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2 years ago

I need help with #4 and #5

Mathematics

1 answer:

GrogVix [38]2 years ago

6 0

I can’t readdddddddddddddddd send it back

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The area of a rectangular garden can be expressed as A = 2x^2 -x -6. The width of the garden is 2x+3. Find an expression for the

zalisa [80]

Answer:

Step-by-step explanation:

2x² -x - 6 = 2x² - 4x + 3x - 3*2

= 2x*(x - 2) + 3*(x - 2)

= (x-2)(2x + 3)

A = 2x² -x - 6

A = (x - 2)(2x + 3)

length*width = (x - 2)(2x + 3)

length * (2x +3) = (x - 2)(2x + 3)

$length=\frac{(x-2)(2x+3)}{(2x+3)}$

length = x - 2

5 0

2 years ago

Highest place value for 54

zvonat [6]

Answer:

tens=5

Step-by-step explanation:

4 0

2 years ago

Select the correct answer.

erica [24]

Answer:

there is 1/2 a chance you will roll an odd number.

8 0

2 years ago

Read 2 more answers

Line m passes through points (-20, 5) and (-4,7). Line n passes through points (-5,5) and (7,4). Are lines m and n perpendicular

LuckyWell [14K]

Answer:b

Step-by-step explanation: it’s right because it is

5 0

2 years ago

Simplify. (Assume all variables represent positive real numbers). Leave answer in radical form.

Flauer [41]

Answer:

$\sqrt{128a^{6}b^{13}} = 8 a^{3} b^{6} \sqrt{2b}$

Step-by-step explanation:

Given

$\sqrt{128a^{6}b^{13}}$

Required

Solve

$\sqrt{128a^{6}b^{13}}$

The expression can be split to:

$\sqrt{128a^{6}b^{13}} = \sqrt{128} * \sqrt{a^{6}} * \sqrt{b^{13}}$

$\sqrt{128a^{6}b^{13}} = \sqrt{64 * 2} * \sqrt{a^{6}} * \sqrt{b^{13}}$

$\sqrt{128a^{6}b^{13}} = \sqrt{64} * \sqrt{2} * \sqrt{a^{6}} * \sqrt{b^{13}}$

$\sqrt{128a^{6}b^{13}} = \sqrt{64} * \sqrt{2} * \sqrt{a^{6}} * \sqrt{b^{12 + 1}}$

$\sqrt{128a^{6}b^{13}} = \sqrt{64} * \sqrt{2} * \sqrt{a^{6}} * \sqrt{b^{12}} * \sqrt{b}$

So, we have:

$\sqrt{128a^{6}b^{13}} = 8 * \sqrt{2} * a^{6/2} * b^{12/2} * \sqrt{b}$

$\sqrt{128a^{6}b^{13}} = 8 * \sqrt{2} * a^{3} * b^{6} * \sqrt{b}$

Rewrite as:

$\sqrt{128a^{6}b^{13}} = 8 * a^{3} * b^{6}* \sqrt{2} * \sqrt{b}$

$\sqrt{128a^{6}b^{13}} = 8 a^{3} b^{6}* \sqrt{2b}$

$\sqrt{128a^{6}b^{13}} = 8 a^{3} b^{6} \sqrt{2b}$

5 0

2 years ago