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sveticcg [70]
1 year ago
5

If m∠NQM=2x+8 and m∠PQR=x+22, find the value of x

Mathematics
2 answers:
Oduvanchick [21]1 year ago
6 0

Answer:

14

Step-by-step explanation:

2x+8=x+22

2x - x= 22 - 8

x=14

Alex777 [14]1 year ago
3 0

Answer:

x = 14

Step-by-step explanation:

∠ NQM and ∠ PQR are congruent ( indicated by the same orange arc within the 2 angles ) , then

2x + 8 = x + 22 ( subtract x from both sides )

x + 8 = 22 ( subtract 8 from both sides )

x = 14

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Answer:

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Step-by-step explanation:

2x² - 10x + 6 = 24x

Collect like terms

2x² + 6 = 24x + 10x

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David's gasoline station offers 4 cents off per gallon if the customer pays in cash and does not use a credit card. Past evidenc
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Answer:

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Katyanochek1 [597]
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Find the derivative.
Aleksandr [31]

Answer:

Using either method, we obtain:  t^\frac{3}{8}

Step-by-step explanation:

a) By evaluating the integral:

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The integral itself can be evaluated by writing the root and exponent of the variable u as:   \sqrt[8]{u^3} =u^{\frac{3}{8}

Then, an antiderivative of this is: \frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}

which evaluated between the limits of integration gives:

\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}

and now the derivative of this expression with respect to "t" is:

\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

g(x)=\int\limits^x_a {f(t)} \, dt

is continuous on [a,b], differentiable on (a,b) and  g'(x)=f(x)

Since this this function u^{\frac{3}{8} is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}

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Step-by-step explanation:

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