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Morgarella [4.7K]

3 years ago

13

Find the perimeter and the area of the figure. Round answers to the nearest tenth.

1 answer:

vlada-n [284]3 years ago

7 0

Answer:

34 rounded is 30

Step-by-step explanation:

15 plus 15 is 30 30 plus 4 is 34

You might be interested in

According to the table, what was the hiker’s total displacement?

shtirl [24]

Answer:

A: 0

Step-by-step explanation:

Important to remember - total displacement is the direct distance between the start and finish points, and isn't always equal to the total distance travelled.

In the case of the table above, the hiker travels the same distance in opposite directions, which actually means they cancel each other out! Traveling 4 km to the west, then 4 km to east means you end up in the same spot, right? So your total displacement is 0! (or 4 - 4, which equals 0). Same happens with the north-south hikes - the hiker ends up in the same spot, as the displacement is equal to 6 - 6 = 0.

So, the total displacement of the hiker is 0. (Even though the hiker travelled 20 km!)

7 0

3 years ago

Pig 1 worked on his house 3 hours for a day for 10 days.How long did he work

WITCHER [35]

He worked 30 hours in total

3 0

3 years ago

99999999999999999+99999999999999

lina2011 [118]

Answer:

1.001e+17

Step-by-step explanation:

8 0

3 years ago

Troyanec [42]

Answer:

Step-by-step explanation:

Lateral surface area of the triangular prism = Perimeter of the triangular base × Height

By applying Pythagoras theorem in ΔABC,

AC² = AB² + BC²

(34)² = (16)² + BC²

BC = $\sqrt{1156-256}$

= $\sqrt{900}$

= 30 in.

Perimeter of the triangular base = AB + BC + AC

= 16 + 30 + 34

= 80 in

Lateral surface area = 80 × 22

= 1760 in²

Total Surface area = Lateral surface area + 2(Surface area of the triangular base)

Surface area of the triangular base = $\frac{1}{2}(\text{Base})(\text{Height})$

= $\frac{1}{2}(30)(16)$

= 240 in²

Total surface area = 1760 + 2(240)

= 1760 + 480

= 2240 in²

Volume = Area of the triangular base × Height

= 240 × 20

= 4800 in³

6 0

3 years ago

The polynomial x^2+3x-1 is a factor of x^4+3x^3-2x^2-3x+1 true or false?

pashok25 [27]

Answer:

Yes, it is true that $x^2+3x-1$ is a factor of $x^4+3x^3-2x^2-3x+1$ .

Step-by-step explanation:

Let us try to factorize $x^4+3x^3-2x^2-3x+1$

$x^4+3x^3-2x^2-3x+1\\\Rightarrow x^4-2x^2+1-3x+3x^3$

Let us try to make a whole square of the given terms:

$\Rightarrow (x^2)^2-2\times x^2 \times 1+1^2+3x^3-3x\\\Rightarrow (x^2-1)^2+3x^3-3x\\$

--------------

Formula used above:

$a^{2} -2 \times a \times b +b^{2} = (a-b)^2$

In the above equation, we had $a = x, b = 1$ .

--------------

Further solving the above equation, taking $3x$ common out of $3x^3-3x$

$\Rightarrow (x^2-1)^2+3x(x^2-1)\\$

Taking $(x^{2} -1)$ common out of the above term:

$\Rightarrow (x^2-1)((x^2-1)+3x)\\\Rightarrow (x^2-1)(x^2+3x-1)$

So, the two factors are $(x^2-1)\ and\ (x^2+3x-1)$ .

$\therefore$ The statement that $x^2+3x-1$ is a factor of $x^4+3x^3-2x^2-3x+1$ is <em>True.</em>

7 0

3 years ago