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3 years ago

Multiply. give your answer in standerd form. (3n^2+2n+4)(2n-1)

Mathematics

2 answers:

astra-53 [7]3 years ago

6 0

The answer is 6n^3+n^2+6n-4 :)

mr Goodwill [35]3 years ago

5 0

The answer is 6n^3+n^2+6n-4 I hope I helped you!

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the legs of a right triangle are 18 cm and 80 cm. what is the length of the hypotenuse? 82 cm 85 cm 89 cm 98 cm

GREYUIT [131]

A^2 + b^2 = c^2
324 + 6400 = c^2
6,724 = c^2
c = 82

82 cm.

4 0

3 years ago

Read 2 more answers

What is the solution for the inequality y > |2x-3| +1?

liubo4ka [24]

1/2x-3................

5 0

3 years ago

The product (x) of two numbers is 24 and their sum (+) is 10. What is the value of the largest of the two numbers?

RideAnS [48]

let the two numbers be x and y

From the first sentence,

xy=24

x+y=10

Then make y in equation 2 the subject of the formular and substitute in equation 1

x+y=10

y=10-x

substituting in equation 2

x(10-x)=24

open the bracket

10x-x^2=24

=-x^2+10x=24

Transfer the constant to the left hand side

=-x^2+10x-24=0

Then factorise completely

Look at the photo above

4 0

3 years ago

Subtract and simplify 20 2/7 - 26/35

marissa [1.9K]

20& 10/35 - 26/35

19& 16/35 simplified to it's fullest.

4 0

3 years ago

One more time!

CaHeK987 [17]

Since q(x) is inside p(x), find the x-value that results in q(x) = 1/4

$\frac{1}{4} = 5 - x^2\ \Rightarrow\ x^2 = 5 - \frac{1}{4}\ \Rightarrow\ x^2 = \frac{19}{4}\ \Rightarrow \\
x = \frac{\sqrt{19} }{2}$

so we conclude that
$q(\frac{\sqrt{19} }{2} ) = 1/4$

therefore

$p(1/4) = p\left( q\left(\frac{ \sqrt{19} }{2} \right) \right)$

plug $x=\sqrt{19}/2$ into p( q(x) ) to get answer

$p(1/4) = p\left( q\left( \frac{ \sqrt{19} }{2} \right) \right)\ \Rightarrow\ \dfrac{4 - \left( \frac{\sqrt{19} }{2}\right)^2 }{ \left( \frac{\sqrt{19} }{2}\right)^3 } \Rightarrow \\ \\ \dfrac{4 - \frac{19}{4} }{ \frac{19\sqrt{19} }{8}} \Rightarrow \dfrac{8\left(4 - \frac{19}{4}\right) }{ 8 \cdot \frac{19\sqrt{19} }{8}} \Rightarrow \dfrac{32 - 38}{19\sqrt{19}} \Rightarrow \dfrac{-6}{19\sqrt{19}} \cdot \frac{\sqrt{19}}{\sqrt{19}}\Rightarrow$

$\dfrac{-6\sqrt{19} }{19 \cdot 19} \\ \\ \Rightarrow -\dfrac{6\sqrt{19} }{361}$

$p(1/4) = -\dfrac{6\sqrt{19} }{361}$

3 0

3 years ago