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3 years ago

What other information is needed to prove that the two triangles congruent by SAS?

1 answer:

6 0

If we have two triangles:
Triangle GLT and Triangle NMQ and we are given that LT = MQ and L = M, then the remaining condition to prove that the two triangles are similar by SAS, is
D. LG = MN
since LG and LT make up angle L and MQ and MN make up angle M

If we have two triangles,
triangle RST and UVW, then to prove that they are similar by SAS, these conditions must be met:
ST = UV
A. S = V
E. RS = WU

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Write the equation of the line containing the points (3, 6) and (7, - 2) in standard form.

Rama09 [41]

Answer:

$y = mx + c \\ 6 = 3m + c...(1) \\ - 2 = 7m + c...(2) \\ subtract \: (2) - (1) \\ 4m = - 8 \\ \therefore \: m = - 2 \: sub \: in \: (1) \\ 6 = 3( - 2) + c \\ \therefore \: c = 12 \\ \boxed{y = - 2x + 12}$

7 0

2 years ago

Read 2 more answers

A central angle of a circle measures 1.5 radians. If the radius of the circle is 3 cm, what is the area of the related sector?

soldi70 [24.7K]

Answer:

Option B. $6.75\ cm^{2}$

Step-by-step explanation:

we know that

The area of a circle is equal to

$A=\pi r^{2}$

we have

$r=3\ cm$

substitute

$A=\pi (3^{2})=9 \pi\ cm^{2}$

Remember that

$2\pi$ radians subtends the complete circle of area $9 \pi\ cm^{2}$

by proportion

Find the area of the related sector for a central angle of $1.5$ radians

Let

x------> the area of the related sector

$\frac{9 \pi}{2\pi}\frac{cm^{2}}{radians} =\frac{x}{1.5}\frac{cm^{2}}{radians}\\ \\x=9*1.5/2\\ \\x= 6.75\ cm^{2}$

7 0

3 years ago

Let f (x) = 3x − 1 and ε > 0. Find a δ > 0 such that 0 < ∣x − 5∣ < δ implies ∣f (x) − 14∣ < ε. (Find the largest

s344n2d4d5 [400]

Answer:

This proves that f is continous at x=5.

Step-by-step explanation:

Taking f(x) = 3x-1 and $\varepsilon>0$ , we want to find a $\delta$ such that $|f(x)-14|$

At first, we will assume that this delta exists and we will try to figure out its value.

Suppose that $|x-5|$ . Then

$|f(x)-14| = |3x-1-14| = |3x-15|=|3(x-5)| = 3|x-5|< 3\delta$ .

Then, if $|x-5|$ , then $|f(x)-14|$ . So, in this case, if $3\delta \leq \varepsilon$ we get that $|f(x)-14|$ . The maximum value of delta is $\frac{\varepsilon}{3}$ .