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3 years ago

In the pythagorean theorem what are the a and b referred to as

Mathematics

1 answer:

wlad13 [49]3 years ago

8 0

A and b are legs, c is hypotenuse

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Determine any data values are missing from the table assuming that the data represents a linear function.

IgorC [24]

Answer:

Missing x:0

Missing y:1

Hence, option b is correct.

Step-by-step explanation:

Given the table

x y

-9 3

-6 2

-3 missing

missing 0

As we know that the linear function has a constant common difference between the x values and y values.

From the table, it is clear that the x values have the same common difference, and the consecutive x values are separated by -3.

i.e. -9-(-6)=-3

-6-(-3)=-3

So the missing x value will be:

-3-(x)=-3

-3-(0)=-3

Therefore, the missing x value wil be: 0

From the table, it is clear that the x values have the same common difference, and the consecutive x values are separated by 1.

i.e.

3-2=1

2-y=1

y=1

Therefore, the missing y-value will be 1.

Therefore,

Missing x:0

Missing y:1

Hence, option b is correct.

3 0

3 years ago

Consider the relation y = 4|x + 2| + 7. What are the coordinates of the vertex?

aev [14]

To answer, we set the expression inside the absolute value symbol (II) equal to 0. That is,
x + 2 = 0
Then, we solve for the value of x.
x + 2 - 2 = 0 - 2
Hence, the value of x from the equation above is equal to -2.

Then, substitute the value of x to the equation and drop the absolute value symbol.
y = 4(x + 2) + 7
Substituting,
y = 4(-2 + 2) + 7
y = 4(0) + 7
y = 7

Thus, the vertex of the absolute value equation is equal to (-2,7). The answer is the last choice.

5 0

3 years ago

I need help with this question :-/

Karolina [17]

3s = 72 (Multiply 8 both side).

s = 24 (divide by 3 both side)

24 students

6 0

4 years ago

The width of a rectangle is 6 cm less than its length. If its perimeter is 80 cm then its lengths (in cm) and area (in cm²) are

mr_godi [17]

Start by stating the given info as mathematical expressions:

1. "<span>The width of a rectangle is 6 cm less than its length."
$w=l-6$

2. "I</span><span>ts perimeter is 80 cm."
$P=80$

Using the formula for perimeter ( $P=2l+2w$ ) we can show:
$80=2l+2w$ .

We now have 2 equations in 2 unknowns and can solve by substitution:
</span> $80=2l+2w$
$80=2l+2(l-6)$
$80=2l+2l-12)$
$92=4l$
$l=23$

The length is 23cm.

Having the length w can then find the width:
$w=l-6$
$w=23-6$
$w=17$

The width is 17cm.

And finally the area:
$A=lw$
$A=(23)(17)$
$A=391$

The area is $391cm^{2}$ .

As a check we can use the known perimeter:
$80=2l+2w$
$80=2(23)+2(17)$
$80=46+34$
$80=80$ Checks true.