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3 years ago

Factor out the gcf from each polynomial 30x^2-18x^3+6x^4

Mathematics

1 answer:

inn [45]3 years ago

4 0

Answer:

Step-by-step explanation:

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What’s the answer for r

natka813 [3]

Answer:

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3 years ago

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Tony spent $60.54 and bought 3 packs of notebooks and a calculator. How much does a pack of notebooks cost if a calculator costs

Cerrena [4.2K]

Let the price of a notebook = X.

Multiply the cost of a notebook by the quantity, so you get 3x.

Add that to the price of a calculator to get the total spent:

60.54 = 45.87 + 3x

Now solve for x:

Subtract the cost of the calculator from both sides:

3x = 14.67

Divide both sides by 3:

x = 14.67 / 3

x = 4.89

A notebook cost $4.89

8 0

3 years ago

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Q7Which statement is incorrect for the graph of thefunction y = -5 cosGcos( 3 (x – 3)) +62The period is 4.The amplitude is 3.The

guajiro [1.7K]

The given function is,

$y=-5\cos (\frac{\pi}{2}(x-3))+6$

The graph can be drawn as,

The period will be distance between two consequetive maximas,

$\text{Period}=1-(-3)=4$

Thus, period is correct.

The amplitude can be determined as,

$\begin{gathered} Amplitude=\frac{\max -\min }{2} \\ =\frac{11-1}{2} \\ =5 \end{gathered}$

Thus, amplitue is incorrect.

The range is,

$y\in\lbrack1,11\rbrack$

Thus, the range is correct.

The midline is,

$\begin{gathered} \text{midline}=\frac{\max +\min }{2} \\ =\frac{1+11}{2} \\ =6 \end{gathered}$

Thus, the midline is correct.

Thus, option (b) is the solution.

5 0

1 year ago

Solve x and y please need it for tomorrow

Evgen [1.6K]

Solve for x by cross multiplying then isolating the variable:
$\frac{9}{15}= \frac{x}{10}$ cross multiply
15x=90 Isolate the variable
x=6
Solve for y by cross multiplying then isolating the variable:
$\frac{9}{15} = \frac{6}{y}$ cross multiply
9y=90 Isolate the variable
y=10

4 0

3 years ago

Read 2 more answers

Laws Of Indicies<br>Can someone please help me with this?

nasty-shy [4]

Answer:

See below

Step-by-step explanation:

$1) : {a}^{5} {b}^{4} {c}^{3} \div {a}^{2} {b}^{3} c \\ \\ = {a}^{5 - 2} {b}^{4 - 3} {c}^{3 - 1} \\ \\ = {a}^{3} {b}^{1} {c}^{2} \\ \\ = {a}^{3} {b} {c}^{2} \\ \\ \\2)\: (x^2 y^3)^3\\\\= x^{2\times 3}y^{3\times 3}\\\|= x^{6}y^{9}\\\\\\3)\:5 {a}^{4} \div ( {a}^{2} \times 3a) \\ \\ = 5 {a}^{4} \div (3 {a}^{2 + 1} ) \\ \\ = 5 {a}^{4} \div (3 {a}^{3} ) \\ \\ = \frac{5}{3} {a}^{4 - 3} \\ \\ = \frac{5}{3} {a} \\ \\ \\ 4)\: \frac{14 {a}^{5} }{2 {a}^{3} \times 7 {a}^{4} } \\ \\ = \frac{14 {a}^{5} }{2 \times 7 {a}^{3 + 4} } \\ \\ = \frac{ \cancel{14} {a}^{5} }{ \cancel{14}{a}^{3 + 4} } \\ \\ = \frac{ {a}^{5} }{ {a}^{7} } \\ \\ = {a}^{5 - 7} \\ \\ = {a}^{ - 2} \\ \\ = \frac{1}{{a}^{2} } \\ \\$

8 0

2 years ago

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