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Mariana [72]
4 years ago
15

How do I solve this using the Pythagorean theorem and leaving it in radical form ?

Mathematics
1 answer:
scoray [572]4 years ago
7 0
If the legs of a right triangle are a and b and the hypotnuse is c then
a²+b²=c²

so we are given
the legs are x and √7
and the hypotnuse is √19
so

x²+(√7)²=(√19)²
x²+7=19
minus 7 both sides
x²=12
sqrt both sides
x=√12
x=√(4*3)
x=(√4)(√3)
x=2√3
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4 5x−4≥12 OR 12x+5≤−45
Oksanka [162]

Answer:

its 98

Step-by-step explanation:

first add the n sub.

6 0
3 years ago
Please help me with number 14 name the propriety illustrated by the following
ASHA 777 [7]
A. .x -x=0
x+ -x=0
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all real numbers are solutions
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6 0
4 years ago
Which table represents a linear function?<br><br> Please help I have to take my final
Ede4ka [16]

Answer:

#3

Step-by-step explanation:

You can tell if a table is linear, if the slope is constant within the graph. #3 is the only option that has a constant slope. In the others, if you pick different points from that same table the slope changes. That is how you know that is not a linear function.

3 0
2 years ago
Don’t tell me the steps just answer please!!!
r-ruslan [8.4K]

Answer:

A, D, F

Step-by-step explanation:

4 0
3 years ago
Prove that they are equal
Westkost [7]

Step-by-step explanation:

(1-tanx)/(1+tanx) = (1-sin2x)/cos2x

Replace tanx with sinx/cosx:

(1-sinx/cosx)/(1+sinx/cosx)

Multiply numerator and denominator by cosx:

(cosx-sinx)/(cosx+sinx)

Multiply numerator and denominator by cosx-sinx:

(cos^2x-2sinxcosx+sin^2x)/(cos^2x-sin^2x)

For the numerator, since sin2x = 2sinxcosx, and sin^2x + cos^2x = 1, we have 1-sin2x

For the denominator, cos^2x - sin^2x = cos2x

So (1-tanx)/(1+tanx) = (1-sin2x)/cos2x

4 0
4 years ago
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