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3 years ago

I need help on this.

Mathematics

2 answers:

ch4aika [34]3 years ago

8 0

The right answer is 6.

Please see the attached picture for full solution.

Hope it helps...

Good luck on your assignment

kirza4 [7]3 years ago

7 0

Answer:

i believe the answer is 6

Step-by-step explanation:

comment if u want an explanation cus i cba to write one for no reason

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Please help me. I suck at math

serg [7]

Answer:

Answer is ...Obtuse

Step-by-step explanation:

$\frac{m$

$\frac{5x + 4}{2} = \frac{3}{2} x + 21$

$\frac{5x + 4}{2} = \frac{3x + 42}{2}$

$5x + 4 = 3x + 42$

$5x - 3x = 42 - 4$

$2x = 38$

$x = 19$

$substituting \: x = 19 \: \:in \: \: \: \: \: \: \: \\ m$

We get,

m<ABC =

$(5 \times 19) + 4 = 99$

Hence <ABC is obtuse

4 0

3 years ago

Which expression is equivalent to (4g3h2k4)3

RSB [31]

Answer:

d) $8g^{6}h^{4} k^{12} - (h^{25} k^{15} )$

$\frac{(4g^{3} h^{2}k^{4} )^{3} }{8g^{3}h^{2} } - (h^{5} k^{3} )^{5}$ $= 8g^{6}h^{4} k^{12} - (h^{25} k^{15} )$

Step-by-step explanation:

Explanation

Given expression

= $\frac{(4g^{3} h^{2}k^{4} )^{3} }{8g^{3}h^{2} } - (h^{5} k^{3} )^{5}$

By using

(ab)ⁿ = aⁿbⁿ

$\frac{a^{m} }{a^{n} } = a^{m-n}$

= $\frac{(4)^{3} g^{9} h^{6}k^{12} ) }{8g^{3}h^{2} } - (h^{5} k^{3} )^{5}$

After simplification , we get

$= 8g^{9}g^{-3} h^{6} h^{-2} k^{12} - (h^{5} k^{3} )^{5}$

$= 8g^{9-3}h^{6-2} k^{12} - (h^{5} k^{3} )^{5}$

$= 8g^{6}h^{4} k^{12} - (h^{25} k^{15} )$

3 0

2 years ago

Please help with number 6 ASAP I will mark branlist

DerKrebs [107]

Answer:

<h2>12</h2><h2>Step-by-step explanation:</h2>

$4\sqrt{9}\\\\\mathrm{Factor\:the\:number:\:}\:9=3^2\\=4\sqrt{3^2}\\\\\mathrm{Apply\:radical\:rule}:\quad \sqrt{a^2}=a,\:\quad \:a\ge 0\\\sqrt{3^2}=3\\\\=4\times\:3\\\\=12$

5 0

3 years ago

Where would you insert parentheses to make the statement true?

otez555 [7]

No you had it right, it is around the 18 / 3 and the 5 + 1

6 0

3 years ago

Read 2 more answers

Whats the square root of 39 to the nearest whole

Rasek [7]

The square root of 39 is 6.224

5 0

3 years ago

I need help on this.​

I need help on this.