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3 years ago

-1 + 6(-1 - 3x) > -39 - 2x. - A.X<2 B. x < 3 C. x <-3 D. x < -10

Mathematics

1 answer:

Mnenie [13.5K]3 years ago

5 0

Answer:

x < 2

Step-by-step explanation:

→ Expand out brackets

-1 - 6 - 18x > -39 - 2x

→ Simplify

- 7 -18x > -39 - 2x

→ Add -7 to both sides

-18x > -32 - 2x

→ Add -2x to both sides

-16x > -32

→ Solve for x

x < 2

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valentina_108 [34]

By comparing all the options with the formula of sum of cubes and difference of cubes we get that $(x+4)\left(x^{2}-4 x+16\right)$ is sum of cubes

<h3>What is formula of sum of cubes and difference of cubes?</h3>

Formula of sum of cubes and difference of cubes are following

$(a+b)(a^2 -ab +b^2) = a^3+ b^3 \\\\\\(a - b)(a^2+ ab+ b^2) = a^3 - b^3$

(a)

$(x-4) \left(x^{2}+4 x-16\right)\\=(x-4) \left(x^{2}+4 x-4^2\right)\\\\$

by comparison with formula we can say that this is neither sum or nor difference of cubes.

(b)

$(x-1)\left(x^{2}-x+1\right)\\=(x-1)\left(x^{2}-x+1^2\right)$

by comparison with formula we can say that this is neither sum nor difference of cubes.

(c)

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

we can say that this is neither sum nor difference of cube as it's difference of squares.

(d)

$(x+4)\left(x^{2}-4 x+16\right)\\\\(x+4)\left(x^{2}-4 x+4^2\right)$

by comparison with formula this is equal to

$x^{3}+4^{3}\\$

Hence this is sum of cubes

(e)

$(x+4)\left(x^{2}+4 x+16\right)\\(x+4)\left(x^{2}+4 x+4^2\right)$

by comparison with formula we can say that this is neither sum nor difference of cubes.

By comparing all the options with the formula of sum of cubes and difference of cubes we get that $(x+4)\left(x^{2}-4 x+16\right)$ is sum of cubes

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2 years ago

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The school originally used $20 for the purchase of the candy.
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Sub in the 15 bars for t
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Therefore the school made $10, because they earned an extra $10 in comparison to the amount they bought the candy bars for, which was $20.
Hope this helps!

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3 years ago

In the graph below, Point A represents Owen's house, Point B represents David's house and Point C represents the school. Who liv

ycow [4]

Answer:

• David

• 4 miles

Explanation:

In the graph:

The given locations are:

• Owen's House, A(11,3)

• David's House, B(15,13)

• School, C(3,18)

We determine both Owen's and David's distance from the school using the distance formula.

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Owen's distance from school (AC)

$\begin{gathered} AC=\sqrt[]{(3-11)^2+(18-3)^2} \\ =\sqrt[]{(-8)^2+(15)^2} \\ =\sqrt[]{64+225} \\ =\sqrt[]{289} \\ AC=17\text{ miles} \end{gathered}$

David's distance from school (BC)

$\begin{gathered} BC=\sqrt[]{(3-15)^2+(18-13)^2} \\ =\sqrt[]{(-12)^2+(5)^2} \\ =\sqrt[]{144+25} \\ =\sqrt[]{169} \\ BC=13\text{ miles} \end{gathered}$

We see from the calculations that David lives closer to the school, and by 4 miles.

The graph below is attached for further understanding: