Ask question

Ask question

All categories

3 years ago

12

Need help !!!!!!!!!!

1 answer:

Sidana [21]3 years ago

3 0

the data in the table is linear

You might be interested in

Do all animals have the same life span. Explain your answer

Talja [164]

No, because every animal is different.

5 0

3 years ago

A right triangle has two equal sides. The hypotenuse is 10 cm. What is the perimeter of the triangle?

Scorpion4ik [409]

Answer:

Isosceles Right Triangle Example

Step-by-step explanation:

Find the area and perimeter of an isosceles right triangle whose hypotenuse side is 10 cm. Therefore, the length of the congruent legs is 5√2 cm. Therefore, the perimeter of an isosceles right triangle is 24.14 cm.

Hope this answer helps ^^

4 0

3 years ago

. <br><br><br> Solve the equation. <br><br> 2/3 x – 6 = 9

Allisa [31]

Hey there!

Okay let's get cracking :D

The equation is.... (2x/3) - 6 = 9

2x/3 = 15
2x = 45
<u>x = 22.5</u><u /> <----- And that is your answer!

I hope it helped!

<u>Please vote for me as Brainliest if there is a second answer!</u>

4 0

3 years ago

Read 2 more answers

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

3 years ago

Is this a function?<br>

zhuklara [117]

No I don’t believe it is

4 0

2 years ago