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aev [14]
3 years ago
5

Which points are on the graph of ƒ(x) = 2(0.5)x?

Mathematics
2 answers:
gogolik [260]3 years ago
7 0
Hello,

f(x)=2/2^x
f(0)=2/1=2        A:True

f(1)=2/2^1=2/2=1  B:True

f(-2)=2/2^(-2)=2/(1/4)=8    C:false (negative 8)

f(2)= 2/2^2=2/4=1/2  D: False

f(-1)=2/2^(-1)=2/1/2=2*2=4  E: false
Points are A and B

dimaraw [331]3 years ago
3 0
First set of points (0,2) (1,1)
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Answer: 2/3.

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Write the mathematical expression for the following verbal expression: the product of 12 and a number is more than 190
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Hello,

12(x) > 190

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3 years ago
Kara is building a sandbox shaped like a kite for her nephew. The top two sides of the sandbox are 29 inches long. The bottom tw
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The wording is unclear without a diagram. As such, there are two possible cases and two possible answers.

Case 1: Diagonal \overline{DB} i<span><span>s formed by connecting the vertices formed by the meeting points of a 25-inch side and a 29-inch side. </span>
</span>Call the intersection point of \overline{DB} and \overline{AC} E. \overline{AC} bisects \overline{DB}, so DE=BE=20\text{ inches}. Since the diagonals of a kite are perpendicular to each other, \triangle AED and \triangle CED are both right triangles. One has a hypotenuse of 29, and the other has a hypotenuse of 25, but both share a leg of 20. Using the Pythagorean Theorem, we can get that the length of the other leg in the triangle with a hypotenuse of 29 is 21. Similarly, for the triangle with a hypotenuse of 25, the other leg has a length of 15. Together, these legs make up \overline{AC}, meaning AC=21+15=36 \text{ inches}, our final answer.

Case 2: Diagonal \overline{DB} i<span>s formed by connecting the vertices formed by the meeting points of the sides with equal lengths. 
</span>Call the intersection point of \overline{DB} and \overline{AC} E. We will focus on two triangles, namely \triangle ADE and \triangle ABE. Since diagonals intersect perpendicularly, these triangles are right triangles. One of them has a hypotenuse of 29, and the other has a hypotenuse of 25. They both share a leg that is half of \overline{AC} because \overline{DB} bisects \overline{AC}. Let AE=y and the non-shared leg of the right triangle with a hypotenuse of 29 equal x. Since DB=40, the non-shared leg of the other right triangle (the one with a hypotenuse of 25) has a length of 40-x. Using the Pythagorean Theorem, we can get the equations x^2+y^2=29^2 and (40-x)^2+y^2=25^2. These can simplify to x^2+y^2=841 and 1600-80x+x^2 + y^2=625. Isolating the term y^2, we can get y^2=841-x^2 and y^2=625-x^2+80x-1600. The latter can simplify to y^2=-975-x^2+80x. Using substitution, we can combine the two equations into one and get 841-x^2=-975-x^2+80x. We can simplify that to 80x=1816, meaning x=22.7. However, we are looking for 2y (y is only half of \overline{AC}). We can solve for y using the Pythagorean Theorem and the triangle with a hypotenuse of 29 and a leg of 22.7. We get y \approx 18.05, meaning \overline{AC}=2y \approx 36.1 \text{ inches}, our final answer.
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