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adelina 88 [10]
3 years ago
8

20 - 9x = x + 10x What is X?

Mathematics
2 answers:
Bingel [31]3 years ago
8 0
I got x= 1 too hope it helped
Molodets [167]3 years ago
4 0
Answer: x=1

Explantion: 20+-9x=x+10x
20+-9x=11x
20+-9x+-11x=11x+-11
20+-20x=11x+-11x
20+-20x=0
20+-20+-20=0+-20
-20x=-20
x=1
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Write an expression for the calculation the difference of the products 5 and 2 and 5 and 1
algol13

Answer: 5

Step-by-step explanation: The first piece of information that you need to know is that the product is the answer when two numbers are multiplied. Therefore, you need to multiply the two sets of numbers, and then subtract.

The formula will be:

(5 x 2) - (5 x 1)=

First, do what is inside the parentheses:

10 - 5=

And solve...

5

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3 years ago
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Simplify. Round to the nearest tenth. 5.4% of 65 *
satela [25.4K]

Answer:

65 times 5.4 divided by 100

351/100

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3.5

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mr. huber bought a block of fudge that weighed 2/5 of a pound. he cut the fudge into 6 equal pieces. what was the weight of each
Marina86 [1]

Answer:

\large \boxed{\frac{1}{15} \text{ lb}}

Step-by-step explanation:

You are starting with ⅖ lb of fudge. Then you cut it into six pieces.

You want to evaluate ⅖ ÷ 6

1. Treat the integer as a fraction

\dfrac{2}{5}\div{\dfrac{6}{1}}

2. Invert the divisor and change divide to multiply

\dfrac{2}{5}\times{\dfrac{1}{6}}

3. Divide numerator and denominator by 2

\dfrac{1}{5}\times{\dfrac{1}{3}}

4. Multiply numerators and denominators

\dfrac{1}{5}\times{\dfrac{1}{3}} = \mathbf{\dfrac{1}{15}}\\\\\text{Each piece of fudge weighs $\large \boxed{\mathbf{\frac{1}{15}} \textbf{ lb}}$}

7 0
4 years ago
Find the integral of √(x² +4) W.R.T x​
Allushta [10]

Answer:

\frac{x}{2} *\sqrt{x^{2} +4} +\frac{1}{2}*LN(|\frac{x+\sqrt{x^{2} +4} }{2}|) +C

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we will have to do a trig sub for this

use x=a*tanθ for sqrt(x^2 +a^2) where a=2

x=2tanθ, dx= 2 sec^2 (θ) dθ

this turns \int\limits {\sqrt{x^{2}+4 } } \, dx into integral(sqrt( [2tanθ]^2 +4) * 2sec^2 (θ) )dθ

the sqrt( [2tanθ]^2 +4) will condense into 2sec^2 (θ) after converting tan^2(θ) into sec^2(θ) -1

then it simplifies into integral(4*sec^3 (θ)) dθ

you will need to do integration by parts to work out the integral of sec^3(θ) but it will turn into (1/2)sec(θ)tan(θ) + (1/2) LN(|sec(θ)+tan(θ)|) +C

then you will need to rework your functions of θ back into functions of x

tanθ will resolve back into \frac{x}{2} (see substitutions) while secθ will resolve into \frac{\sqrt{x^{2} +4} }{2}

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after resolving back into functions of x, substitute ratios for trig functions:

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3 0
3 years ago
What is (6-4)63 + 6 =
xxTIMURxx [149]

Answer:

132 is the answer

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3 years ago
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