Answer:
I hope that help
Step-by-step explanation:
Subtract
2
from both sides of the equation.
y
=
−
3
x
4
−
3
4
−
2
To write
−
2
as a fraction with a common denominator, multiply by
4
4
.
y
=
−
3
x
4
−
3
4
−
2
⋅
4
4
Combine
−
2
and
4
4
.
y
=
−
3
x
4
−
3
4
+
−
2
⋅
4
4
Combine the numerators over the common denominator.
y
=
−
3
x
4
+
−
3
−
2
⋅
4
4
Simplify the numerator.
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y
=
−
3
x
4
+
−
11
4
Move the negative in front of the fraction.
y
=
−
3
x
4
−
11
4
Use the slope-intercept form to find the slope and y-intercept.
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Slope:
−
3
4
y-intercept:
−
11
4
Any line can be graphed using two points. Select two
x
values, and plug them into the equation to find the corresponding
y
values.
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x
y
0
−
11
4
3
−
5
Graph the line using the slope and the y-intercept, or the points.
Slope:
−
3
4
y-intercept:
−
11
4
9/100
Alternative form : 0.09
<h3>
Answer: The other two sides are 4 inches each</h3>
Rule: If x is the leg of a 45-45-90 triangle, then x*sqrt(2) is the hypotenuse.
We can prove this by showing that x^2+x^2 = (x*sqrt(2))^2 is an identity. It's from the pythagorean theorem a^2+b^2 = c^2.
Put another way, if we know that a = x and b = x are the two congruent legs, then x^2+x^2 = c^2 leads to c = x*sqrt(2). This triangle is known as an isosceles right triangle.
So if 4*sqrt(2) is the hypotenuse, we match that up with the template x*sqrt(2) and see that x = 4 must be the case.
Answer:
The order of the start of the proof seems fine; we're to choose the next steps I guess.
segment UV is parallel to segment WZ Given
Points S, Q, R, and T all lie on the same line. Given
m∠SQT = 180° Definition of a Straight Angle
m∠SQV + m∠VQT = m∠SQT Angle Addition Postulate
m∠SQV + m∠VQT = 180° Substitution Property of Equality
That's all valid up to here. It seems to me sort of the hard way to get to linear supplements but here we are.
ZRS is mentioned in the rest of the lines; let's find the one that comes first.
III m∠VQT + m∠ZRS = 180° Same-Side Interior Angles Theorem
Now we have two things equal to 180 degrees, so they're equal to each other.
II m∠SQV + m∠VQT = m∠VQT + m∠ZRS Substitution Property of Equality
Now comes
I m∠SQV + m∠VQT − m∠VQT = m∠VQT + m∠ZRS − m∠VQT
m∠SQV = m∠ZRS Subtraction Property of Equality
And we conclude,
∠SQV ≅ ∠ZRS Definition of Congruency