = 5a - 12

Subtracting 2a from both sides of the equation,

2a - 5a = 5a - 12 - 5a

-3a = -12

a = -12/-3

a = 4

Therefore, the length of XY is 2a = 2×4 = 8

5 0
3 years ago
Pls help !!! :))<br> 17. If m_1 = 98º and mz2 = 19º,<br> what is mz3?
Feliz [49]

Answer:

angle 3 = 79 degree

Step-by-step explanation:

given:

angle 1 = 98 degree

angle 2 : 19 degree

angle 3 =?

angle 2 + angle 3 =angle 1 (sum of two interior angle opposite angles is equal to the exterior angle formed)

19 + angle 3 = 98

angle 3 = 98 - 19

angle 3 = 79 degree

4 0
3 years ago
Read 2 more answers
A coin is tossed 1000 times and heads showed up 625 and tails 375
Mnenie [13.5K]

Answer:

Heads: 5/8

Tails: 3/8

Step-by-step explanation:

The probability of getting tails is 375/1000 OR 3/8.

The probability of getting heads is 625/1000 OR 5/8.

4 0
3 years ago
Please help me <br> Thank you x
polet [3.4K]

Answer:

l think that the answer is 56576

8 0
3 years ago
QUESTION 11.1
lawyer [7]

The group paid $ 5250 at first city and $ 6250 at second city

<u>Solution:</u>

Let x = the charge in 1st city before taxes

Let y = the charge in 2nd city before taxes

The hotel charge before tax in the  second city was $1000 higher than in the first

Then the charge at the second hotel before tax will be x + 1000

y = x + 1000 ----- eqn 1

The tax in the first city was 8.5% and the  tax in the second city was 5.5%

The total hotel tax paid for the two cities was $790

<em><u>Therefore, a equation is framed as:</u></em>

8.5 % of x + 5.5 % of y = 790

\frac{8.5}{100} \times x + \frac{5.5}{100} \times y = 790

0.085x + 0.055y = 790 ------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

<em><u>Substitute eqn 1 in eqn 2</u></em>

0.085x + 0.055(x + 1000) = 790

0.085x + 0.055x + 55 = 790

0.14x = 790 - 55

0.14x = 735

<h3>x = 5250</h3>

<em><u>Substitute x = 5250 in eqn 1</u></em>

y = 5250 + 1000

<h3>y = 6250</h3>

Thus the group paid $ 5250 at first city and $ 6250 at second city

8 0
3 years ago
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