All categories

3 years ago

Sketch seven points. Then, connect the points to form a closed plane figure. What kind of polygon did you draw?

Mathematics

2 answers:

weqwewe [10]3 years ago

7 0

It’s a heptagon, which has seven points.

denis-greek [22]3 years ago

6 0

Answer:

A Polygon that has 7 points would have to be a heptagon.

Step-by-step explanation:

You might be interested in

What is900 times 8 please help me

ddd [48]

900 x 8= 7200 is your answer

6 0

4 years ago

Read 2 more answers

On a certain hot summer's day, 581 people used the public swimming pool. The daily prices are $1.50 for children and $2.50 for

Tanya [424]

Answer:

A = 219

C = 362

Step-by-step explanation:

Given,

581 = C + A

and

$1090.50 = 1.5C + 2.5A

solve 1st for A

A =581 - C

substute into 2nd

1090.50 = 1.5C + 2.5(581 - C)

solve for C

1090.50 = 1.5C + 1452.50 - 2.5C

1090.50 = -C + 1452.50

1090.50 - 1452.50 = -C

-362 = -C

C = 362

then

581 = C + A

581 = 362 + A

219 = A

A = 219

8 0

3 years ago

B/c =a what is c ????????????

timama [110]

If you only want to know what it means ¨c¨ then it is the Denominator, cause you are referring a fraction.

6 0

3 years ago

I need help with this question ASAP

zubka84 [21]

Answer:

A, D, and I think E

I hope this helps :)

8 0

3 years ago

. Evaluate tan(α + β) = sin(????+????) / cos(????+????) to show tan(???? + ????) = tan(????)+tan(????????) / 1−tan(????) tan(???

grandymaker [24]

Answer with Step-by-step explanation:

We are given that

$tan(\alpha+\beta)$

$\frac{sin(\alpha+\beta}{cos(\alpha+\beta)}$

By using formula $tan x=\frac{sin x}{cos x}$

$\frac{sin\alpha cos\beta+sin\beta cos\alpha}{cos\alpha cos\beta-sin\alpha sin\beta}$

By using property: $sin(x+y)=sin x cosy+cos x sin y$

$cos(x+y)=cos x cosy-sin x siny$

Divide numerator and denominator by $cos\alpha cos\beta$

Then, we get

$\frac{\frac{sin\alpha}{cos\alpha}+\frac{sin\beta}{cos\beta}}{1-\frac{sin\alpha sin\beta}{cos\alpha cos\beta}}$

$tan(\alpha+\beta)=\frac{tan\alpha+tan\beta}{1-tan\alpha tan\beta}$

Hence, proved

Substitute $\alpha=\beta$

Then we get

$tan 2\alpha=\frac{tan\alpha+tan\alpha}{1-tan^2\alpha}$

$tan(2\alpha)=\frac{2tan\alpha}{1-tan^2\alpha}$

Hence, proved.

3 0

4 years ago