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2 years ago

The rectangle ABCD shown in the diagram is reflected across the line y = 2. What

Mathematics

1 answer:

kobusy [5.1K]2 years ago

3 0

Answer:

Since you haven't attached the diagram here, I'll go over how to do this in general. In order to reflect the rectangle across the horizontal line drawn at y = 2, consider the number of units of each coordinate from this line of reflection. Now count the same number of units on the other side of the line to plot the coordinates of the reflected rectangle. For example, if point A was at (1, 5), the point A' would then be at (1, -1). Notice that the x-coordinate remains the same.

Hope that answers the question, have a great day!

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Explain how complex fractions can be used to solve problems involving ratios

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When you have ratios and some unknowns you can create complex fractions from them.Bring them to the same denominator and solve for X.

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Carmella wants to attend Bayside Community College. She'll need to have $25,000 six years from today. Carmella is wondering what

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2 years ago

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List the first five terms of the sequence. a1 = 4, an+1 = 5an − 3

RUDIKE [14]

The recursive formula

$a_{n+1} = 5a_n-3$

is telling you that the next term in the sequence is five times the previous one, minus 3.

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$a_2 = 5a_1-3 = 5\cdot 4 - 3 = 20-3 = 17$

And similarly we can obtain all the next terms:

$a_3 = 5a_2-3 = 5\cdot 17 - 3 = 85-3 = 82$

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2 years ago

What is the balance on an amortized loan of $110,000 after the first payment if the interest rate is 5.5% with a monthly P&I

marishachu [46]

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2 years ago

A basketball player is shooting a basketball toward the net. The height, in feet, of the ball t seconds after the shot is modele

tresset_1 [31]

Answer:

Yes, between 0.84 seconds and 0.85 seconds after the shot is launched.

Step-by-step explanation:

$\text{The equation for ball height} : 6+30\cdot t-16\cdot t^{2}\\\text{The equation for shot blocker's height} : 9+25\cdot t-16\cdot t^{2}$

But, the shot is made before two tenths of a second or 0.2 seconds therefore modified equation for ball height is :

$6+30\cdot (t-0.2)-16\cdot (t-0.2)^{2}$

Now for the shot to be blocked,the height of shot blocker must be greater than the height of the ball which is shot before 0.2 seconds :

$\implies 9+25\cdot t-16\cdot t^{2}\geq 6+30\cdot (t-0.2)-16\cdot (t-0.2)^{2}\\\implies 9+25\cdot t-16\cdot t^{2}\geq 6+30\cdot t-6-16\cdot (t^{2}-0.4\cdot t+0.04)\\\implies9+25\cdot t-16\cdot t^{2}\geq 30\cdot t-16\cdot t^{2}+6.4\cdot t-0.64\\\implies 9+25\cdot t\geq 36.4\cdot t-0.64\\\implies 9.64\geq 11.4\cdot t\\\\\implies t\leq \frac{9.64}{11.4}\approx 0.846$

Hence, the shot was blocked between 0.84 and 0.85 seconds after the shot is launched.

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2 years ago

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