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3 years ago

200+25x=150+35x. Can someone please help me

Mathematics

1 answer:

cluponka [151]3 years ago

8 0

The answer is x=5 .

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1 Minutes, 30 seconds

Serggg [28]

I think it would be (15-6)*5 because when it says 5 times you would cross out both B and D but then you have to find out what large as 15 minus 6 which i think would be (15-6) so your answer would be C. (15-6)*5

6 0

3 years ago

Read 2 more answers

SOLVE EQUATION BY TAKING SQUARE ROOTS.<br> -100+81a^2=0

IceJOKER [234]

Answer:

a = ±10/9

Step-by-step explanation:

$-100+81a^2=0\\81a^2=100\\a^2=\frac{100}{81}\\\sqrt{a^2}=\sqrt{\frac{100}{81} }\\a=10/9\\a=-10/9$

7 0

3 years ago

Read 2 more answers

Please help on number 11 just like number 10

Alja [10]

Attachment Unavailable

7 0

3 years ago

3.Find the area of Triangle and Parallelogram having base 13.5cm and height is 18cm.

patriot [66]

Base=13.5cm
Height=18cm

$\\ \sf\longmapsto Area=\dfrac{1}{2}Base\times Height$

$\\ \sf\longmapsto Area=\dfrac{1}{2}(18)(13.5)$

$\\ \sf\longmapsto Area=9(13.5)$

$\\ \sf\longmapsto Area=121.5cm^2$

3 0

3 years ago

The length of the hypotenuse of a right angled triangle exceeds the length of the base by 2cm and exceeds twice the length of th

insens350 [35]

Answer: a = 8; b = 15; c = 17

Step-by-step explanation:

c = b + 2 and c = 2a + 1

b = c − 2 and a = $\frac{c-1}{2}$

c² = b² + a²

$c^{2} = (c-2)^{2} + \frac{c-1}{2}\\$

$c^{2}= \frac{4 (c-2)^{2}+(c - 1)^{2} }{4}$

$4c^{2}={4 (c^{2} + 4 -4c)+c ^{2} + 1-2c$

$4c^{2}={4 c^{2} + 16 -16c)+c ^{2} + 1-2c$

$c^{2}-18c+17=0$

$c(c -17)-1(c-17)=0$

$(c -1)(c-17)=0$

c = 1 or c = 17

If c = 1 then b = 1 - 2 = -1, that's not possible

So x = 17

b = 17 - 2 = 15

a = $\frac{17-1}{2} = 8$

Length sides of the triangle are 17 cm, 15 cm and 8 cm.

I saw some of this on a site but I want to make sure its correct so we're gonna apply the pythagorean theory and see if its correct.

$a^{2} + b^{2} = c^{2} \\8^{2} + 15^{2} = 17^{2} \\64 + 225 = 289\\289 = 289\\$

This shows that this is correct.

Hope this helped! Again, some of this was from a site, not from me.

3 0

3 years ago