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3 years ago

9x - 3 + x² , when x = 4

Mathematics

2 answers:

Fiesta28 [93]3 years ago

8 0

Answer:

Step-by-step explanation:

hope this helps

9(4)-3+4^2

36-3+4^2

33+4^2

33+16

Step2247 [10]3 years ago

5 0

Step-by-step explanation:

=9x-3+x²

=9×4-3+4²

=36-3+16

=49

hope it helps.

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Rewrite the function by completing the square f(x)=x^2-4x-26

leonid [27]

Answer:

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Step-by-step explanation:

To complete the square

add/subtract ( half the coefficient of the x- term )² to x² - 4x

f(x) = x² + 2(- 2)x + 4 - 4 - 26 = (x - 2)² - 30

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3 years ago

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kumpel [21]

Answer:

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Step-by-step explanation:

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3 years ago

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dimulka [17.4K]

Answer:

33.47

Step-by-step explanation:

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3 years ago

The length and width of a rectangle are measured as 55 cm and 49 cm, respectively, with an error in measurement of at most 0.1 c

valentina_108 [34]

Answer:

The maximum error in the calculated area of the rectangle is $10.4 \:cm^2$

Step-by-step explanation:

The area of a rectangle with length $L$ and width $W$ is $A= L\cdot W$ so the differential of <em>A</em> is

$dA=\frac{\partial A}{\partial L} \Delta L+\frac{\partial A}{\partial W} \Delta W$

$\frac{\partial A}{\partial L} = W\\\frac{\partial A}{\partial W}=L$ so

$dA=W\Delta L+L \Delta W$

We know that each error is at most 0.1 cm, we have $|\Delta L|\leq 0.1$ , $|\Delta W|\leq 0.1$ . To find the maximum error in the calculated area of the rectangle we take $\Delta L = 0.1$ , $\Delta W = 0.1$ and $L=55$ , $W=49$ . This gives

$dA=49\cdot 0.1+55 \cdot 0.1$

$dA=10.4$

Thus the maximum error in the calculated area of the rectangle is $10.4 \:cm^2$

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State the x- and y-intercepts of

Makovka662 [10]

The x intercept is (5,0)
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3 years ago