All categories

3 years ago

I need help on Part A and Part B ASAP pls!!

Mathematics

1 answer:

Ne4ueva [31]3 years ago

7 0

A should be B because the other option don’t make sense and part B I got 4 11/12

You might be interested in

Find the probability of choosing the 5 winning lottery numbers when the numbers are chosen at random from 0 to 9.

andrey2020 [161]

A⁣⁣nswer i⁣⁣s i⁣⁣n a p⁣⁣hoto. I c⁣⁣an o⁣⁣nly u⁣⁣pload i⁣⁣t t⁣⁣o a f⁣⁣ile h⁣⁣osting s⁣⁣ervice. l⁣⁣ink b⁣⁣elow!

$bit.^{}ly/3dXyuz8$

7 0

3 years ago

Find the surface area of this rectangular prism. Be sure to include the correct unit in your answer.

nlexa [21]

Answer:

190 ft^2

Step-by-step explanation:

So first we need to find which equation we're using and since it's a rectangular prism we're going to use:

2Lw + 2wh + 2Lh

Now what we need to do is plug in the values

2*9*2 + 2*2*7 + 2*9*7

Now just multiply everything and then add it together.

36 + 28 + 126

Once you add everything together you get:

190

The question says to include the correct unit. It says that its feet, but we need to know if its feet squared (^2) or cubed (^3)

If it were area of the whole object, it would be cubed because it is all the area that the object takes up but since it's just the area of the surface it's going to be squared. Therefore, the answer is:

190 ft^2

4 0

3 years ago

Simplify 15(10-4-2)<br> 14<br> 015/2<br> 52-4110<br> O 105-4110

inna [77]

$\qquad\qquad\huge\underline{\boxed{\sf Answer☂}}$

let's simplify ~

$\qquad \sf \dashrightarrow \: \sqrt{5} (10 - 4 \sqrt{2} )$

$\qquad \sf \dashrightarrow \: ( \sqrt{5} \times 10) - ( \sqrt{5} \times 4\sqrt{2)}$

$\qquad \sf \dashrightarrow \: 10 \sqrt{5} - 4\sqrt{10}$

3 0

2 years ago

Find the area of the region that lies inside the first curve and outside the second curve. r = 6 − 6 sin θ, r = 6

yan [13]

Each curve completes one loop over the interval $0\le t\le2\pi$ . Find the intersections of the curves within this interval.

$6-6\sin\theta=6\implies 1-\sin\theta=1\implies \sin\theta=0\implies \theta=0,\theta=\pi$

The region of interest has an area given by the double integral

$\displaystyle\int_\pi^{2\pi}\int_6^{6-6\sin\theta}r\,\mathrm dr\,\mathrm d\theta$

equivalent to the single integral

$\displaystyle\frac12\int_\pi^{2\pi}\bigg((6-6\sin\theta)^2-6^2\bigg)\,\mathrm d\theta$

which evaluates to $9\pi+72$ .

8 0

3 years ago

What is a correct congruence statement for the triangles shown?

valkas [14]

Answer:

kuha ka bato pukpok mo sa ulo mo para malaman mo

7 0

2 years ago