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sattari [20]
3 years ago
15

What number is 30% of 150

Mathematics
2 answers:
mrs_skeptik [129]3 years ago
8 0
Find 30% of 150.  To do this, mult. 150 by 0.30.  Result:  45.
vova2212 [387]3 years ago
6 0
\displaystyle 30\% \ of \ 150 = \\ \\  \frac{30}{10\not0} \cdot 15\not0= \\ \\\\  \frac{30}{10}  \cdot 15= \\ \\  \\\frac{450}{10} =45
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What is the common difference for this arithmetic sequence?<br> -6, -2, 2, 6, 10, ...
FrozenT [24]

Answer:

common difference: t2-t1

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-8

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3 years ago
I keep on reading 9 over and over and I can’t seem to get it.. Anyone mind helping?
belka [17]
Part A: 15MPH
Part B: 22.5MP

I could provide an explanation if needed.
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5 0
3 years ago
Hurry ASAP I need help on 3 and 4
attashe74 [19]

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3.B

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Step-by-step explanation:

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Read 2 more answers
The coordinates of the vertices of ΔPQR are (–2, –2), (–6, –2), and (–6, –5).
melamori03 [73]
<h3>Answer:</h3>

Yes, ΔPʹQʹRʹ is a reflection of ΔPQR over the x-axis

<h3>Explanation:</h3>

The problem statement tells you the transformation is ...

... (x, y) → (x, -y)

Consider the two points (0, 1) and (0, -1). These points are chosen for your consideration because their y-coordinates have opposite signs—just like the points of the transformation above. They are equidistant from the x-axis, one above, and one below. Each is a <em>reflection</em> of the other across the x-axis.

Along with translation and rotation, <em>reflection</em> is a transformation that <em>does not change any distance or angle measures</em>. (That is why these transformations are all called "rigid" transformations: the size and shape of the transformed object do not change.)

An object that has the same length and angle measures before and after transformation <em>is congruent</em> to its transformed self.

So, ... ∆P'Q'R' is a reflection of ∆PQR over the x-axis, and is congruent to ∆PQR.

6 0
3 years ago
A rectangle is twice as long as it is wide. If its length and width are both
puteri [66]

Answer:

a rectangle is twice as long as it is wide . if both its dimensions are increased 4 m , its area is increaed by 88 m squared make a sketch and find its original dimensions of the original rectangle

Step-by-step explanation:

Let l = the original length of the original rectangle

Let w = the original width of the original rectangle

From the description of the problem, we can construct the following two equations

l=2*w (Equation #1)

(l+4)*(w+4)=l*w+88 (Equation #2)

Substitute equation #1 into equation #2

(2w+4)*(w+4)=(2w*w)+88

2w^2+4w+8w+16=2w^2+88

collect like terms on the same side of the equation

2w^2+2w^2 +12w+16-88=0

4w^2+12w-72=0

Since 4 is afactor of each term, divide both sides of the equation by 4

w^2+3w-18=0

The quadratic equation can be factored into (w+6)*(w-3)=0

Therefore w=-6 or w=3

w=-6 can be rejected because the length of a rectangle can't be negative so

w=3 and from equation #1 l=2*w=2*3=6

I hope that this helps. The difficult part of the problem probably was to construct equation #1 and to factor the equation after performing all of the arithmetic operations.

5 0
3 years ago
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