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jok3333 [9.3K]
3 years ago
9

Really easy i just don't feel like doing it, 10 extra points, (find x)

Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
6 0
X=24 should be the correct answer
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ASAP!
ANTONII [103]

Two figures are known as similar figures if there the corresponding angles are equal and the corresponding sider is in ratio. The value of x is 10. Thus, the correct option is B.

<h3>What are Similar Figures?</h3>

Two figures are known as similar figures if there the corresponding angles are equal and the corresponding sider is in ratio. It is denoted by the symbol "~".

Since the two prisms are similar, their sides will be in a common ratio. The value of the common ratio will be,

Common ratio = 4/2 = 5/2.5 = 3/1.5 = 2

Since the common ratio of every side is equal to 2, the value of x can be written as,

Common Ratio = 20/x = 2

20/x = 2

x = 10

Hence, the value of x is 10. Thus, the correct option is B.

Learn more about Similar Figures:

brainly.com/question/11315705

#SPJ1

3 0
2 years ago
WILL MARK BRAINLEST
Snowcat [4.5K]

Answer:

109.0125 dollars, rounded is 109.01$

Step-by-step explanation:

6 0
3 years ago
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Find the derivative of f(t)=2√t - 2/√t
Semmy [17]
The answer is t=1.

Hope this helps.
8 0
3 years ago
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At which x value will f(x)=3x+3 exceed g(x)=3x+10
Anvisha [2.4K]
We write an inequality:

f(x) > g(x)

3^x + 3 > 3x + 10

3^x > 3x + 7

This equation cannot be solved using trivial methods found in high-school classes, so we resort to graphical examination.  3x+7 is a linear function while 3^x is an exponential one (with limit zero as x approaches - \infty).  We see that 3^x = 3x+7 at approximately x=2.4 and x=-2.3.

Indeed, using a computer algebra system such as the ones on modern TI calculators and on many internet sites gives equality at x=2.42, -2.31.  By observing our graph, we see that f(x) > g(x) when x > 2.42 or x < -2.31.
3 0
2 years ago
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Twice the square of an integer is 3 less than 7 times the integer. Find the integer!!
Solnce55 [7]
The integer you’re looking for is 3
3 0
3 years ago
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