All categories

3 years ago

HELP ME PLEASE!!!

Mathematics

1 answer:

Xelga [282]3 years ago

8 0

Answer:7/2

Step-by-step explanation:

You might be interested in

There are 295 students in 6th grade at Central Middle School. Must take science. The maximum number of students in the class is

vesna_86 [32]

14 classes must be planned you get this by dividing 295 by 22 and you get 13.4 but you have to round up because we are talking about people

7 0

3 years ago

Work out<br> 5 1/4+ 1/2 x2/3<br> Give your answer as a mixed number

Alika [10]

Answer:

5 10/12

Step-by-step explanation:

using BODMAS

1/2×2/3=1/3

THEN; 5 1/4+1/3

Change 5 1/4 to 21/4 then add to 1/3

you get 70/12

divide 70÷12

answer 5 10/12

hope it helped

6 0

3 years ago

A bookmark is shaped like a rectangle with a semicircle attached at both ends. The rectangle is 16 cm long and 4 cm wide. The di

Sedaia [141]

Answer: 76.56sqcm

Step-by-step explanation:

Area of rectangle= length * width

= 16 * 4 = 64sqcm

Area of semicircle= (1/2)(r^2)π

r= d/2= 4/2= 2

Area of semicircle= (1/2)(2^2)(3.14)

= 6.28sqcm

Area of bookmark= Area of rectangle + 2*Area of semicircle

Area of bookmark= 64 + 2(6.28)

=64 + 12.56 = 76.56sqcm

3 0

3 years ago

What does paragraph 26 reveal about Rachel's character?

4vir4ik [10]

Answer:

your answer is A hope this helps

Step-by-step explanation:

5 0

3 years ago

On a coordinate plane, triangle A B C is shown. Point A is at (0, 0), point B is at (3, 4), and point C is at (3, 2). What is th

Elena L [17]

Answer:

The area of triangle for the given coordinates is 1.5 $\sqrt{4.6}$

Step-by-step explanation:

Given coordinates of triangles as

A = (0,0)

B = (3,4)

C = (3,2)

So, The measure of length AB = a = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, a = $\sqrt{(3-0)^{2}+(4-0)^{2}}$

Or, a = $\sqrt{9+16}$

Or, a = $\sqrt{25}$

∴ a = 5 unit

Similarly

The measure of length BC = b = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, b = $\sqrt{(3-3)^{2}+(2-4)^{2}}$

Or, a = $\sqrt{0+4}$

Or, b = $\sqrt{4}$

∴ b = 2 unit

And

So, The measure of length CA = c = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, c = $\sqrt{(3-0)^{2}+(2-0)^{2}}$

Or, c = $\sqrt{9+4}$

Or, c = $\sqrt{13}$

∴ c = $\sqrt{13}$ unit

Now, area of Triangle written as , from Heron's formula

A = $\sqrt{s\times (s-a)\times (s-b)\times (s-c)}$

and s = $\frac{a+b+c}{2}$

I.e s = $\frac{5+2+\sqrt{13}}{2}$

Or. s = $\frac{7+\sqrt{13}}{2}$

So, A = $\sqrt{(\frac{(7+\sqrt{13})}{2})\times ((\frac{(7+\sqrt{13})}{2})-5)\times (\frac{7+\sqrt{13}}{2}-2)\times (\frac{7+\sqrt{13}}{2}-\sqrt{13})}$

Or, A = $\sqrt{(\frac{(7+\sqrt{13})}{2})\times (\frac{(\sqrt{13}-3)}{2})\times (\frac{4+\sqrt{13}}{2})\times (\frac{7-\sqrt{13}}{2})}$

Or, A = $\frac{3}{2}$ × $\sqrt{1+\sqrt{13} }$

∴ Area of triangle = 1.5 $\sqrt{4.6}$

Hence The area of triangle for the given coordinates is 1.5 $\sqrt{4.6}$ Answer

7 0

3 years ago

Read 2 more answers