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1 year ago

Simplify: -16b4 + (-18b4)

Mathematics

1 answer:

GuDViN [60]1 year ago

4 0

Answer:

-34b4

Step-by-step explanation:

Since the variables are the same b4, you just add the constants (-16 & -18).

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Class 8 squares and square roots without adding find the sum of a) 1+3+5+7+9+11+13+15+17.

Nata [24]

Answer:

given

n=9

a=1

d=3-1=2

Step-by-step explanation:

now , by formula,

sn = n/2(2a+(n-1)d)

= 9/2(2×1+(9-1)×2)

=9/2×18

=81

8 0

2 years ago

What is the cubic units of 2ft by 2 ft by 2ft

Blababa [14]

2ft × 2ft × 2ft is 8ft cubed

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3 years ago

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Gail baked some cookies. She sold 2/7 of the cookies on Monday. She sold 1/3 more of the cookies on Tuesday than on Monday. What

lorasvet [3.4K]

2/7 x 1/3 = 2/ 21

In simplest form 2/21 = 2/21

Answer + 2/21

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2 years ago

Triangle ABC is an isosceles triangle in which side

Shkiper50 [21]

Answer:

Option B.

Step-by-step explanation:

Consider the below figure attached with this question.

It is given that triangle ABC is an isosceles triangle.

From the below figure it is clear that the vertices of triangle are A(-2,-4), B(2,-1) and C(3,-4).

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using this formula, we get

$AB=\sqrt{(2-(-2))^2+(-1-(-4))^2}=\sqrt{16+9}=5$

$BC=\sqrt{(3-2)^2+(-4-(-1))^2}=\sqrt{1+9}=\sqrt{10}$

$AC=\sqrt{(3-(-2))^2+(-4-(-4))^2}=\sqrt{25}=5$

Now,

Perimeter of triangle ABC = AB + BC + AC

$=5+\sqrt{10}+5$

$=10+\sqrt{10}$

So, perimeter of triangle ABC is $10+\sqrt{10}$ units.

Therefore, the correct option is B.

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3 years ago

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Evaluate the cosine if the angle of rotation which contains the point (9, -3) on its terminal side

Liono4ka [1.6K]

so we know the terminal point is at (9, -3), now, let's notice that's the IV Quadrant

$\bf (\stackrel{x}{9}~~,~~\stackrel{y}{-3})\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{9^2+(-3)^2}\implies c=\sqrt{81+9}\implies c=\sqrt{90} \\\\[-0.35em] ~\dotfill$

$\bf cos(\theta )=\cfrac{\stackrel{adjacent}{9}}{\stackrel{hypotenuse}{\sqrt{90}}}\implies \stackrel{\textit{rationalizing the denominator}}{\cfrac{9}{\sqrt{90}}\cdot \cfrac{\sqrt{90}}{\sqrt{90}}\implies \cfrac{9\sqrt{90}}{90}}\implies \cfrac{\sqrt{90}}{10}\implies \cfrac{3\sqrt{10}}{10}$

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2 years ago