All categories

2 years ago

I need Help fast Please

Mathematics

1 answer:

Sedbober [7]2 years ago

6 0

Answer: the answer is D because it is repeated many times

You might be interested in

Cos^2A+cos^2A*cot^2A=cot^2A

strojnjashka [21]

LHS: cos^2 A( 1+cot^2A)
=cos^2A(cosec^2A) ,1+cot^2A=cosec^2A
=cos^2A( 1/sin^2A)
=(cos^2A/ sin^2A)
=cot^2A
=RHS

Hope this can help.

4 0

2 years ago

AB and PQ are lines of length 6 cm a) AX: AB 1:3 Mark where point X would be on the line, with a cross. A B b) PX: XQ 1 :3 Mark

Law Incorporation [45]

Answer:

a) X would be marked 2 cm away from point A towards the point B

b) X would be marked 1.5 cm away from point P towards the point Q

Step-by-step explanation:

Data provided in the question:

Length of AB = 6 cm

Length of PQ = 6 cm

a) AX : AB 1 : 3

This means

$\frac{AX}{AB}=\frac{1}{3}$

on substituting the value of AB, we get

$\frac{AX}{6\ cm}=\frac{1}{3}$

AX = $\frac{6}{3}\ cm$

AX = 2 cm

Hence,

X would be marked 2 cm away from point A towards the point B

b) PX: XQ 1 :3

This means

$\frac{PX}{XQ}=\frac{1}{3}$

⇒ 3PX = XQ .............(1)

also,

PX + XQ = PQ

PX + XQ = 6 cm

substituting the value of XQ from (1)

PX + 3PX = 6 cm

4PX = 6 cm

PX = $\frac{6}{4}\ cm$

PX = 1.5 cm

Hence,

X would be marked 1.5 cm away from point P towards the point Q

3 0

2 years ago

Read 2 more answers

Prove. sin³A-cos³A/sinA-cosA = 1+sinAcosA

katovenus [111]

Answer:

$\frac{ \sin {}^{3} A - \cos {}^{3} A }{ \sin A - \cos A} \\ \\ { \sf{ = \frac{ {( \sin A - \cos A)}^{3} + 3 \sin A \cos A( \sin A - \cos A)}{ \sin A - \cos A} }} \\ \\ = { {( \sin A - \cos A)}^{2} + 3 \sin A \cos A} \\ { \sf{ = ( \sin {}^{2} A + \cos {}^{2} A) - 2 \sin A \cos A + 3\sin A \cos A}} \\ = { \sf{1 +\sin A \cos A }}$

#hence proved

7 0

2 years ago

Angle Relationships

leva [86]

Answer:

B. 14 ft

Step-by-step explanation:

I calculated it logically

6 0

2 years ago

Read 2 more answers

Which is the distance from point A to B?

Aleks04 [339]

Answer:

D) radical 26

Step-by-step explanation:

So this is a question where you need to find the hypotenuse of a triangle

the sides are 1 and 5

you can use Pythagoras (a^2+b^2=c^2)

substitute the values in

1^2+5^2=c^2

simplify the exponents

1+25=c^2

add them together

26=c^2

take the square root of 26

the answer is radical 26

5 0

2 years ago

I need Help fast Please ​

I need Help fast Please