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3 years ago

10

Pls answer ASAP!! due 6:00 pm on 9/25/2020, but I would really appreciate a late reply too. pls answer, will give 25 points and

brainiest!
Replace ∗ with a monomial so that the trinomial may be represented by a square of a binomial:

a) ∗ +28a+49
b) 36–24x + ∗
c) 6.25a^2 + ∗ + 1/4 b^2
d)∗ +2bc+100c^2

Thank you so much if you answered. I tried a lot, but couldn't get. PLEASE SHOW YOUR WORK I REALLY DONT GET IT!! ;-;

1 answer:

malfutka [58]3 years ago

3 0

Answer:

Im answering so i wont lose this convo... lol

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Another math problem, i really, really need help with this one -4x^2-3+4+3-5y^2+3y^2-5y^2

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-4x*2 - 7y*2 + 4

Step-by-step explanation:

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Kyle drew three line segments with lengths: 2/4 inch 2/3inch, and 2/6 inch list the fractions in order from least to greatest

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Method 1
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Since the numerators are the same, the smaller the denominators, the greater the fraction is.

Arranging from the least to the greatest
$\dfrac{2}{6} \ , \ \dfrac{2}{4} \ , \ \dfrac{2}{3}$

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Method 2
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Lets change all to the same denominators

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Now that all the denominators are the same, we can arrange the fractions by comparing the numerators. The bigger the numerators, the greater the fraction.

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3 years ago

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Pani-rosa [81]

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The formula for the area of a trapezoid is shown below.

$Area =\frac{1}{2} (^{b} 1 + ^{b}2)h$

The <em>b's</em> represent the bases which are the parallel sides and <em>h</em> is the height.

So in the trapezoid shown, the bases are 6 cm and 12 cm and the height is 6 cm. Plugging this information into the formula, we have $\frac{1}{2} (6 cm +12cm)(6 cm)$ .

Next, the order of operations tell us that we must simplify inside the parentheses first. 6 cm + 12 cm is 18 cm and we have $\frac{1}{2}(18 cm)(6 cm)$ .

$\frac{1}{2} (18 cm)$ is 9 cm and we have 9 cm · 6 cm of 54 cm²

So the area of the trapezoid shown is 54 cm².

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3 years ago

Montano1993 [528]

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Referring to that finding, if we set initial value less than zero, which means that we are solving 0.5x+1<0 and taking a number in the interval of the solution, which means x ∈ (- ∞, -20). Setting x=-19, we find that $F(x)=0.5x+1=-19$ and x=-40. In the next iteration, $F(x+1)=0.5(x+1)+1=0.5(1-40)+1=-18.5$ . In the next iteration, $F(x+2)=0.5(x+2)+1=0.5(2-40)+1=-18$ . By this way, we find that even if the initial value is less than zero, value of the successive iterations is increasing.

Using the function $g(x)=-x+2$ and taking the initial value equal to 4, we find that $g(x)=-x+2=4$ and x=-2. In the next iteration, $g(x+1)=-(x+1)+2=-(-2+1)+2=3$ . If we continue the iterations we'll see that they are decreasing.
Setting the initial value equal to 2, we find that $g(x)=-x+2=2$ and x=0. The next iteration is $g(x+1)=-(x+1)+2=1$ . In this case, the interations are also decreasing.
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