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2 years ago

Help me pls I need this grade

Mathematics

2 answers:

Ivahew [28]2 years ago

5 0

yes lol bigger hah trigger lol susush

sladkih [1.3K]2 years ago

5 0

Yes if the circle is bigger than the quadrilateral.

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Solve for x: −4x+6(6−2x)=−7x−36

ValentinkaMS [17]

Answer:

x = 8

Step-by-step explanation:

-4x + 6(6 - 2x) = -7x - 36

-4x + 36 - 12x = -7x - 36

-16x + 36 = -7x - 36

+36 +36

-16x + 72 = -7x

+16x +16x

72 = 9x

divide by 9

x = 8

4 0

2 years ago

Read 2 more answers

13/26 = x/24 solve for x PLEASE ANSWER

EastWind [94]

Answer:

x= 12

Step-by-step explanation:

hope this helps!!

3 0

2 years ago

Square root 15*3sqyare root 81y

Deffense [45]

No simplification exists, Root remains :
sqrt(15y)

7 0

3 years ago

X+6=x can you help please thank you

Anna35 [415]

x + 6 = x

there is literally no solution.

7 0

2 years ago

What's the answer for this question? (The numbers after the letters are indexes btw) 27a9 x 18b5 x 4c2 Over 18a4 x 12b2 x 2c

zysi [14]

Answer:

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c} = \frac{9}{2}a^5b^3c$

Step-by-step explanation:

Given

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c}$

Required

Simplify

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c}$

Cancel out 18

$\frac{27a^9 * b^5 * 4c^2 }{a^4 * 12b^2 * 2c}$

Divide 4 and 2

$\frac{27a^9 * b^5 * 2c^2 }{a^4 * 12b^2 *c}$

Divide 27 and 12 by 3

$\frac{9a^9 * b^5 * 2c^2 }{a^4 * 4b^2 *c}$

Apply law of indices

$\frac{9a^{9-4} * b^{5-2} * 2c^{2-1} }{4}$

$\frac{9a^5 * b^3 * 2c }{4}$

Divide 2 and 4

$\frac{9a^5 * b^3 * c}{2}$

$\frac{9a^5b^3c}{2}$

Rewrite as:

$\frac{9}{2}a^5b^3c$

Hence:

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c} = \frac{9}{2}a^5b^3c$

4 0

2 years ago