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3 years ago

PLEASE I BEG YOU SOMEONE HELP PLEASE AHHHH

Mathematics

1 answer:

sattari [20]3 years ago

6 0

I’m pretty sure that it is “the same area as”

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Help pretty pretty please :(

Evgen [1.6K]

Answer:

27/20

Step-by-step explanation:

The absolute value of any expression is always positive or zero

3/5 + 3/4

Add the fractions

27/20

Solution

27/20

Alternate form

1 7/20, 1.35

8 0

3 years ago

Read 2 more answers

What is 20(-1.5r + 0.75)

mylen [45]

Answer:

-30r+15

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Find the area of a regular hexagon if given a side length of 1.9 to the square root of 3.

dalvyx [7]

Answer:

28.137 sq. units.

Step-by-step explanation:

Area of a regular hexagon is

$A=\dfrac{3\sqrt{3}}{2}a^2$

where, a is the side length.

It is given that the side length of the regular hexagon is $1.9\sqrt{3}$ .

Substitute $a=1.9\sqrt{3}$ in the above formula.

$A=\dfrac{3\sqrt{3}}{2}(1.9\sqrt{3})^2$

$A=(1.5\sqrt{3})(10.83)$

$A=16.245\sqrt{3}$

$A=28.137$

Therefore, the area of regular hexagon is 28.137 sq. units.

6 0

3 years ago

Airida [17]

Answer:

$308.50

Step-by-step explanation:

3702 ÷ 12 = 308.5 is her monthly payment

5 0

3 years ago

Help Me! I suck with vectors.<br> Look at the attached pictures for the problem.

Westkost [7]

Multiplying a vector $x$ by a constant $k$ gives you a new vector whose components are the same as $x$ , but scaled by $k$ :

$x=\langle x_1,x_2\rangle\implies x=\langle kx_1,kx_2\rangle$

So we have

$u=\langle5,-7\rangle\implies-6u=\langle-30,42\rangle$

$v=\langle-11,3\rangle\implies2v=\langle-22,6\rangle$

Vector addition is just a matter of adding the corresponding components together:

$x=\langle x_1,x_2\rangle,y=\langle y_1,y_2\rangle\implies x+y=\langle x_1+y_1,x_2+y_2\rangle$

Then

$2v-6u=\langle-22,6\rangle+\langle-30,42\rangle=\langle-22-30,6+42\rangle$

$\implies\boxed{2v-6u=\langle-52,48\rangle}$

$\|x\|$ denotes the norm/magnitude of the vector $x$ :

$x=\langle x_1,x_2\rangle\implies\|x\|=\sqrt{{x_1}^2+{x_2}^2}$

We have

$\|2v-6u\|=\sqrt{(-52)^2+48^2}=4\sqrt{313}$

$\implies\boxed{\|2v-6u\|\approx70.77}$

7 0

4 years ago