3 years ago

Can you explain how greatest common factor of 8,14?

Mathematics

1 answer:

Nuetrik [128]3 years ago

4 0

Answer:

There are 2 common factors of 8 and 14, that are 1 and 2. Therefore, the greatest common factor of 8 and 14 is 2.

Step-by-step explanation:

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the function g(x) Is the graph below resulted from shifting the graph of f(x) down 2 units. If f’(2)=8, what is g’(2)?

Sauron [17]

We know that the graph of g(x) is the result of a shift of 2 units down from f(x).

Then, f(x) and g(x) are related by:

$g(x)=f(x)-2$

Taking the derivative:

$g^{\prime}(x)=f^{\prime}(x)$

This is because the derivative of a constant number is 0. Finally, evaluating at x = 2:

$\begin{gathered} g^{\prime}(2)=f^{\prime}(2) \\ \\ \therefore g^{\prime}(2)=8 \end{gathered}$

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1 year ago

The area of a circle is 80.86cm2. Find the length of the radius rounded to 2 DP.

weeeeeb [17]

Answer:

Solution given:

The area of a circle =80.86cm²

we have:

The area of a circle =πr²

substituting value of area of circle we get

80.86cm=3.14*r²

dividing both side by 3.14

80.86/3.14=3.14*r²/3.14

25.75=r²

doing square root on both side

$\sqrt{25.75}=\sqrt{r²}$

r=5.07cm

The length of radius is 5.07cm

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3 years ago

A scale drawing of a patio is shown here. The scale is 4 inches : 6.8 feet. What is are length and width of the patio? Find the

mars1129 [50]

Answer:

Step-by-step explanation:

Divide 6.8ft by 4in then you get 1.7 and the multiply 15 an 14 by 1.7 then you get 25.5 and 23.8 then multiply 25.5 with 23.8 and get 606.9

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3 years ago

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In the picture above, line BX is the bisector of A, angle B and C. Line CX is the bisector of A, angle C and B. If angle A = 68°

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Answer:

Step-by-step explanation:

answer. is in the image

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3 years ago

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A function f has derivatives of all orders for all real numbers, x. Assume f(2) = −3, f ′(2) = 5, f ″(2) = 3 and the third deriv

nadezda [96]

The Taylor polynomial would be

$T_3(x)=f(2)+f'(2)(x-2)+\dfrac{f''(2)}2(x-2)^2+\dfrac{f'''(2)}6(x-2)^3$

$T_3(x)=-3+5(x-2)+\dfrac32(x-2)^2-\dfrac43(x-2)^3$

Then

$f(1.5)\approx T_3(1.5)=-\dfrac{119}{24}\approx-4.96$

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3 years ago