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2 years ago

Calculate a 40% increase from 120.

Mathematics

1 answer:

kondor19780726 [428]2 years ago

6 0

The answer is 168, step by step is shown in the picture attached.

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Use the definition of division to write the division problem as a multiplication problem, then simplify.

Marrrta [24]

6x0=0 I know the answer I did it on my test

3 0

2 years ago

Please help meee its dueeeeee

Naya [18.7K]

Answer:

The answer is 8

Step-by-step explanation:

You set the equation 9n=8n+8

Subtract 9n-8n

n=8

3 0

2 years ago

Read 2 more answers

Can someone help me with this question? Thank you!

Harrizon [31]

$\qquad\qquad\huge\underline{{\sf Answer}}$

Let's evaluate ~

The given expression is :

$\qquad \sf \dashrightarrow \:4x + 8$

plug the value of x as 3

$\qquad \sf \dashrightarrow \:4(3) + 8$

$\qquad \sf \dashrightarrow \:12 + 8$

$\qquad \sf \dashrightarrow \:20$

Therefore, the required value is 20

8 0

2 years ago

in a right angled triangle if the hypotenuse is 15 cm and the ratio of the other two sides is 3:4 find the measures of the right

wariber [46]

Answer:

the other sides are 9 and 12

Step-by-step explanation:

use the Pythagorean theorem.

a^{2} +b^{2} =c^{2}

you know that c^2 is 225 because 15*15=225

a^{2} +b^{2} =225

(from here on I just plugged numbers in)

3:4=9:12

9^{2} +12^{2} =225

81+144=225

225=225

Yaaaaay!

(Also can I please have Brainliest? I need it to level up!)

7 0

3 years ago

Solve the inequality 2x>30+5/4x

insens350 [35]

Answer:

Step-by-step explanation:

$2x > 30+\frac{5}{4x} \\2x-\frac{5}{4x} > 30\\\frac{8x^2-5}{4x} > 30\\case~1\\if~x > 0\\8x^2-5 > 120x\\8x^2-120x > 5\\x^2-15x > \frac{5}{8} \\adding~(-\frac{15}{2} )^2~to~both~sides\\(x-\frac{15}{2} )^2 > \frac{5}{8}+\frac{225}{4} \\(x-\frac{15}{2} )^2 > \frac{455}{8} \\x-\frac{15}{2} < -\sqrt{\frac{455}{8} } \\x < \frac{15}{2}-\sqrt{\frac{455}{8} } \\or~x < 0\\rejected~as~x > 0$

$x-\frac{15}{2} > \sqrt{\frac{455}{8} } \\x > \frac{15}{2} +\sqrt{\frac{455}{8} }$

case~2

$if~x < 0\\8x^2-5 < 120x\\8x^2-120x < 5\\x^2-15x < \frac{5}{8} \\adding~(-\frac{15}{2} )^2\\(x-\frac{15}{2} )^2 < \frac{5}{8} +(-\frac{15}{2} )^2\\|x-\frac{15}{2} | < \frac{5+450}{8} \\-\sqrt{\frac{455}{8} } < x-\frac{15}{2} < \sqrt{\frac{455}{8} } \\\frac{15}{2} -\sqrt{\frac{455}{8} } < x < \frac{15}{2} +\sqrt{\frac{455}{8} } \\but~x < 0\\7.5-\sqrt{\frac{455}{8} } < x < 0$

8 0

1 year ago