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Vitek1552 [10]
2 years ago
13

What is the area of a triangle with vertices at (−2, 1), (2, 1), and ​(3, 4)​ ?

Mathematics
1 answer:
castortr0y [4]2 years ago
5 0

Answer:

6 Units.

Step-by-step explanation:

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Does 3 (x-1) = 2x + 9 have one solution, no solution, or infinite solutions?
Allushta [10]

Answer:

one solution

x = 12

Step-by-step explanation:

3 (x-1) = 2x + 9

3x - 3 = 2x + 9

subtract 2x from each side of the equation:

x - 3 = 9

add 3 to each side:

x = 12

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3 years ago
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I know how to set it up but just don’t remember how to solve it. V stands for volume.
Tasya [4]

Answer:

29.5788

Step-by-step explanation:

Multiply all of the terms together

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3 years ago
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Which expression is a possible leading term for the polynomial function graphed below?
Anon25 [30]

Answer:

D, 22x^9

Step-by-step explanation:

Here's a key for end behavior: Look at the leading term

x^even = x -> neg inf, f(x) -> pos inf; x -> pos inf, f(x) -> pos inf

-x^even = x -> neg inf, f(x) -> neg inf; x -> pos inf, f(x) -> neg inf

x^odd = x -> neg inf, f(x) -> neg inf; x -> pos inf, f(x) -> pos inf

-x^odd = x -> neg inf, f(x) -> pos inf; x -> pos inf, f(x) -> neg inf

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3 years ago
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Colleen McHugh wishes to deposit checks for $13.75, $92.08, and $8.21 into her account. She would like to receive $15.00 in cash
CaHeK987 [17]

her total deposit is 98.94

4 0
3 years ago
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A line tangent to the curve f(x)=1/(2^2x) at the point (a, f(a)) has a slope of -1. What is the x-intercept of this tangent?
kirza4 [7]

Answer:

x-intercept = 0.956

Step-by-step explanation:

You have the function f(x) given by:

f(x)=\frac{1}{2^{2x}}   (1)

Furthermore you have that at the point (a,f(a)) the tangent line to that point has a slope of -1.

You first derivative the function f(x):

\frac{df}{dx}=\frac{d}{dx}[\frac{1}{2^{2x}}]  (2)

To solve this derivative you use the following derivative formula:

\frac{d}{dx}b^u=b^ulnb\frac{du}{dx}

For the derivative in (2) you have that b=2 and u=2x. You use the last expression in (2) and you obtain:

\frac{d}{dx}[2^{-2x}]=2^{-2x}(ln2)(-2)

You equal the last result to the value of the slope of the tangent line, because the derivative of a function is also its slope.

-2(ln2)2^{-2x}=-1

Next, from the last equation you can calculate the value of "a", by doing x=a. Furhtermore, by applying properties of logarithms you obtain:

-2(ln2)2^{-2a}=-1 \\\\2^{2a}=2(ln2)=1.386\\\\log_22^{2a}=log_2(1.386)\\\\2a=\frac{log(1.386)}{log(2)}\\\\a=0.235

With this value you calculate f(a):

f(a)=\frac{1}{2^{2(0.235)}}=0.721

Next, you use the general equation of line:

y-y_o=m(x-x_o)

for xo = a = 0.235 and yo = f(a) = 0.721:

y-0.721=(-1)(x-0.235)\\\\y=-x+0.956

The last is the equation of the tangent line at the point (a,f(a)).

Finally, to find the x-intercept you equal the function y to zero and calculate x:

0=-x+0.956\\\\x=0.956

hence, the x-intercept of the tangent line is 0.956

5 0
3 years ago
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