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3 years ago

Evaluate each expression below. Click on "Undefined" as needed. 0/9= 3÷0 =

Mathematics

1 answer:

damaskus [11]3 years ago

6 0

0/9 = 0 and 3/0 = undefined. Any number divided by 0 is undefined, while 0 divided by any number is just 0.

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Farmers often sell fruits and vegetables at farmers’ markets during the summer. Each tomato stand at the Bentonville farmers’ ma

olga nikolaevna [1]

Answer:

the probability that all tomatoes are sold is 0.919 (91.9%)

Step-by-step explanation:

since the random variable X= number of tomatoes that are demanded, is normally distributed we can make the standard random variable Z such that:

Z=(X-μ)/σ = (83 - 125)/30 = -1.4

where μ= expected value of X= mean of X (since X is normally distributed) , σ=standard deviation of X

then all tomatoes are sold if the demand surpasses 83 tomatos , therefore

P(X>83) = P(Z>-1.4) = 1- P(Z≤-1.4)

from tables of standard normal distribution →P(Z≤-1.4)=0.081 , therefore

P(X>83) = 1- P(Z≤-1.4) = 1 - 0.081 = 0.919 (91.9%)

thus the probability that all tomatoes are sold is 0.919 (91.9%)

6 0

3 years ago

Find the missing side. Need help please. Need to show how to get the answer.

Nutka1998 [239]

Answer:

Step-by-step explanation:

7 0

3 years ago

Identify the x-intercepts of the function below f(x)=x^2+12x+24

damaskus [11]

ANSWER:

x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

SOLUTION:

Given, $f(x)=x^{2}+12 x+24$ -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

$X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here, for (1) a = 1, b= 12 and c = 24

$X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}$

$\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}$

Hence the x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

8 0

3 years ago

Prove that the diagonals of a parallelogram bisect each other

Nady [450]

Answer:

[ See the attached picture ]

The diagonals of a parallelogram bisect each other.

✧ Given : ABCD is a parallelogram. Diagonals AC and BD intersect at O.

✺ To prove : AC and BD bisect each other at O , i.e AO = OC and BO = OD.

Proof : $\begin{array}{ |c| c | c | } \hline \tt{SN}& \tt{STATEMENTS} & \tt{REASONS}\\ \hline 1& \sf{In \: \triangle ^{s} \:AOB \: and \: COD } \\ \sf{(i)}& \sf{ \angle \: OAB = \angle \: OCD\: (A)}& \sf{AB \parallel \: DC \: and \: alternate \: angles} \\ \sf{(ii)} &\sf{AB = DC(S)}& \sf{Opposite \: sides \: of \: a \: parallelogram} \\ \sf{(iii)} &\sf{ \angle \: OBA= \angle \: ODC(A)} &\sf{AB \parallel \:DC \: and \: alternate \: angles} \\ \sf{(iv)}& \sf{ \triangle \:AOB\cong \triangle \: COD}& \sf{A.S.A \: axiom}\\ \hline 2.& \sf{AO = OC \: and \: BO = OD}& \sf{Corresponding \: sides \: of \: congruent \: triangle}\\ \hline 3.& \sf{AC \: and \: BD \: bisect \: each \: other \: at \: O}& \sf{From \: statement \: (2)}\\ \\ \hline\end{array}. Proved ✔$

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5 0

3 years ago

What is the solution to the inequality a - 1 > 11

Vanyuwa [196]

It would be a > 12, move all terms to one side and then solve for x. make sure if you divide then you flip the term

8 0

3 years ago