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Fantom [35]
3 years ago
8

Pls help w these thx <33

Mathematics
1 answer:
stepladder [879]3 years ago
7 0

Answer:

36.81233927

13.02532494

Step-by-step explanation:

You need to use the law of sine for both of them

the law of sine is as follows:

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

Question 1:

\frac{15}{\sin(64)}=\frac{10}{\sin(y)}\\y=36.81233927

Question 2:

Let x= AB

Start by solving for the missing angle, 53

\frac{8.4}{\sin(31)}=\frac{x}{\sin(53)}

x= 13.02532494

I will leave you to round the answers

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The function g(x) = x2 is transformed to obtain function h:
Rudik [331]

Answer:

Option C

The graph of g is vertically shifted 5 units down

4 0
3 years ago
Solve the equation using square roots.
anastassius [24]

Answer:

A : ±2

Step-by-step explanation:

Add 14 and you have ...

... x² = 4

Take the square root and you have ...

... x = ±√4 = ±2

3 0
3 years ago
Anybody know this ?????
sammy [17]

Answer:

The correct answer is  option C

They are complementary angles

Step-by-step explanation:

From the figure we can see two angles <ABC and <LMN,

<u>To find the correct option</u>

From figure we get,

m<ABC = 20°° and m<LMN = 70°

m<ABC + m<LMN = 20 + 70 = 90°

Therefore <ABC and <LMN are complementary angles

The correct answer is option C

They are complementary angles

7 0
3 years ago
In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= root 2 cm.
Nata [24]

Given CD is an altitude such that AD=BC , AB=3 cm and CD= √2 cm.

Let AD=x, Since given AB=3

                                     AD+DB=3

                                       x+DB = 3

                                        DB = 3-x

Since ΔBCD is rght angle triangle, let's apply Pythagoras theorem

BC^{2} = DB^{2} +CD^{2}

BC^{2} = (3-x)^{2} +(\sqrt{2} )^{2}

BC^{2} =(3-x)^{2} +2

Since given AD=BC,let us plugin BC=x in above step.

x^{2} =(3-x)^{2} +2

x^{2} =9-6x+x^{2} +2

6x=11

x=\frac{11}{6}

Now we know AD=x=\frac{11}{6} and given CD=√2.

Let us apply Pythagoras theorem for ΔACD

AC^{2} =AD^{2} +DC^{2}

AC^{2} = (\frac{11}{6} )^{2} +(\sqrt{2} )^{2}

AC^{2} =\frac{193}{36}

AC=\sqrt{\frac{193}{36} } = 2.315cm

3 0
3 years ago
Can someone tell me if i did this right? im also not sure which one is the scale factor.
PIT_PIT [208]
If you are moving the center of circle 2 to the the center of circle 1, then the translation rule is (x,y) ---> (x+4, y+10). 
Note how x = 1 turns into x = 5. So we add 4
Also, y = -2 turns into y = 8. We add 10

The scale factor to turn the radius r = 4 into r = 8 is 2. Basically we double the radius. We can divide the two radii to see 8/4 = 2

----------------------------------

Summary:
To go from circle 2 to circle 1, we apply these two transformations
translation: (x,y) ---> (x+4, y+10)
dilation: scale factor 2

note: 
if you want to go backwards, go from circle 1 to circle 2, then undo the transformations above
3 0
3 years ago
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