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ANTONII [103]
3 years ago
13

An English class has read the first 384 pages of their novel.

Mathematics
1 answer:
Pavel [41]3 years ago
6 0
X - the total number of pages

75\%x=384 \\
\frac{75}{100}x=384 \\
\frac{3}{4}x=384 \\
x=384 \times \frac{4}{3} \\
x=128 \times 4 \\
x=512

There are 512 pages in the novel.
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Write an equation for a line parallel to y=2x -1 that passes through (-1, 6).
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I think it’s Y = 2x + 8

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3 years ago
Select the fraction equivalent of 0.06. reduce to the lowest terms
viva [34]

Answer: 3/50

Step-by-step explanation:

0.06 = 6/100 , 100 would be the denominator because we have two figures after the decimal point. Each figures can also be represented by 10,

Again,

0.06 = 6 × 10-²

Now 0.06 = 6/100

= 3/50.

Therefore, the fractional form = 3/50 in its lowest term.

3 0
3 years ago
The function f(x) = (x - 4)(x - 2) is shown.
agasfer [191]

Answer:

all real numbers greater than or equal to -1

Step-by-step explanation:

In order to solve this problem we need to know the vertex and the direction its pointing.

First we expand,

x^2-6x+8

To find the x value of the vertex we use this formula (-b/2a).

-(-6)/2(1) = 3

Now we plug 3 in the equation to get the y value,

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The vertex is (3,-1)

We know the graph points up because x^2 is positive

The vertex is the lowest point, so we now know that -1 is the starting range and if the graph is pointing up, that means all values greater than -1.

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6 0
4 years ago
Which system has no solutions?
Vesnalui [34]
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7 0
3 years ago
Please find the exact length of the midsegment of trapezoid JKLM with vertices J(6, 10), K(10, 6), L(8, 2), and M(2, 2). Thank y
I am Lyosha [343]

Answer:

the exact length of the midsegment of trapezoid JKLM  = \mathbf{ = 3 \sqrt{5} } i.e 6.708 units on the graph

Step-by-step explanation:

From the diagram attached below; we can see a graphical representation showing the mid-segment of the trapezoid JKLM. The mid-segment is located at the line parallel to the sides of the trapezoid. However; these mid-segments are X and Y found on the line JK and LM respectively from the graph.

Using the expression for midpoints between two points to determine the exact length of the mid-segment ; we have:

\mathbf{ YX = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }

\mathbf{ YX = \sqrt{(8-5)^2+(8-2)^2} }

\mathbf{ YX = \sqrt{(3)^2+(6)^2} }

\mathbf{ YX = \sqrt{9+36} }

\mathbf{ YX = \sqrt{45} }

\mathbf{ YX = \sqrt{9*5} }

\mathbf{ YX = 3 \sqrt{5} }

Thus; the exact length of the midsegment of trapezoid JKLM  = \mathbf{ = 3 \sqrt{5} } i.e 6.708 units on the graph

8 0
3 years ago
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