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3 years ago

I need help with all the problems in this photo I need help now

Mathematics

2 answers:

NeX [460]3 years ago

5 0

The answer to 6 is repeating I believe

Anarel [89]3 years ago

3 0

I'm pretty sure number 2 is Negative and number 4 is Absolute Value

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Find the number of sides of a polygon if the sum of interior angles is 10 right angles.

ludmilkaskok [199]

USE SOCRATIC ITS REALLY HELPS

7 0

3 years ago

Read 2 more answers

Please help me!!!!!! I really neeeeeeed

Butoxors [25]

Answer:

A-(x+4,y-5)

B- move 1 unit left and 7 units down

Step-by-step explanation:

A- 4 units right is x so

(x+4,y__)

5 units down is y so

(x+4,y-5)

B- move 1 unit left and 7 units down

because (x-1,y__) is going left because of the subtraction sign and its the x-coordinate, and (x-1,y-7) is down because its a Y-coordinate and sub so its going down.

<h2>TIP- Going left/down is always subtract/negative going right/up is positive/addition.</h2>

6 0

3 years ago

Please someone help me to prove this.

dem82 [27]

<h3>Answer :</h3>

We know that,

$\dag\bf\:sin^2A=\dfrac{1-cos2A}{2}$

$\dag\bf\:sin2A=2sinA\:cosA$

___________________________________

Now, Let's solve !

$\leadsto\:\bf\dfrac{sin^2A-sin^2B}{sinA\:cosA-sinB\:cosB}$

$\leadsto\:\sf\dfrac{\frac{1-cos2A}{2}-\frac{1-cos2B}{2}}{\frac{2sinA\:cosA}{2}-\frac{2sinB\:cosB}{2}}$

$\leadsto\:\sf\dfrac{1-cos2A-1+cos2B}{sin2A-sin2B}$

$\leadsto\:\sf\dfrac{2sin\frac{2A+2B}{2}\:sin\frac{2A-2B}{2}}{2sin\frac{2A-2B}{2}\:cos\frac{2A+2B}{2}}$

$\leadsto\:\sf\dfrac{sin(A+B)}{cos(A+B)}$

$\leadsto\:\bf{tan(A+B)}$

5 0

3 years ago

Read 2 more answers

What is y=-x-14 y=-8x

Marina86 [1]

Substitute y in the first equation.

-8x = -x - 14
+x +x

-7x = -14
-7x/-7 = -14/-7
x = 2

Now solve for y.

y = -18(2)
y = -36

Answer: (2, -36) or y = -36 and x = 2

6 0

3 years ago

Ill give you brainiest

White raven [17]

Answer:

D and C

Step-by-step explanation:

because EF is diameter

then radius diameter divided by 2

8 0

3 years ago