2 years ago

What is the vertex? (2,0) (0,-6) (-2,-8) (-6,0)

Mathematics

1 answer:

Ksenya-84 [330]2 years ago

5 0

Hi,

The picture you published shows a parabola, if you look at its lowest possible point, that is the vertex, the coordinates are (--2, --8).

Have you understood ?

Green eyes.

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This is line segments

lianna [129]

Step-by-step explanation:

$here \: given \: that \: total \: lenth = bd = 67 \\ then \: bd = bc + cd \\ so \: we \: can \: write \: that \\ 67 = 3x - 2 + 4x + 13 \\67 = 3x + 4x - 2 + 13 \\ 67 = 7x + 11 \\ 7x = 67 - 11 \\ 7x = 56 \\ x = \frac{56}{7} \\ x = 8 \\ thankyou$

8 0

3 years ago

Read 2 more answers

-2/3-5/6= write in simplest form

sleet_krkn [62]

-2/3 - 5/6

We need common denominators so, since 3 can go into 6, we only need to multiply the first fraction by 2.

= -2 x 2 / 3x2 - 5/6
= -4 / 6 - 5/6

We only subtract the numbers that are in the numerators,

= -4-5 / 6
= -9/6

Both nine and six are divisible by 3, so to put into lowest terms...

=-9÷3 / 6÷3
= -3/2 <--- Final Answer

3 0

3 years ago

Write as a product: ax^2+ay^2–bx^2–by^2+b–a

Vesna [10]

Answer:

$\large\boxed{ax^2+ay^2-bx^2-by^2+b-a=(a-b)(x^2+y^2-1)}$

Step-by-step explanation:

$ax^2+ay^2-bx^2-by^2+b-a\\\\=(ax^2-bx^2)+(ay^2-by^2)+(b-a)\qquad\text{use the distributive property}\\\\=x^2(a-b)+y^2(a-b)-1(a-b)\\\\=(a-b)(x^2+y^2-1)$