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Mazyrski [523]
3 years ago
13

STVU is an isosceles trapezoid. If SV = 6x - 5 and TU = 2x + 7, find the value of x.x=3

Mathematics
1 answer:
Thepotemich [5.8K]3 years ago
7 0
Because <span>STVU is an isosceles trapezoid
so SV = TU
</span><span>6x - 5 = 2x + 7
6x - 2x = 7 + 5
4x = 12
  x = 3

answer
x = 3</span>
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Consider the inverse function. Which conclusions can be drawn about f(x) = x2 + 2? Select three options. f(x) has a limited rang
PolarNik [594]

Answer:

f(x) has a limited range

f(x) has a maximum at the point (0, 2)

f(x) has a y-intercept at the point (0, 2).

Step-by-step explanation:

Given the function;

f(x) = x^2+2

The domain is the value of the input variables for which the function will exist. According to the expression given, the function exists on all real values of x. The same goes with range which deals with the output values. It also exists on all real values from 2 and above.

Hence f(x) have a limited range (since values less than 2 are not included compare to domain that exists on all real values) and does not have a restricted domain.

For the x intercept, x intercept occur at y = 0

substitute y = 0 into the function and get y

if y = f(x)

y = x^2+2

0 = x^2 + 2

x^2 = -2

x = 2i

Hence  f(x) does not have an x-intercept of (2, 0)

For the y intercept, y intercept occur at x = 0

substitute x = 0 into the function and get y

if y = f(x)

y = x^2+2

y = 0^2 + 2

y = 2

Hence  f(x) has a y-intercept at point (0, 2)

f(x) is at maximum if d(fx))/dx = 0

d(fx))/dx  = 2x

since  d(fx))/dx  = 0

0 = 2x

x = 0

substitute x = 0 into the function

f(x) = x^2 + 2

y = 0^2+2

y = 2

Hence f(x) has a maximum at the point (0, 2)

5 0
3 years ago
Read 2 more answers
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gayaneshka [121]

Answer:

E is not a subspace of \mathbb{R}^2

Step-by-step explanation:

E is not a subspace of  \mathbb{R}^2

In order to see this, we must find two points (a,b), (c,d) in  E such that (a,b) + (c,d) is not in E.

Consider

(a,b) = (1,1)

(c,d) = (-1,-1)

It is easy to see that both (a,b) and (c,d) are in E since 1*1>0 and (1-)*(-1)>0.  

But (a,b) + (c,d) = (1-1, 1-1) = (0,0)

and (0,0) is not in E.

By the way, it can be proved that in any vector space all sub spaces must have the vector zero.

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3 years ago
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Firdavs [7]

Answer:

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2 years ago
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Answer:

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3/400, 0.0075

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