All categories

3 years ago

Diagonal of a rectangle Pythagorean theorem

Mathematics

1 answer:

kykrilka [37]3 years ago

4 0

Answer:

a squared + b squared = c squared

find the base, that's a, find the other side, that's b

a*a+b*b=c*c

You might be interested in

Find the area of a parallelogram with these dimensions.

aev [14]

The area is 14! hope this helps

3 0

3 years ago

During 4 days, the price of the stock of PEV Corporation went up 1/4 of a point, down 1/3 of a point, down 3/4 of a point, and u

HACTEHA [7]

The price of the stock of PEV Corporation varies as shown. we have to determine the net charge. Net change is just how much it changed overall, so we will look for all the ups minus all the downs.

Here, "went up" goes with a + sign, and "went down" goes with a - sign.

So, Net charge = $\frac{1}{4} - \frac{1}{3}-\frac{3}{4} + \frac{7}{10}$

= $\frac{1}{4}-\frac{3}{4}- \frac{1}{3} + \frac{7}{10}$

= $-\frac{2}{4}- \frac{1}{3} + \frac{7}{10}$

= $-\frac{1}{2}- \frac{1}{3} + \frac{7}{10}$

LCM of 2,3 and 10 is 30

= $\frac{-15-10+21}{30}$

= $\frac{-4}{30}$

= $\frac{-2}{15}$

So, the net charge is $\frac{-2}{15}$

8 0

2 years ago

Let f be the function given by f(x)= (x-1)(x^2-4)/x^2-a. For what positive values of a is f continuous for all real numbers x?

9966 [12]

Answer:

a =1 and a=4.

Step-by-step explanation:

The function is

$f(x)=\frac{(x-1)(x^2-4)}{x^2-a}$

If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.

Now, for a a positive real we have that $x=\sqrt{a}$ will annulate the denominator, i.e

$(\sqrt{a})^2-a = a-a = 0$ . But, if a = 1 we have:

$f(x)=\frac{(x-1)(x^2-4)}{x^2-1} = \frac{(x-1)(x^2-4)}{(x-1)(x+1)}=\frac{(x^2-4)}{x+1}$

so, the value $x=\sqrt{a} = \sqrt{1} = 1$ won't annulate the denominator.

Now, for a = 4 we have:

$f(x)=\frac{(x-1)(x^2-4)}{x^2-4} = x-1$

so, the value $x=\sqrt{a} = \sqrt{4} = 2$ won't annulate the denominator.

In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.

4 0

3 years ago

Give an example of an addition problem in which you would and would not group the addends differently to add

madam [21]

<span>Changing the grouping of the addends should not change the sum, according to the associative property of addition. You might group them differently with (50 + 3) + 47, so that you have 50 + (3 + 47). You might not regroup them with (16 + 4) + 5 rather than 16 + (4 + 5).</span>

8 0

2 years ago

5. Jocelyn needs to study two stars for her astronomy research. She has narrowed her choices down to four stars that are located

user100 [1]

The four stars are located in star m

7 0

3 years ago