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Arlecino [84]
3 years ago
12

What is the solution to x2 = 225?

Mathematics
2 answers:
Blababa [14]3 years ago
6 0

Answer:

x = ± 15

Step-by-step explanation:

Given

x² = 225 ( take the square root of both sides )

x = ± \sqrt{225} = ± 15

That is x = - 15 0r x = 15

Sauron [17]3 years ago
3 0

Answer:

Step-by-step explanation:

X² = 225

Finding the square root of both side

√X² = √225

X = 15 or -15

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Could someone please help ;(
Iteru [2.4K]

Answer:

a. 25/33

b. 36

c. 1/15

d. 35/12

e. 19/12

7 0
3 years ago
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Wgat am I supposed to do here
svetoff [14.1K]
There is nothing there what am I supposed to
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if there is 27 marbles in a bag and the amount of red marbles is 5 less than 3 times the amount of blue marbles how many marbles
alexira [117]

Answer:

54

Step-by-step explanation:

5-3=2

so you multiply 2 times 27.

2 times 27 is 54

8 0
3 years ago
Find the equation of the line which passes through the point (−2,−5) and is parallel to the line −3x+5y=−4.
leonid [27]
-3x+5y=-4
5y=3x-4
Y=3/5x-4/5


When it parallel to the line y=3/5x-4/5

It can be y=-3/5x+b

-5=6/5+b

B=-5-6/5=-25/5-6/5=-31/5

Y=-3/5x-31/5

3/5x+y=-31/5

Multiply 5

3x+5y=-31
4 0
3 years ago
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WILL MARK BRAINLIEST!
Artist 52 [7]

Answer:

Following are the responses to the given points:

Step-by-step explanation:

Dilation implies the triangle \Delta XYZ stretched through the factor "2".  

m\angle Y = m\angle C = 90^{\circ} \leftarrow given \\\\  

right triangles:

\Delta  XYZ and \Delta  ACB \\\\\angle X \cong \angle A \leftarrow given\\\\

complementary angles:  

\angle X \ and\  \angle Z

m\angle Z = 90^{\circ} - m\angle X \\\\

complementary angles:

\angle A \ and\  \angle B

m\angle B = 90^{\circ} - m\angle A \\\\\therefore \\\\ \angle Z \cong \angle B \\\\\Delta  XYZ \sim  \Delta ACB \\\\

Same triangles: proportional side are corresponding, congruence angles are respective.

\sin \angle X = \frac{5}{5.59} \leftarrow given \\\\\sin\angle X = \frac{YZ}{XY} \\\\YZ = 5 (units) \rightarrow leg in \Delta XYZ \\\\XZ = 5.59 (units) \rightarrow hypotenuse \ in\  \Delta XYZ\\\\

Calculating the length of leg XY:

XY = \sqrt{((5.59)^2 - 52)} \cong 2.4996 (units)\\\\\frac{CB}{YZ} = 2 \leftarrow given \\\\CD = 2YZ = 2 \times 5 = 10 \ (units)\\\\\frac{AC}{XY} = 2 \leftarrow  \ given\\\\AC = 2XY \cong 2 \times  2.4996 \cong 4.999\  (units)\\\\

3 0
3 years ago
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