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3 years ago

Fractions that are the same number are blank

Mathematics

1 answer:

maks197457 [2]3 years ago

4 0

Answer:

plebs

Step-by-step explanation:

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PLEASE HELP I NEED ANSWER

Lubov Fominskaja [6]

You have to use the y=mx+b formula to find the equation

7 0

3 years ago

Who ever gets this wright first gets a thanks

bulgar [2K]

32 people because 24 adding 8 is 32 and 8 times 4 is 32

7 0

4 years ago

In trapiziod ABCD with bases AB and DC , diagonals intersect at point O. Find the length of diagonal BD, if BO=6 cm and A/O=3/1

ElenaW [278]

Answer:

the length of the diagonal BD is 8 cm

Step-by-step explanation:

The computation of the length of diagonal BD is as follows

Here the trapezoid diagonals divides each other in an equivalent ratio

So, equation would be

AO ÷ OC = OB ÷ OD

Now put there values to the above equation

3 ÷ 1 = 6 ÷ OD

3OD = 6

Now divided it by 3 in both the sides

OD = 2

Now the BD would be

= BO + OD

= 6 + 2

= 8 cm

hence, the length of the diagonal BD is 8 cm

3 0

3 years ago

Write the equation of the conic section satisfying the given conditions. focus at the pole, e = 3/4, horizontal directrix 2 unit

hjlf

Answer:

The equation in the polar form is;

$r = \dfrac{6}{4 + 3 \cdot sin(\theta)}$

Step-by-step explanation:

e = 3/4 > 1, we have an hyperbola

The polar equation of a conic is of the form;

For vertical directrix

$r = \dfrac{k \cdot e}{1\pm e \cdot cos (\theta)}$

For horizontal directrix

$r = \dfrac{k \cdot e}{1\pm e \cdot sin(\theta)}$

Where;

k = Distance from the focus to the directrix = 2

We have;

$r = \dfrac{2 \cdot \dfrac{3}{4} }{1 + \dfrac{3}{4} \cdot sin(\theta)}$

$r = \dfrac{\dfrac{3}{2} }{1 + \dfrac{3}{4} \cdot sin(\theta)}$

Which gives the equation in the polar form as follows;

$r = \dfrac{6}{4 + 3 \cdot sin(\theta)}$ .

8 0

3 years ago

Which equation is represented by the table?

MatroZZZ [7]

Answer:

$y=-|x-3|-3$

Step-by-step explanation:

The table shows values of $y$ for different values of $x$ .

Let us test each equation with the values given in the table.

Equation 1: $y=|x+3|-3$

For $x$ = 1, we get,

$y=|x+3|-3=|1+3|-3=|4|-3=4-3=1$

So, we got 1 for $y$ . But as per table, value of $y$ should be -5.

So, equation 1 is incorrect.

Equation 2: $y=-|x-3|-3$

For $x$ = 1, we get,

$y=-|x-3|-3=-|1-3|-3=-|-2|-3=-(2)-3=-5$

So, we got -5 for $y$ . It matches with the value of table.

Let us check for another value of $x$ in order to confirm.

For $x =2$ , we get,

$y=-|x-3|-3=-|2-3|-3=-|-1|-3=-(1)-3=-4$

So, we got -4 for $y$ . It matches with the value of table.

So, equation 2 is the correct answer.

We can be completely sure by checking the remaining equations too.

Equation 3: $y=|x-3|-3$

For $x$ = 1, we get,

$y=|x-3|-3=|1-3|-3=|-2|-3=2-3=-1$

So, we got -1 for $y$ . But as per table, value of $y$ should be -5.

So, equation 3 is incorrect.

Equation 4: $y=-|x+3|-3$

For $x$ = 1, we get,

$y=-|x+3|-3=-|1+3|-3=-|4|-3=-4-3=-7$

So, we got -7 for $y$ . But as per table, value of $y$ should be -5.

So, equation 4 is incorrect.

3 0

3 years ago