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3 years ago

5

Can you use Pythagorean Theorem to find the missing side? Why or why not?

1 answer:

Dvinal [7]3 years ago

4 0

No. You cannot use the Pythagorean theorem to find the missing side, because you can only use The Pythagorean theorem when you are dealing with a right triangle.

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Svet_ta [14]

Answer:

the answer is 5.1 killograms

Step-by-step explanation:

hope this helps

8 0

2 years ago

How do you graph y=7/2x-2

RoseWind [281]

You can either make a table using any numbers you would like (I would suggest -5 to 5) and then graphing the rule

ex. (7/2)(-5)-2 = -19.5 (I multiplied (7/2) by -5 and then subtracted 2)

Or you can put the rule in a graphing calculator and check the points from there

8 0

3 years ago

How do you simplify this?<br>x²y+xy² / y²+2/5 × xy

polet [3.4K]

$\huge \boxed{\mathbb{QUESTION} \downarrow}$

How do you simplify this?
x²y+xy² / y²+2/5 × xy

$\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}$

$\sf\frac{ { x }^{ 2 } y+x { y }^{ 2 } }{ { y }^{ 2 } + \frac{ 2 }{ 5 } \times xy } \\$

Factor the expressions that are not already factored.

_____

<u>How </u><u>to</u><u> factorise</u><u> </u><u>:</u><u>-</u>

<u>NUMERATOR</u> $\downarrow$

$\sf \: x ^ { 2 } y + x y ^ { 2 }$

Factor out xy.

$\sf \: xy\left(x+y\right)$

<u>DENOMINATOR</u> $\downarrow$

$\sf{ y }^{ 2 } + \frac{ 2 }{ 5 } \times xy \\$

Factor out 1/5.

$\sf \: {\frac{1}{5}y\left(2x+5y\right)} \\$

_____

Continuing...

$\sf\frac{xy\left(x+y\right)}{\frac{1}{5}y\left(2x+5y\right)} \\$

Cancel out y in both the numerator and denominator.

$\sf\frac{x\left(x+y\right)}{\frac{1}{5}\left(2x+5y\right)} \\$

Expand the expression.

$\sf\frac{x^{2}+xy}{\frac{2}{5}x+y} \\$

This can further simplified to as $\downarrow$

$= \boxed{\boxed{\bf\frac{5x\left(x +y\right)}{2x+5y}}}$

3 0

3 years ago

Consider the accompanying data on flexural strength (MPa) for concrete beams of a certain type.

Vinvika [58]

Answer:

The point estimate for population mean is 8.129 Mpa.

Step-by-step explanation:

We are given the following in the question:

Data on flexural strength(MPa) for concrete beams of a certain type:

11.8, 7.7, 6.5, 6.8, 9.7, 6.8, 7.3, 7.9, 9.7, 8.7, 8.1, 8.5, 6.3, 7.0, 7.3, 7.4, 5.3, 9.0, 8.1, 11.3, 6.3, 7.2, 7.7, 7.8, 11.6, 10.7, 7.0

a) Point estimate of the mean value of strength for the conceptual population of all beams manufactured

We use the sample mean, $\bar{x}$ as the point estimate for population mean.

Formula:

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$\bar{x} = \dfrac{\sum x_i}{n} = \dfrac{219.5}{27} = 8.129$

Thus, the point estimate for population mean is 8.129 Mpa.

4 0

3 years ago

Answer the question below

Lina20 [59]

Answer:

kckgkfjjfufufuf8f8ff88r8r8r8r8r8rrffufuufuff

3 0

3 years ago