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ASHA 777 [7]
3 years ago
13

PLEASE HELP!!!!!!!! a2 - 10ab + 3b^2

Mathematics
1 answer:
Zielflug [23.3K]3 years ago
5 0
Let's simplify step-by-step.

<span><span><span>a2</span>−<span><span>10a</span>b</span></span>+<span>3<span>b2

</span></span></span>There are no like terms.

Answer:

<span>=<span><span><span>a2</span>−<span><span>10a</span>b</span></span>+<span>3<span>b<span>2</span></span></span></span></span>
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What is the standard form of 1.46e-6
GalinKa [24]
Standard form would be 1.46e-6 because the variable has one degree and the number 6 does not have a degree.
3 0
3 years ago
What percent of increase for a population that changed from 25,000 to 30,000
tiny-mole [99]

percent of increase for a population that changed from 25,000 to 30,000

Increase in population = 30,000 - 25,000 = 5000

So, the increase in population is 5000

To find percent of increase , we divide the increase in population by initial population.

So percent of increase = \frac{5000}{25000} * 100

= 0.2 * 100

= 20 %

20% of increase for a population that changed from 25,000 to 30,000.

7 0
3 years ago
Read 2 more answers
Helpppppppppppppppppppppppppppppppppppp
Vadim26 [7]

Answer:

m = - 1

Step-by-step explanation:

9m + 13 = 4

subtract 13 from both sides

9m = 4 - 13

9m = - 9

divide both sides by 9

m = - 1  (negative 1)

5 0
2 years ago
7. Write an inequality that matches the graph shown.
IrinaK [193]

Answer:

To make an inequality its in the part that is shaded so

Step-by-step explanation:

(-2,-2) (-4,1) ETC

8 0
2 years ago
D<br> Evaluate<br> arcsin<br> (6)]<br> at x = 4.<br> dx
sineoko [7]

Answer:

\frac{1}{2\sqrt{5} }

Step-by-step explanation:

Let, \text{sin}^{-1}(\frac{x}{6}) = y

sin(y) = \frac{x}{6}

\frac{d}{dx}\text{sin(y)}=\frac{d}{dx}(\frac{x}{6})

\frac{d}{dx}\text{sin(y)}=\frac{1}{6}

\frac{d}{dx}\text{sin(y)}=\text{cos}(y)\frac{dy}{dx} ---------(1)

\frac{1}{6}=\text{cos}(y)\frac{dy}{dx}

\frac{dy}{dx}=\frac{1}{6\text{cos(y)}}

cos(y) = \sqrt{1-\text{sin}^{2}(y) }

          = \sqrt{1-(\frac{x}{6})^2}

          = \sqrt{1-(\frac{x^2}{36})}

Therefore, from equation (1),

\frac{dy}{dx}=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}

Or \frac{d}{dx}[\text{sin}^{-1}(\frac{x}{6})]=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}

At x = 4,

\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}

\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}

                   =\frac{1}{6\sqrt{\frac{36-16}{36}}}

                   =\frac{1}{6\sqrt{\frac{20}{36} }}

                   =\frac{1}{\sqrt{20}}

                   =\frac{1}{2\sqrt{5}}

4 0
2 years ago
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