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Anika [276]
3 years ago
6

Can someone help? Explain and give work in how you got the answer.

Mathematics
1 answer:
kvasek [131]3 years ago
4 0

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

We know the side relations of a right angled Triangle (that is Pythagoras therom)

now, let the missing side be x

  • Hypotenuse² = sum of squares of other legs

that is ~

  • 15 {}^{2}  = 5 {}^{2}  +  {x}^{2}

  • 225 = 25 +  {x}^{2}

  • {x}^{2}  = 225 - 25

  • {x}^{2}  = 200

  • x =  \sqrt[]{200}

  • x = 10 \sqrt{2}

The correct choice is ~ B

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HELP DUE TODAY!
olya-2409 [2.1K]

Answer:

25

Step-by-step explanation:

2 + 2+ 2 + 2 = 8

3 + 3 + 3 = 9

4 + 4 = 8

8

8

9

____+

25

3 0
3 years ago
Find the y-intercept of the line on the graph.
BARSIC [14]
It's a little hard to see, but I'm almost positive the y intercept is -1.
 
Remember, the y intercept is wherever the line intercepts in the y axis 0.

Good luck my man.
5 0
3 years ago
Read 2 more answers
I own a large truck, and my neighbor owns four small trucks that are all identical. My truck can carry a load of at least $600$
erastova [34]

Answer:

  there is no greatest load

Step-by-step explanation:

Let x and y represent the load capacities of my truck and my neighbor's truck, respectively. We are given two relations:

  x ≥ y +600 . . . . . my truck can carry at least 600 pounds more

  x ≤ (1/3)(4y) . . . . . my truck carries no more than all 4 of hers

Combining these two inequalities, we have ...

  4/3y ≥ x ≥ y +600

  1/3y ≥ 600 . . . . . . . subtract y

  y ≥ 1800 . . . . . . . . multiply by 3

My truck's capacity is greater than 1800 +600 = 2400 pounds. This is a lower limit. The question asks for an <em>upper limit</em>. The given conditions do not place any upper limit on truck capacity.

4 0
3 years ago
Read 2 more answers
What’s the answer?and how do you get it
Kay [80]

the solid is made up of 2 regular octagons, 8 sides, joined up by 8 rectangles, one on each side towards the other octagonal face.

from the figure, we can see that the apothem is 5 for the octagons, and since each side is 3 cm long, the perimeter of one octagon is 3*8 = 24.

the standing up sides are simply rectangles of 8x3.

if we can just get the area of all those ten figures, and sum them up, that'd be the area of the solid.

\bf \textit{area of a regular polygon}\\\\ A=\cfrac{1}{2}ap~~ \begin{cases} a=apothem\\ p=perimeter\\[-0.5em] \hrulefill\\ a=5\\ p=24 \end{cases}\implies A=\cfrac{1}{2}(5)(24)\implies \stackrel{\textit{just for one octagon}}{A=60} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{two octagon's area}}{2(60)}~~+~~\stackrel{\textit{eight rectangle's area}}{8(3\cdot 8)}\implies 120+192\implies 312

7 0
4 years ago
What is the unit value of the 6 in 216?
eduard
The 6 is in the ones place the 1 is in the tens place and the 2 is in the hundreds place.
8 0
3 years ago
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