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3 years ago

list the integers that satisfy both these inequalities 2x+9 0 x > -12 put the answer on the first line

Mathematics

1 answer:

rodikova [14]3 years ago

8 0

Answer:

The Answer is 5

Step-by-step explanation:

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17x25i don't know how to do this

shepuryov [24]

Answer:

17×25 is 425 Because u add 7 and 5 together then u add all of them u in get your answer

6 0

3 years ago

Read 2 more answers

. A regular hexagon is composed of 12 congruent 30 o -60 o -90 o triangles. If the length of the hypotenuse of one of those tria

sergey [27]

see the attached figure to better understand the problem

Let

2L---------> lenght side of hexagon

we know that

in one of those triangles

sin 30=1/2

sin 30=L/hypotenuse

solve for L

L=hypotenuse*sin 30-------->L=( 1 8 √ 3 )*(1 /2)------> L=9√3 units

Find the perimeter

perimeter=(2L)*6--------> L*12-------> Perimeter=9√3*12--------->108√3 units

therefore

the answer is

the perimeter of the regular hexagon is 108√3 units

5 0

4 years ago

PLEASE HELP RIGHT AWAY, <br> THANK U SO MUCH!! <3

vredina [299]

Answer:

1. Ans;

${a}^{7} \times {a}^{12} = {a}^{7 + 12} = {a}^{19}$

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2. Ans;

$\frac{ {h}^{6} }{ {h}^{2} } = {h}^{4}$

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3. Ans;

${e}^{6} \times {e}^{6} = {e}^{6 + 6} = {e}^{12}$

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4.Ans;

$\frac{ {w}^{9} }{ {w} } = {w}^{8}$

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5. Ans;

${g}^{2} \times {h}^{3} \times {g}^{2} \times h = {g}^{2 + 2} \times {h}^{3 + 1} = {g}^{4} {h}^{4}$

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6.Ans;

$\frac{10 {a}^{12} }{ 6{a}^{3} } = \frac{5 {a}^{9} }{3}$

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7.Ans;

$({m}^{4} )^{5} = {m}^{4 \times 5} = {m}^{20}$

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8.Ans;

${(3 {a}^{2} })^{3} = ( {3})^{3} ( {a}^{2} ) ^{3} = 27 {a}^{6}$

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9.Ans;

${( \frac{c}{k} })^{4} = \frac{ {c}^{4} }{ {k}^{4} }$

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10.Ans;

${( \frac{8x}{9} })^{2} = \frac{( {8x})^{2} }{ ({9})^{2} } = \frac{64 {x}^{2} }{81}$

__o_o__

11.Ans;

$4 {c}^{2} \times {c}^{3} = 4 {c}^{2 + 3} = 4 {c}^{5}$

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12.Ans;

${( - 9 {h}^{4} {k}^{6} })^{2} = {( - 9)}^{2} {( {h}^{4} })^{2} {( {k}^{6} })^{2} = 81 \: \: {h}^{8} {k}^{12}$

__o_o__

13.Ans;

$( { \frac{2d}{3e} })^{3} = \frac{ {(2d)}^{3} }{ ({3e)}^{3} } = \frac{8 {d}^{3} }{27 {e}^{3} }$

__o_o__

14. Ans;

${( \frac{4}{ {w}^{5} }) }^{3} = \frac{ ({4)}^{3} }{( {w}^{5} ) ^{3} } = \frac{64}{ {w}^{15} }$

_____o___o_____

$\frac{5 {v}^{4} \times 4 {v}^{5} }{2 {v}^{3} } = \frac{20 {v}^{4 + 5} }{2 {v}^{3} } = \frac{20 {v}^{9} }{2 {v}^{3} } = 10 {v}^{6}$

_o_o_

$( \frac{3 {f}^{4}h}{4 {h}^{9} } ) ^{3} = \frac{( {3})^{3} ( {f}^{4})^{3} ({h})^{3} }{ {(4)}^{3} ( {h}^{9})^{3} } \\ \\ \\ = \frac{27 {f}^{4 \times 3} {h}^{3} }{64 {h}^{9 \times 3} } = \frac{27 {f}^{12} {h}^{3} }{64 {h}^{27} } \\ \\ \\ = \frac{27 {f}^{12} }{64 {h}^{24} }$