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telo118 [61]
2 years ago
9

Help please this makes no sense

Mathematics
1 answer:
Degger [83]2 years ago
5 0

Answer:

ST = 7.1 in

Step-by-step explanation:

we can use SOH-CAH-TOA because we've got a right triangle

side ST would be 'adjacent'

side TU would be the 'hypotenuse'

so we use Cosine because that's used when you're dealing with  adjacent and hypotenuse:

cos 42° = ST/9.5

ST = cos 42° × 9.5

ST = 7.1

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Dafna11 [192]

The letter "x" is often used in algebra to mean a value that is not yet known. It is called a "variable" or sometimes an "unknown". In x + 2 = 7, x is a variable, but we can work out its value if we try! so im pretty sure its 20

5 0
3 years ago
Please Help Me! ASAP!
Radda [10]

Answer:

two solutions

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6 0
3 years ago
Drag and drop the reasons for the steps into each box correctly to solve the equation
Viktor [21]

Yes your answers are correct :)

Step by step review:

First line has product of 5 and (x+2) so product property is good.

2nd line has division so quotient property is good.

3rd line has equal log with base 2 so equality property is good.

4th line has multiplication so that is also good

5th line has distribution of 5 and 7 so distributive property is good.

Your answer is correct.

7 0
3 years ago
Eugene had 24$ to spend on three pencils. After buying them he had 18$. How much did each pencil coast?
Helga [31]
Each pencil costed $2.
7 0
3 years ago
In a certain city, the daily consumption of water (in millions of liters) follows approximately a gamma distribution with α = 2
Ne4ueva [31]

Answer:

The probability that on any given day the water supply is inadequate 0.1991

Step-by-step explanation:

Given

α = 2 and β = 3

As per Gamma distribution Function

P(X>9)= 1-P(X\leq 9)\\

Expanding the function and putting the given values, we get -

[tex]1 - \int\limits^9_0 {f(x;2,3)} \, dx \\1- \int\limits^0_0 {\frac{1}{9}xe^{\frac{-x}{3} } \, dx\\\\= 1- \frac{1}{9} [x(-3e^{\frac{-x}{3}}) -\int\limits {(-3e^{\frac{-x}{3}})} \, dx]^9_0= 1- 1/9 [x(-3e^{\frac{-x}{3}}) -9e^{\frac{-x}{3}})} \, dx]^9_0\\1-((\frac{-1}{3} *9*e^({\frac{-9}{3}})- e^(\frac{-9}{3}))- ((\frac{-1}{3} *0*e^(\frac{-9}{3})-e^{\frac{-0}{3}})\\1-((-3e^{\frac{-9}{3} }-e^{\frac{-9}{3}}-(0-1))\\1-(1-4e^{-3})\\1-(1-0.1991)\\1-0.8009\\0.1991

The probability that on any given day the water supply is inadequate 0.1991

5 0
4 years ago
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