Ask question

Ask question

All categories

Keith_Richards [23]

3 years ago

10

What is 1234567890987654321234567890987654321234567890x987654321234567890987654321234567890

1 answer:

ValentinkaMS [17]3 years ago

3 0

The answer is 293742847472927463629293857462919384635151143849492

You might be interested in

Need help how can i do this?

TEA [102]

ANSWER

$\boxed {c. \: \: 33}$

EXPLANATION

The relationship between the number of faces, number of vertices and number of edges of a solid shape is given by the formula,

$F+V-2=E$

It was given that, F=24 and V=11.

We substitute these values into the formula to obtain,

$24+11-2=E$

We solve for E to get,

$33=E$

or

$E = 33$

Therefore the correct answer is C

7 0

3 years ago

Read 2 more answers

I need help with this

zhuklara [117]

Answer:

A

Step-by-step explanation:

ok, without reading through any of the answers lets see what we can get just from looking at that picture.

We can see that the angles being bisected are on a straight line.

A straight line is 180 degrees.

We know that you have to add both angles together to get 180.

The answer has to equal 180.

A is the only option that offers that.

3 0

3 years ago

What is the area of the kite? <br><br> 135 m2<br><br> 108 m2<br><br> 90 m2

Virty [35]

Answer:

Step-by-step explanation:

135 m2

7 0

4 years ago

Find the angle whose complement is 79 degree

Tom [10]

Answer:

Complementary angels= 90⁰

90-79= 11

11⁰

7 0

3 years ago

Read 2 more answers

I need help on this it’s about trig identities

Cerrena [4.2K]

Answer:

The answer to your question is:

Step-by-step explanation:

1.-

$\frac{1 + sin\alpha }{cos\alpha } + \frac{cos\alpha }{1 + sin\alpha } = 2 sec\alpha$

$\frac{(1 + sin\alpha)^{2} + cos^{2} \alpha }{cos\alpha (1 + sin\alpha) }$

$\frac{1 + 2sin\alpha + sin^{2} \alpha+ cos^{2} \alpha }{cos\alpha + sin\alphacos\alpha }$

$\frac{2 + 2sin\alpha }{cos\alpha+ sin\alpha cos\alpha }$

$\frac{2(1 + sin\alpha) }{cos\alpha(1 + sin\alpha ) }$

$\frac{2}{cos\alpha }$

2sec $\alpha$

2.-

sec²x - tanxsecx

$\frac{1}{cos^{2}x } - \frac{sinx}{cosx} \frac{1}{cosx}$

$\frac{1}{cos^{2} x} - \frac{sinx}{cos^{2}x}$

$\frac{1 - sinx}{cos^{2}x }$

$\frac{1 - sinx}{1 - sin^{2}x }$

$\frac{1 - sinx}{(1 - sinx)(1 + sinx)}$

$\frac{1}{1 + sinx}$

3 0

3 years ago