All categories

3 years ago

Is it possible to draw a triangle with sides 150cm, 10cm, 10cm

1 answer:

6 0

It is absolutely impossible to draw a triangle of sides 150 , 10 and 10 cm. With one side so large, it is impossible, which will make the two angles two small to intersect and the angle will be < 0, which is not possible in a triangle.

You might be interested in

By rounding to 1 significant figure, estimate <br><br> 306.9 x 6.4

coldgirl [10]

Answer:2000

Step-by-step explanation:

mutiply

306.9 * 6.4

=1964.16

to one significant figure

=2000

8 0

3 years ago

Alyssa needs to lease out a music studio to record her new album. The studio charges an initial studio-use fee plus an hourly fe

uysha [10]

Answer:

blue is the best color to use in this situation

Step-by-step explanation:

6 0

4 years ago

(y+6)^2-(y-2)^2 <br><br> I got 16y+26 but it is wrong?

Gemiola [76]

Answer:

16y + 32

Step-by-step explanation:

Expand each term.

(y+6)² - (y-2)²

= (y+6)(y+6) - (y-2)(y-2)

= y² + 12y + 36 - (y² - 4y + 4)

Subtract the second group by changing each term's signs

= y² + 12y + 36 - y² + 4y - 4

Collect like terms

= 16y + 32

5 0

3 years ago

What’s the area of 7.4cm and 4.5cm

Vadim26 [7]

Answer:

33.3

Step-by-step explanation:

The formula for area is length*width.

7.4*4.5=33.3

5 0

3 years ago

Read 2 more answers

What is the value of b? IT 3b b+52 Н G j b =

cluponka [151]

b=26

Explanation

Step 1

we have two equals triangles, because

$\begin{gathered} HG=GJ \\ \text{and} \\ GI\text{ is a common side for both triangles} \end{gathered}$

then, the hypotenuses are equal

$3b=b+52$

"is step 1 clear?

Step 2

solve the equation

$\begin{gathered} 3b=b+52 \\ \text{subtract b in both sides} \\ 3b-b=b+52-b \\ 2b=52 \\ \text{divide both sides by 2} \\ \frac{2b}{2}=\frac{52}{2} \\ b=26 \end{gathered}$

4 0

2 years ago