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1 year ago

Whats the answer to a(4a + 3)

Mathematics

2 answers:

NemiM [27]1 year ago

8 0

Answer:

4a² +3a

Step-by-step explanation:

4a x a =4a²

3 x a= 3a

Shtirlitz [24]1 year ago

3 0

Answer: The answer to that question would be 4a^2+3a

Step-by-step explanation:

Multiply using the distributive property.

Search : what's the answer to a(4a + 3)

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What is the greatest common factor of 24, 40 and 32? A. 1 B. 2 C. 4 D. 8

trasher [3.6K]

It’s 8 because 24/8 is 3 and 40/8 is 5 and 32/8 is 4 so the answer is 8 YOUR WELCOME

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2 years ago

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Evaluate cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.

Andre45 [30]

Answer:

Option d) 5 to the power of negative 5 over 6 is correct.

$\dfrac{\sqrt[3]{\bf 5} \times \sqrt{\bf 5}}{\sqrt[3]{\bf 5^{\bf 5}}}= 5^{\frac{\bf -5}{\bf 6}}$

Above equation can be written as 5 to the power of negative 5 over 6.

ie, $5^\frac{\bf -5}{\bf 6}$

Step-by-step explanation:

Given that cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.

It can be written as below

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}} \times 5^{\frac{1}{2}}}{5^{\frac{5}{3}}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}+\frac{1}{2}}}{5^{\frac{5}{3}}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{2+3}{6}}}{5^{\frac{5}{3}}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5}{6}} \times 5^{\frac{-5}{3}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5-10}{6}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{5^5}= 5^{\frac{-5}{6}}$

Above equation can be written as 5 to the power of negative 5 over 6.

7 0

3 years ago

Liam had \$250$250dollar sign, 250. Then, he and his classmates bought a present for their teacher, evenly splitting the \$p$pdo

klasskru [66]

Answer:

250-120/e

Step-by-step explanation:

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2 years ago

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Which equation represents a line which is parallel to the line 2x+y=-8?A. y=− 1/2 x−4B. y=−2x+5C. y=2x+3D. y= 1/2x+2

katrin2010 [14]

Two lines that are parallel have the same slope. In its slope-intersect form, we can write the equation of a line with slope m and y-intercept b as:

$y=mx+b$

Step 1

Write the given equation in slope-intercept form and identify its slope m.

$\begin{gathered} 2x+y=-8 \\ \\ 2x+y-2x=-2x-8 \\ \\ y=-2x-8 \end{gathered}$

Thus:

$m=-2$

Step 2

Find the equation with the same slope m = -2. We need to identify which of them has -2 multiplying the variable x.

Answer

From the given options, the only one with the same slope m = -2, therefore parallel to the given line, is:

$y=-2x+5\text{ (option B)}$

4 0

10 months ago

Help with my algebra homework.

just olya [345]

Q1. The answer is $\frac{(x-4)(x-4)}{(x+3)(x+1)}= \frac{ x^{2}-4x-4x+16}{ x^{2} +x+3x+3} = \frac{ x^{2} -8x+16}{ x^{2} +4x+3}$
$\frac{ x^{2} -16}{ x^{2} +5x+6} / \frac{ x^{2} +5x+4}{ x^{2} -2x-8} = \frac{ x^{2} -16}{ x^{2} +5x+6}* \frac{x^{2} -2x-8}{ x^{2} +5x+4}$
Now, factorise the numerators and denominators:
x² - 16 = x² - 4² = (x + 4)(x - 4)
x² + 5x + 6 = x² + 2x + 3x + 2*3 = x(x+2) + 3(x+2) = (x + 2)(x + 3)
x² - 2x - 8 = x² + 2x - 4x - 2*4 = x(x+2) - 4(x+2) = (x + 2)(x - 4)
x² + 5x + 4 = x² + x + 4x + 4*1 = x(x+1) + 4(x+1) = (x + 1)(x + 4)

