All categories

3 years ago

Question 1

Mathematics

2 answers:

notka56 [123]3 years ago

8 0

The answer would be a.
HOPE THIS HELPS! ^_^

viktelen [127]3 years ago

5 0

60 square centimeters

You might be interested in

Graphing linear inequalities digital escapes puzzle 4

Vadim26 [7]

Answer:

1 - F

2 - B

3 - H

4 - A

Step-by-step explanation:

m = -2

b = 4

y < mx + b

y < -2x + 4

m = 1

b = -2

y < mx + b

y < 1x -2

y < x - 2

m = -4

y <= mx + b

H is the only one that follows those rules

m = 1

y >= mx + b

A is the only one that follows those rules

5 0

3 years ago

How to determine a perfect square trinomial

AfilCa [17]

Answer:

Check whether the first and last terms of the trinomial are perfect squares.

Multiply the roots of the first and third terms together.

Compare to the middle terms with the result in step two

If the first and last terms are perfect squares, and the middle term’s coefficient is twice the product of the square roots of the first and last terms

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

What is the equation of the quadratic graph with a focus of (6, 0) and a directrix of y = −10 ?

Mademuasel [1]

Answer:

$(x-6)^{2}=20(y+5)^{2}$

Step-by-step explanation:

The standard for for the equation of a parabola is $(x-h)^{2}=4p(y-k)$

The focus is given as $(h, k+p)$

The directrix is given by $y=k-p$

Comparing focus given as $(6,0)$ to formula of focus $(h,k+p)$ , we see that $h=6$ and $k+p=0$
Comparing directrix given as $y=-10$ to formula of directrix $y=k-p$ , we can write $k-p=-10$

We can use the two system of equations [ $k+p=0$ and $k-p=-10$ ] to solve for $k$ and $p$ .

Adding the two equations gives us:

$2k=-10\\k=\frac{-10}{2}\\k=-5$

Using this value of $k$ , we can find $p$ by plugging this value in either equation. Let's put it in Equation 1. We have:

$k-p=-10\\-5-p=-10\\-5+10=p\\p=5$

Now, that we know $h=6$ , $k=-5$ , and $p=5$ , we can plug these values in the standard form equation to figure out the parabola's equation. Doing so and rearranging gives us: