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2 years ago

Evaluate the expression. 5 x (-6 / -1 + -10)

Mathematics

1 answer:

kotykmax [81]2 years ago

8 0

Answer:

-20

Step-by-step explanation:

$5 * (-6 / -1+ -10)$

According to PEMDAS, we need to do the math in the parentheses first.

$-6 / -1 + -10$

Applying PEMDAS to the inside of the parentheses, then we have to divide -6 by -1, where we would get 6.

$6 + -10$

Adding a negative is the same as subtracting, so adding negative 10 is the same as subtracting 10.

$6 - 10$

Subtracting, we get 6 - 10 = -4 and we're left with:

$5 * (-4)$

Multiplying, <u>our final answer is -20.</u>

Hope this helps!

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Is a cube a polygon? Why or why not?

myrzilka [38]

No. A cube is a 3D figure and polygon is a 2D one.

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3 years ago

For what value of k will k+9,2k-1 and 2k+7 are the consecutive terms of an A.P

Stolb23 [73]

k = 18

the common difference d of an arithmetic sequence is

d = a₂ - a₁ = a₃ - a₂ = ....

a₂ - a₁ = 2k - 1 - k - 9 = k - 10 and

a₃ - a₂ = 2k + 7 - (2k - 1) = 2k + 7 - 2k + 1 = 8, hence

k - 10 = 8 ⇒ k = 8 + 10 = 18

the 3 consecutive terms are

27, 35, 43

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3 years ago

HELP PLS Simplify an expression (2x-2x^4)-(2x+2x^4)

Naily [24]

2x-2x∧4-2x-2x∧4
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3 years ago

The length of a rectangle is 20 units more than its width. The area of the rectangle is x4−100.

nadezda [96]

Answer:

$1. x^2-10$ because the area expression can be rewritten as $(x^2-10)(x^2+10)$ which equals $(x^2-10)((x^2-10)+20).$

Step-by-step explanation:

Area of the rectangle $=(x^4-100)$

$x^4-100=(x^2)^2-10^2\\$Applying difference of two squares: a^2-b^2=(a-b)(a+b)\\(x^2)^2-10^2=(x^2-10)(x^2+10)$

Since the length of a rectangle is 20 units more than its width.

$Width: x^2-10\\Length=x^2+10=x^2-10+20$

The correct option is therefore 1.

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3 years ago

Read 2 more answers

Find the 8th roots of 3i-3<br> Urgent...please help!!!!

m_a_m_a [10]

$z=-3+3i=3(-1+i)=3\left(\sqrt2e^{i3\pi/4}\right)=3\sqrt2e^{i3\pi/4}$

Let $z=w^8$ . Then the eight roots of $z$ are

$w=\left(3\sqrt2e^{i3\pi/4}\right)^{1/8}$
$w=3^{1/8}2^{1/16}}e^{i(3\pi/4+2\pi k)/8}$

where $k=0,1,\ldots,7$ . So the eighth roots are

$w=\begin{cases}3^{1/8}2^{1/16}e^{i3\pi/32}&\text{for }k=0\\\\3^{1/8}2^{1/16}e^{i11\pi/32}&\text{for }k=1\\\\3^{1/8}2^{1/16}e^{i19\pi/32}&\text{for }k=2\\\\3^{1/8}2^{1/16}e^{i27\pi/32}&\text{for }k=3\\\\3^{1/8}2^{1/16}e^{i35\pi/32}&\text{for }k=4\\\\3^{1/8}2^{1/16}e^{i43\pi/32}&\text{for }k=5\\\\3^{1/8}2^{1/16}e^{i51\pi/32}&\text{for }k=6\\\\3^{1/8}2^{1/16}e^{i59\pi/32}&\text{for }k=7\end{cases}$

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3 years ago