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Kay [80]
2 years ago
15

Solve for c y-(-10-c)=z

Mathematics
2 answers:
raketka [301]2 years ago
8 0
Distribute
y+10+c=z
minus y from both sides
10+c=z-y
minus 10 both sides
c=z-y-10
elena55 [62]2 years ago
6 0
Y+10+c=z
Y=z-10-c
C=z-10-y
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Write an algebraic expression for “12 less than the quotient of 12 and a number z.”
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Step-by-step explanation:

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1) 1, 4, 16, 64, ...<br>CR=L<br>Next 3 Terms:​
avanturin [10]

Answer:

256, 1024, 4096.

Step-by-step explanation:

The rule is x*4.

--> 1*4 = 4.

--> 4*4 = 16.

-->  16*4 = 64.

--> 64*4 = 256.

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20 points!! Please Help!
Galina-37 [17]

Answer:

• x = 9

• y = 6√2

Step-by-step explanation:

The right triangles are all similar, so the ratios of hypotenuse to short side are the same:

27/x = x/3

x^2 = 81 . . . . . multiply by 3x

x = 9 . . . . . . . . take the square root

Then y can be found from the Pythagorean theorem:

x^2 = y^2 + 3^2

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8 0
3 years ago
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Given: Triangle ABC : triangle ADB. AB =24, AD =16 <br> Find: AC
RoseWind [281]

Answer:

<h2>AC = 36.01</h2>

Step-by-step explanation:

Given ΔABC and ΔADB, since both triangles are right angled triangles then the following are true.

From ΔADB, AB² = AD²+BD²

Given AB = 24 and AD = 16

BD² = AB² - AD²

BD² = 24²-16²

BD² = 576-256

BD² = 320

BD = \sqrt{320}

BD = 17.9

from ΔABC, AC² = AB²+BC²

SInce AC = AD+DC and BC² = BD² + DC² (from ΔBDC )we will have;

(AD+DC)² = AB²+ (BD² + DC²)

Given AD = 16, AB = 24 and BD = 17.9, on substituting

(16+DC)² = 24²+17.9²+ DC²

256+32DC+DC² =  24²+17.9²+ DC²

256+32DC = 24²+17.9²

32DC = 24²+17.9² - 256

32DC = 640.41

DC = \frac{640.41}{32}

DC = 20.01

Remember that AC = AD+DC

AC = 16+20.01

AC = 36.01

6 0
3 years ago
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