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Leviafan [203]
2 years ago
12

How to do question 13?

Mathematics
2 answers:
grin007 [14]2 years ago
6 0
You answer on question 13 is 350
elena55 [62]2 years ago
4 0
The correct answer for area is A=246.740109463

Do (pi)3.14159265x5 =

15.70796325 exponent of 2=

246.740109463=A

You divide your circumference by half to get your radius.
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You are going to mix a gallon bucket of window cleaner. The instructions direct you to mix 1 part cleaner to 3 parts water. How
Paul [167]
To mix cleaner and water you need to divide the bucket's capacity into 4.It's easy in this case as gallon is divided into quarts (1 gallon = 4 quarts).We know that one part will be cleaner, it means that you will need to use 1 quart of the cleaner.  
hope thats right
4 0
2 years ago
Read 2 more answers
Angles 1 and 2 are supplementary.
Phantasy [73]

Answer:

m∡1 + m∡2 = 180

Step-by-step explanation:

m∡1 + m∡2 = 180

This is because supplementary means they add up to 180°.

3 0
3 years ago
QUESTION 29<br> Write the equation of the function.
sweet [91]

Answer:

  y = 2 -√(x+1)

Step-by-step explanation:

The square root function is reflected across the x-axis and shifted 1 unit to the left and 2 units up.

  y = -√x . . . . . reflects the function across the x-axis

  y = -√(x+1) . . . shifts the reflected function 1 unit to the left

  y = 2 -√(x +1) . . . shifts the above function 2 units up

The graph is of the equation y = 2 -√(x+1).

3 0
3 years ago
Solve for the value of x: 3x + 25 = 2/3 (6x - 15)​
n200080 [17]

3x + 25 = 2/3 (6x - 15)

9x + 75 = 12x - 30

75 + 30 = 12x - 9x

3x = 105

x = 105 : 3

x = 35

3 0
3 years ago
Given: QR I PT and ZQPR = ZSTR
Rina8888 [55]

Answer:

\triangle PQR \sim \triangle TSR

Step-by-step explanation:

Given :

QR⊥PT

\angle QPR = \angle STR

To Prove : \triangle PQR \sim \triangle TSR

Solution:

Statements                                            Reasons

QR⊥PT                                                Given

∠QRP and ∠SRT are right angles       Def of perpendicular

∠QPR≅∠STR                                     Given

∠QRP = ∠SRT                                    All right angles are equal

ΔPQR≈ΔTSR                                     AA similarity

Hence  \triangle PQR \sim \triangle TSR

5 0
3 years ago
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