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2 years ago

What is the median of the following number: 40, 49, 62, 56, 68, 39, 50, 61, 54, 44

Mathematics

1 answer:

Karo-lina-s [1.5K]2 years ago

6 0

Answer:

Step-by-step explanation:

39, 40, 44, 49, 50, 54, 56, 61, 62, 68

39, 40, 44, 49, 50, 54, 56, 61, 62, 68

$\frac{50 + 54}{2}$ = $\frac{104}{2}$ = 52

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Determine the values of k for which the equation has no solution

Naddik [55]

To solve this, we would need to figure out the values of $k$ for which an absolute value is negative. From basics, you should know that an absolute value is always greater than or equal to 0.

If we rearrange this equation a little bit, we get,

$|2x+7|\leq 5+k$

So $5+k$ cannot be negative. So,

$5+k\geq 0\\k\geq -5$

For this to have a solution, k needs to be greater than or equal to $-5$ .

For no solution, k is less than -5. So, values of k which will make the equation to not have any solution is $k$ .

ANSWER: $k$

5 0

2 years ago

Any help would be appreciated. Thank you!

trapecia [35]

Answer:

Area $=62.5\sqrt{6}$ square units

$AB=5\sqrt{15}$ units

$BC=5\sqrt{10}$ units

Step-by-step explanation:

In a right triangle the altitude drawn to the hypotenuse is the geometric mean of the segments at which this altitude divides the hypotenuse.

So,

$BD^2=15\cdot 10\\ \\BD^2=150\\ \\BD=\sqrt{150}=5\sqrt{6}\ units$

a. The area of the triangle ABC is

$A_{ABC}=\dfrac{1}{2}\cdot BD\cdot AC=\dfrac{1}{2}\cdot 5\sqrt{6}\cdot (15+10)=\dfrac{125\sqrt{6}}{2}=62.5\sqrt{6}\ un^2.$

b. The legs of the right triangle are geometric means of the segment adjacent to this leg and the hypotenuse, so

$AB^2=AD\cdot AC=15\cdot 25\Rightarrow AB=5\sqrt{15}\ units\\ \\BC^2=CD\cdot AC=10\cdot 25\Rightarrow BC=5\sqrt{10}\ units$

5 0

3 years ago

A square patio has an area of 122 square feet. How long is each side of the patio to the nearest 0.05?

jasenka [17]

Should be About 25 feet as your answer

3 0

2 years ago

What is the midpoint of (-7,-9) (6,3)

Advocard [28]

Answer :

Let the first term of both the terms be $\: \sf \: x _1 \: \: \& \: \: x _2$ and last term be $\: \sf \: y _1 \: \: \& \: \: y_2\\$

Now, by using the mid point formula to find the mid point of the segment -

$\\ \sf \bigg( \frac{x_1 + x_2}{2} , \: \frac{y_1 + y_2}{2} \bigg) \\$

Now, by substituting the values of both x and y -

$\\ \sf \bigg( \frac{ - 7 + 6}{2} , \: \frac{ - 9 + 3}{2} \bigg) \\$

Adding -7 and 6 -

$\\ \sf \bigg( \frac{ - 1}{2} , \frac{ - 9 + 3}{2} \bigg) \\$

Now, move the negative in front of the fraction -

$\\ \sf \bigg( \frac{ - 1}{2} , \frac{ - 6}{2} \bigg) \\ \\ \\ \sf \bigg( \frac{ - 1}{2} , - 3\bigg) \\$

6 0

2 years ago

Help: Simplifying inside parenthesis first

Ne4ueva [31]

Answer:

4th one

Step-by-step explanation:

a^-2 goes down and become a^2 and

${a}^{2} \times {a}^{2} = {a}^{4}$

b^-1 goes up and become b

${b}^{2} \times {b}^{1} = {b}^{3}$

that why answer is 4th...mark me brainliest please

3 0

2 years ago