$\frac{ x^{2} -16}{ x^{2} +5x+6}* \frac{x^{2} -2x-8}{ x^{2} +5x+4}= \frac{(x+4)(x-4)}{(x+2)(x+3)} * \frac{(x+2)(x-4)}{(x+1)(x+4)}$
Now, cancel out some factors:
$\frac{(x+4)(x-4)}{(x+2)(x+3)} * \frac{(x+2)(x-4)}{(x+1)(x+4)}= \frac{(x-4)(x-4)}{(x+3)(x+1)}= \frac{ x^{2}-4x-4x+16}{ x^{2} +x+3x+3} = \frac{ x^{2} -8x+16}{ x^{2} +4x+3}$

Q2. The answer is $\frac{7(a-7)}{(a-8)(a+8)}$
Since a² - b² = (a-b)(a+b), then a²- 64 = a² - 8² = (a-8)(a+8).
$\frac{7}{a+8} + \frac{7}{ a^{2} -64} = \frac{7}{a+8} + \frac{7}{ (a+8)(a-8)}= \frac{7(a-8)}{(a+8)(a-8)} + \frac{7}{ (a+8)(a-8)}= \frac{7(a-8)+7}{ (a+8)(a-8)}$
$= \frac{7(a-8)+7*1}{(a+8)(a-8)} =\frac{7(a-8+1)}{(a+8)(a-8)} =\frac{7(a-7)}{(a+8)(a-8)}$

Q3. The answer is $\frac{7(3a-4)}{(a-6)(a+8)}$
$\frac{ a^{2} -2a-3}{ a^{2}-9a+18 }- \frac{a^{2} -5a-6}{ a^{2}+9a+8 } = \frac{a^{2}+a-3a-3*1}{a^{2}-3a-6a+3*6} - \frac{a^{2}-a-6a-6*1}{a^{2}+a+8a+8*1}$
$= \frac{a(a+1)-3(a+1)}{a(a-3)-6(a-3)}- \frac{a(a+1)-6(a+1)}{a(a+1)+8(a+1)}= \frac{(a+1)(a-3)}{(a-6)(a-3)} - \frac{(a+1)(a-6)}{(a+1)(a+8)}$
Now, cancel out some factors:
$\frac{(a+1)(a-3)}{(a-6)(a-3)} - \frac{(a+1)(a-6)}{(a+1)(a+8)}= \frac{a+1}{a-6} - \frac{a-6}{a+8}$
$\frac{a+1}{a-6} - \frac{a-6}{a+8}= \frac{(a+1)(a+8)}{(a-6)(a+8)} -\frac{(a-6)(a-6)}{(a-6)(a+8)} =\frac{(a+1)(a+8)-(a-6)(a-6)}{(a-6)(a+8)}$
$= \frac{ a^{2} +9a+8- a^{2} +12-36}{(a-6)(a+8)} =\frac{9a+8+12-36}{(a-6)(a+8)} =\frac{21a-28}{(a-6)(a+8)} =\frac{7(3a-4)}{(a-6)(a+8)}$

Q4. The answer is $\frac{4x}{(x+3)(1+3x)}=\frac{4x}{ x^{2} +10x+3}$
$\frac{4}{x+3} / (\frac{1}{x}+3 )=\frac{4}{x+3} / (\frac{1}{x}+ \frac{3x}{x})=\frac{4}{x+3} / (\frac{1+3x}{x})= \frac{4}{x+3} * \frac{x}{1+3x} = \frac{4x}{(x+3)(1+3x)}$
$\frac{4x}{(x+3)(1+3x)}= \frac{4x}{x+3 x^{2} +3+9x}= \frac{4x}{ x^{2} +10x+3}$

Q5. The answer is x = 6
$\frac{-2}{x} +4= \frac{4}{x} +3$
$4-3= \frac{4}{x}- \frac{-2}{x}$
$1 = \frac{4-(-2)}{x}$
$1= \frac{4+2}{x}$
$1= \frac{6}{x}$
$x = 6$
Let's check the solution:
Since: $\frac{-2}{x} +4= \frac{4}{x} +3$
Then: $\frac{-2}{6}+4 = \frac{4}{6} +3$
$- \frac{1}{3}+ \frac{4*3}{3}= \frac{2}{3} + \frac{3*3}{3}$
$- \frac{1}{3} + \frac{12}{3} = \frac{2}{3} + \frac{9}{3}$
$\frac{-1+12}{3} = \frac{2+9}{3}$
$\frac{11}{3} = \frac{11}{3}$
Thus, the solution is correct

6 0

2 years ago