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Oksanka [162]
3 years ago
12

Find b for the following right angled triange where a=5 and c=13

Mathematics
1 answer:
lilavasa [31]3 years ago
4 0
You would use the Pythagorean there on which is a^2+b^2=c^2, from there you plug int what they gave you and solve for b, and b=12
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What is 9x8+43 free brainliest
quester [9]

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115

Step-by-step:

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2 years ago
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Solve 2x2 + 20x = −38. (1 point)
RoseWind [281]

Answer:

The solutions are x=-5+\sqrt{6}  and x=-5-\sqrt{6}


Step-by-step explanation:

we have

2x^{2} +20x=-38

Divide by 2 both sides

x^{2} +10x=-19 ------> x^{2} +10x+19=0

we know that


The formula to solve a quadratic equation of the form ax^{2} +bx+c=0 is equal to


x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}


in this problem we have


x^{2} +10x+19=0

so


a=1\\b=10\\c=19


substitute

x=\frac{-10(+/-)\sqrt{10^{2}-4(1)(19)}}{2(1)}


x=\frac{-10(+/-)\sqrt{100-76}}{2}


x=\frac{-10(+/-)\sqrt{24}}{2}


x=\frac{-10(+/-)2\sqrt{6}}{2}


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x2=\frac{-10(-)2\sqrt{6}}{2}=-5-\sqrt{6}


8 0
3 years ago
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Stodoła składa się z 1259 cegieł. Druga stodoła ma o 347 cegieł mniej. Trzecia ma o 567 cegieł więcej od pierwszej stodoły a czw
mel-nik [20]

Answer:

Second barn: 912 bricks

Third barn: 1826 bricks

Fourth barn: 1402 bricks

Step-by-step explanation:

(English question: The barn consists of 1259 bricks. The second barn has 347 bricks less. The third has 567 bricks more than the first barn and the fourth barn has 490 bricks more than the second barn. How many bricks consists of the second, third and fourth barn.)

If the second barn has 347 bricks less than the first barn, so the second barn has:

1259 - 347 = 912 bricks

The third barn has 567 bricks more than the first barn, so the third barn has:

1259 + 567 = 1826 bricks

The fourth barn has 490 bricks more than the second barn, so the fourth barn has:

912 + 490 = 1402 bricks

8 0
3 years ago
GIVEN THAT TITAN HAS A RADIUS OF 2575 KM WHICH IS COVERED BY 600 KM ATMOSPHERE. WHAT
Ber [7]

Answer:

  87.4%

Step-by-step explanation:

The radius to the top of the atmosphere as a fraction of the moon's radius is ...

  (2575 +600)/2575 ≈ 1.23301

The ratio of the moon's volume with atmosphere to the moon's volume without is the cube of this, or 1.87456

Then the ratio of the atmosphere's volume to the moon's volume is ...

  (1.87456 -1)/1 = 0.87456

Atmospheric haze is about 87.4% of the moon's volume.

_____

We have assumed that the desired ratio is to the solid moon's volume, not the volume of moon and atmosphere together. The latter ratio would be 0.875 to 1.875 or about 46.7%.

3 0
4 years ago
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