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3 years ago

Factor by grouping 2a^2+ab+5ab+5b^2

Mathematics

2 answers:

masha68 [24]3 years ago

4 0

2a^2+ab+5ab+5b^2
2a(a+b) +5b(a+b)
(a+b)(2a+5b)

True [87]3 years ago

3 0

$=>2a^2+ab+5ab+5b^2 \\ \\ 
=>2a(a+b) +5b(a+b) \\ \\ 
=>(a+b)(2a+5b) \\ \\ \\ \framebox[1.1\width]{Good Luck ! !} \par


$

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How many edges does a square pyramid have?

Paul [167]

Answer:

Step-by-step explanation:

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4 years ago

Read 2 more answers

Pls help I need the answer quick

VashaNatasha [74]

Answer:
49
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3 years ago

7d – 4d – 3d – 5 = 6d

vazorg [7]

Answer:

$d = -\frac{5}{6}$

Step-by-step explanation:

7d – 4d – 3d – 5 = 6d

(7d - 4d = 3d)

3d - 3d -5 = 6d

(3d - 3d = 0)

0 - 5 = 6d

-5 = 6d

-5/6 = 6d/6

d = $\frac{-5}{6}$

Hope this helps ^-^

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3 years ago

PLEASE HELP ME BY SHOWING YOUR WORK TO THIS ANSWER!

ArbitrLikvidat [17]

Hey Dammy17,
Ok so what we need to do is "plug in" 2oz for x in the equation,

So, 0.6 (2) + 11 =
0.12 + 11= 11.12 oz is the finishing weight.

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3 years ago

1. Find the vertices and locate the foci for the hyperbola whose equation is given.

Irina18 [472]

$\bf \cfrac{(x-{{ h}})^2}{{{ a}}^2}-\cfrac{(y-{{ k}})^2}{{{ b}}^2}=1
\qquad center\ ({{ h}},{{ k}})\qquad
 vertices\ ({{ h}}\pm a, {{ k}})\\\\
-----------------------------\\\\
\textit{now let's take a look at yours}
\\\\\\
49x2 - 16y2 = 784\implies \cfrac{49x^2}{784}-\cfrac{16y^2}{784}=1
\\\\\\
\cfrac{x^2}{16}-\cfrac{y^2}{49}=1\implies \cfrac{(x-0)^2}{4^2}-\cfrac{(y-0)^2}{7^2}=1
\\\\\\
recall\implies center\ ({{ h}},{{ k}})\qquad vertices\ ({{ h}}\pm a, {{ k}})
\\\\\\
$

$\bf \textit{now, for the foci, the foci are "c" distance from the center point}\\\\\
whereas\qquad c=\sqrt{a^2+b^2}\qquad \textit{ that is }\qquad h\pm \sqrt{a^2+b^2}$

notice your "a" and "b" components, to get the distance "c" from the center to either foci and the vertices, of course, are h + a, k and h - a, k

5 0

3 years ago