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3 years ago

What is the answer please

Mathematics

1 answer:

il63 [147K]3 years ago

8 0

Answer:

submit search form of attract new you are doing well in the question

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Express the vector (4,-6) in terms of the standard unit vectors i and j

Aliun [14]

Four, -6 in terms of standard unit factors I NJ makes no sense

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2 years ago

Simplify the expression.<br> −(−x)3 − x3

Olenka [21]

Answer:

Step-by-step explanation:

-(-x) ^3-x^3

x^3-x^3

4 0

3 years ago

Identify the x-intercept and y-intercept of the line 2x−5y=202x−5y=20.

tensa zangetsu [6.8K]

   2x - 5y = 20

At the x-intercept, y=0 .

   2x = 20 ==> x = 10

At the y-intercept, x=0.

   -5y = 20 ==> y = -4

7 0

3 years ago

a pie chart is divided into four sectors in fig. 12.42. Each sector represents a percentage of the whole. The two larger sectors

Novosadov [1.4K]

Answer:

Angle formed by the sector measuring x% will be 126°.

Step-by-step explanation:

Since, sum of all sectors formed in a circle is 100%.

By adding the measures of all the sectors,

x + x + 21 + 9 = 100

2x + 30 = 100

2x = 70

x = 35%

Now we know sum of all the central angles formed at the center of a circle = 360°

Therefore, angle formed by x% = 360° × 35%

= $\frac{360\times 35}{100}$

= 126°

8 0

3 years ago

plain why the function is discontinuous at the given number a. (Select all that apply.) f(x) = x2 − 2x x2 − 4 if x ≠ 2 1 if x =

ElenaW [278]

Answer with Step-by-step explanation:

We are given that

$f(x)=\left\{\begin{matrix}\dfrac{x^2-2x}{x^2-4}&,if\ \ x\neq 2 \\ 1&,if\ \ x=2\end{matrix}\right.$

We have to explain that why the function is discontinuous at x=2

We know that if function is continuous at x=a then LHL=RHL=f(a).

$f(x)=\frac{x(x-2)}{(x+2)(x-2)}=\frac{x}{x+2}$

LHL=Left hand limit when x <2

Substitute x=2-h

where h is small positive value >0

$\lim_{h\rightarrow 0}f(x)=\lim_{h\rightarrow 0}\frac{2-h}{2-h+2}$

$\lim_{h\rightarrow 0}\frac{2-h}{4-h}=\frac{2}{4}=\frac{1}{2}$

Right hand limit =RHL when x> 2

Substitute

x=2+h

$\lim_{h\rightarrow 0}f(x)=\lim_{h\rightarrow 0}\frac{2+h}{2+h+2}=\lim_{h\rightarrow 0}\frac{2+h}{4+h}$

$=\frac{2}{4}=\frac{1}{2}$

LHL=RHL= $\frac{1}{2}$

f(2)=1

$LHL=RHL\neq f(2)$

Hence, function is discontinuous at x=2

4 0

3 years ago