All categories

2 years ago

79=(710)-(7*▲) What is the value of ▲ in the equation?

2 answers:

8 0

Can't type thta trignale thingie so subsitute with x
7*9=(7*10)-(7*x)
we can multipily what we know
7*9=7 times 9=63

7*10=7 times 10=70
now we have
63=70-(7*x)
add (7*x) to both sides
(7*x)+63=70
subtract 63 from boths ides
(7*x)=7
divide both sides by 7
x=1

the answe ris traignle is equal to 1

Nuetrik [128]2 years ago

3 0

7x9= (7x10)-(7xa)
63=70-7a
-7=-7a
a(triangle)=1

You might be interested in

Please answer the following question:

ludmilkaskok [199]

Answer:

99 cm²

Step-by-step explanation:

1/2 BH

1/2 18 x 11

9 x 11

3 0

2 years ago

PLEASE HELP!!

erik [133]

$2.=c.\\\\2\sin4x\cos4x=2\sin(2\cdot4x)=2\sin8x\\\\Used:\\\sin2\alpha=2\sin\alpha\cos\alpha$

$1.=b.\\\\\csc x-\sin x=\dfrac{1}{\sin x}-\dfrac{\sin^2x}{\sin x}=\dfrac{1-\sin^2x}{\sin x}=\dfrac{\cos^2x}{\sin x}\\\\=\dfrac{\cos x\cos x}{\sin x}=\cos x\cdot\dfrac{\cos x}{\sin x}=\cos x\cot x\\\\Used:\\\csc x=\dfrac{1}{\sin x}\\\\\sin^2x+\cos^2x=1\to\cos^2x=1-\sin^2x\\\\\cot x=\dfrac{\cos x}{\sin x}$

$3.=a.\\\\\dfrac{\sin x-1}{\sin x+1}=\dfrac{\sin x-1}{\sin x+1}\cdot\dfrac{\sin x+1}{\sin x+1}=\dfrac{\sin^2x-1^2}{(\sin x+1)^2}=\dfrac{\sin^2x-1}{(\sin x+1)^2}\\\\=\dfrac{-(1-\sin^2x)}{(\sin x+1)^2}=\dfrac{-\cos^2x}{(\sin x+1)^2}\\\\Used:\\(a-b)(a+b)=a^2-b^2\\\\\sin^2x+\cos^2x=1\to \cos^2x=1-\sin^2x$

8 0

2 years ago

A Biycle wheel has a diameter of 52cm.

schepotkina [342]

Answer:

26cm

Step-by-step explanation:

because 2 radius equal 1 diameter

8 0

2 years ago

Help with this, please! It's about the volume of a cone

Lena [83]

Answer:

Step-by-step explanation:

7 0

2 years ago

How many time 2/3 can go into 16 Explanation pls!!

tensa zangetsu [6.8K]

Answer:

24 times

Step-by-step explanation:

16 divided by 2/3 is 24

hope this helped!

7 0

1 year ago

7*9=(7*10)-(7*▲) What is the value of ▲ in the equation?

79=(710)-(7*▲) What is the value of ▲ in the equation?