Which set of sides will make a triangle?10cm, 5cm, 9cm
10cm, 3cm, 4cm
1cm, 3cm, 1cm
5cm, 2cm, 3cm

Answers

Answer 1

Answer:

Answer:

10cm 5cm and 9cm

Step-by-step explanation:

just try this

take any two measurements and add them together if they are larger than the 3rd measurement it will work and all you have to do is do that to each measurement

10+5 is greater than 9

9+5 is greater than 10

9+10 is greater than 5

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What's an extraneous solution?

In your own words

Answers

Answer:

You get this solution while solving your problem but it still is not the correct answer to your question...

(I think)

Line A: x + y = 2Line B: 2x + y = 4

Which statement is true about the solution to the set of equations? (5 points)

There are infinitely many solutions.
There are two solutions.
There is one solution.
There is no solution.

Answers

If you would like to know which statement is true about the solution to the set of equations, you can calculate this using the following steps:

x + y = 2 ... y = 2 - x
2x + y = 4
_________________
2x + (2 - x) = 4
2x - x = 4 - 2
x = 2

y = 2 - x = 2 - 2 = 0

(x, y) = (2, 0)

The correct result would be: There is one solution.

AEFGH is a rhombus. Given EG = 22 and FH = 20, what is the length of one side of the rhombus?
Please help :/

Answers

Applying the properties of a rhombus and the Pythagorean Theorem, the length of one side of the rhombus is: 14.9

Recall:

The diagonals of a rhombus bisect each other at right angles, thereby forming 4 right triangles.

Half of a diagonal and half of the other diagonal make up a right triangle.

Thus, given:

EG = 22 (diagonal)
FH = 20 (diagonal).

Find the length of one side using Pythagorean theorem as shown below:

$HG = \sqrt{((1)/(2)EG)^2 + ((1)/(2)FH)^2 } \n\n$

Substitute

$HG = \sqrt{((1)/(2) * 22)^2 + ((1)/(2) * 20)^2 } \n\nHG = √(11^2 + 10^2) \n\n\mathbf{HG = 14.9}$

Therefore, applying the properties of a rhombus and the Pythagorean Theorem, the length of one side of the rhombus is: 14.9

Learn more here:

brainly.com/question/24618146

How do you solve 13y^2-5yx-8x^2

Answers

Answer:

(13y+8)(y-x)

Step-by-step explanation:

13 $y^(2)$ - 5yx -8 $x^(2)$

we can solve above equation by using splitting the middle term

we can factorize it by splitting the middle term

13 $y^(2)$ - 5yx -8 $x^(2)$

=13 $y^(2)$ - 13yx+8yx -8 $x^(2)$

=13y(y-x)+8x(y-x)

=(13y+8)(y-x)

There are the two factors because the degree of the polynomial is 2

Hence, the correct answer is (13y+8)(y-x)

Step 1 :Simplify 13y2-5yx - 23x2
1.1 Factoring 13y2 - 5yx - 8x2 Try to factor this multi-variable trinomial using trial and error Found a factorization : (y - x)•(13y + 8x)
Final result : (y - x) • (13y + 8x)

This figure shows the nose cone of a rocket used for launching satellites. The nose cone houses the satellite until the satellite is placed in orbit. What is the volume of the nose cone?45 cubic feet
60 cubic feet
90 cubic feet
180 cubic feet

Answers

see the attached figure to better understand the problem

we know that

The volume of the cone is equal to

$V=(1)/(3) \pi r^(2) h$

in this problem

$r=(6)/(2) =3\ ft\n \n h=15\ ft$

Substitute the values in the formula above

$V=(1)/(3)* \pi* 3^(2)* 15\n \n V=45\pi \ ft^(3)$

therefore

the answer is

The volume of the nose cone is $45\pi \ ft^(3)$

The formula to get the volume of a cone is
V = (1/3)πr²h
where V is the volume
r is the radius at the base
h is the height

No values are given but simple substitution should give the correct answer.

Jane has been growing two bacteria farms. Bacteria farm Rod has a starting population of 2 bacteria, while Bacteria farm Sphere has a starting population of 8 bacteria. However, Jane starts growing Rod five hours before she starts growing Sphere. At 8 p.m., Jane checks her farms and finds that they have the exact same population. If the population of Rod doubles every hour, but the population of Sphere is quadrupled every hour, how many hours ago did she start growing Sphere?