2 years ago

Can someone help me on this mess...

Mathematics

2 answers:

serg [7]2 years ago

5 0

Answer:

sas

Step-by-step explanation:

lord [1]2 years ago

3 0

Answer:

one attachment ex picture

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Add. (5y - 1) + (-2y + 4) A) 3y - 5 B) 3y + 3 C) 7y + 3 D) 7y - 3

Ivan

(5y - 1) + (-2y + 4) = 5y - 1 - 2y + 4
= 5y - 2y - 1 + 4
= 3y + 3 [OPTION B]

8 0

2 years ago

if x and y are two overlapping subsets of a universal set U, write the relation between n(U),n(X U Y), and n(xūy).

Andrew [12]

Answer:

the relation between them are

n(U) = n(X U Y) + n(xūy).

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2 years ago

Question 34 please help me

tino4ka555 [31]

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7 0

3 years ago

To find the measure of ∠A in ∆ABC, use the___(Pythagorean Theorem, Law of Sines, Law of Cosines). To find the length of side HI

nadya68 [22]

Part 1) To find the measure of ∠A in ∆ABC, use

we know that

In the triangle ABC

Applying the law of sines

$\frac{a}{sin\ A}=\frac{b}{sin\ B}=\frac{c}{sin\ C}$

in this problem we have

$\frac{a}{sin\ A}=\frac{b}{sin\ theta}\\ \\a*sin\ theta=b*sin\ A\\ \\ sin\ A=\frac{a*sin\ theta}{b} \\ \\ A=arc\ sin (\frac{a*sin\ theta}{b})$

therefore

the answer Part 1) is

Law of Sines

Part 2) To find the length of side HI in ∆HIG, use

we know that

In the triangle HIG

Applying the law of cosines

$g^{2}=h^{2}+i^{2}-2*h*i*cos\ G$

In this problem we have

g=HI

G=angle Beta

substitute

$HI^{2}=h^{2}+i^{2}-2*h*i*cos\ Beta$

$HI=\sqrt{h^{2}+i^{2}-2*h*i*cos\ Beta}$

therefore

the answer Part 2) is

Law of Cosines

3 0

2 years ago

Read 2 more answers

Find the coordinates of the other endpoint of the segment, given its midpoint and one endpoint. (Hint: Let (x,y) be the unkno

melomori [17]

Mid point formula =
(x1/2+x2/2, y1/2+y2/2)

x=
x+0 = 8
/2
x+0 = 8 x 2
x = 16

y=
y+(-2) = 1
/2
y-2 =1 x2
y=2+2
y=4

(16,4)

(In this case, the midpoint (x ,y) will be represent as the answer)

7 0

3 years ago